CHEM 133 Ch. 06-08

0.0(0)
Studied by 0 people
call kaiCall Kai
learnLearn
examPractice Test
spaced repetitionSpaced Repetition
heart puzzleMatch
flashcardsFlashcards
GameKnowt Play
Card Sorting

1/95

encourage image

There's no tags or description

Looks like no tags are added yet.

96 Terms

1
New cards

Thermodynamics

is the study of energy and its transformations

2
New cards

Thermochemistry

is the study of the relationships between chemical reactions and energy changes involving heat

3
New cards

Kinetic Energy

is the energy an object possesses by virtue of its own motion

  • K.E. = ½ mv2


4
New cards

Potential Energy

is the energy and object possesses by virtue of its position of chemical composition

  • P.E. = mgh


5
New cards

Energy

is the capacity to do work or transfer heat

6
New cards

Work

is the energy used to cause an object with mass to move

  • w = F x d


7
New cards

Thermal Energy

is kinetic energy associated with the random motion of atoms and molecules

8
New cards

Temperature

is a quantitative measure of “hot” or “cold”

<p>is a quantitative measure of “hot” or “cold”</p>
9
New cards

Heat (q)

is the transfer of thermal energy between two bodies at different temperatures

heat flows from high temperature to low temperature

<p>is the transfer of thermal energy between two bodies at different temperatures</p><p>heat flows from high temperature to low temperature</p>
10
New cards

Most Substances Expand and Contract when…

as their temperature increases and contract as their temperature decreases

<p>as their temperature increases and contract as their temperature decreases</p>
11
New cards

System

the portion of the universe that we single out to study

  • Example: if we are interested in the interaction between hydrogen and oxygen in a cylinder, then the H2 and O2 in the cylinder form a system


<p>the portion of the universe that we single out to study</p><ul><li><p>Example: if we are interested in the interaction between hydrogen and oxygen in a cylinder, then the H<sub>2</sub> and O<sub>2</sub> in the cylinder form a system</p></li></ul><p></p>
12
New cards

Surroundings

are the rest of the universe (here, the cylinder, piston and eveythign beyond)

<p>are the rest of the universe (here, the cylinder, piston and eveythign beyond)</p>
13
New cards

Types of Systems

  • A system may be open, closed, or isolated

    • open system: both matter and energy may be exchanged with the surroundings

    • closed system: energy, but not matter, may be exchanged with the surroundings

    • isolated system: neither matter nor energy may be exchanged with the surroundings


<ul><li><p>A system may be open, closed, or isolated</p><ul><li><p>open system: both matter and energy may be exchanged with the surroundings</p></li><li><p>closed system: energy, but not matter, may be exchanged with the surroundings</p></li><li><p>isolated system: neither matter nor energy may be exchanged with the surroundings</p></li></ul></li></ul><p></p>
14
New cards

First Law of Thermodynamics

  • The first law of thermodynamics states that energy cannot be created or destroyed

  • The first law of thermodynamics is the law of conservation of energy

    • △EUNIVERSE = △ESYSTEM + △ESURROUNDINGS = 0

  • That is, the energy of system + surrounding is constant

  • Thus, any energy transferred from a system must be transferred to the surroundings (and vise versa)

  • Energy can be converted from one form to another


15
New cards

Internal Energy

  • The total energy, E, of a system is called the internal energy

    • It is the sum of all the kinetic and potential energies of all components of the system

  • Absolute internal energy cannot be measured, only changes in internal energy

    • Changes in internal energy: △E = Efinal - Einitial

    • Thermodynamic quantities such as △E always have three parts:

      • a number

      • a unit

      • and a sign that gives direction

  • Example: a mixture of H2(g) and O2(g) has a higher internal energy than H2O(g)

  • Going from H2(g) and O2(g) to H2O(g) results in a positive change in internal energy, indicating that the system has gained (absorbed) energy from the surroundings: 2H2O(g) → H2(g) + O2(g) △E > 0, Efinal > Einitial


<ul><li><p>The total energy, <em>E</em>, of a system is called the internal energy</p><ul><li><p>It is the sum of all the kinetic and potential energies of all components of the system</p></li></ul></li><li><p>Absolute internal energy cannot be measured, only changes in internal energy</p><ul><li><p>Changes in internal energy: △<em>E</em> = <em>E</em><sub>final</sub> - <em>E</em><sub>initial</sub></p></li><li><p>Thermodynamic quantities such as △<em>E</em> always have three parts:</p><ul><li><p>a number</p></li><li><p>a unit</p></li><li><p>and a sign that gives direction</p></li></ul></li></ul></li><li><p>Example: a mixture of H<sub>2</sub>(<em>g</em>) and O<sub>2</sub>(<em>g</em>) has a higher internal energy than H<sub>2</sub>O(<em>g</em>)</p></li><li><p>Going from H<sub>2</sub>(<em>g</em>) and O<sub>2</sub>(<em>g</em>) to H<sub>2</sub>O(<em>g)</em> results in a positive change in internal energy, indicating that the system has gained (absorbed) energy from the surroundings: 2H<sub>2</sub>O(<em>g</em>) → H<sub>2</sub>(<em>g</em>) + O<sub>2</sub>(<em>g</em>) △<em>E</em> &gt; 0, <em>E</em><sub>final</sub> &gt; <em>E</em><sub>initial</sub></p></li></ul><p></p>
16
New cards

Changes in Internal Energy Relating △E to Heat and Work

  • When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w)

  • That is, from the first law of thermodynamics:

    • △E = q + w


<ul><li><p>When <strong>energy</strong> is exchanged between the system and the surroundings, it is exchanged as either heat (<em>q</em>) or work (<em>w</em>)</p></li><li><p>That is, from the first law of thermodynamics:</p><ul><li><p>△<em>E</em> = <em>q</em> + <em>w</em></p></li></ul></li></ul><p></p>
17
New cards

△E, Q, W, and their Signs

Heat

  • Heat flowing from the surroundings to the system is positive, q > 0

  • Heat flowing from system to surroundings is negative, q < 0

Work

  • Work done by the surroundings on the system is positive, w > 0

  • Work done by the system on the surroundings is negative, w < 0


<p>Heat</p><ul><li><p>Heat flowing from the surroundings to the system is positive, <em>q</em> &gt; 0</p></li><li><p>Heat flowing from system to surroundings is negative, <em>q</em> &lt; 0</p></li></ul><p>Work</p><ul><li><p>Work done by the surroundings on the system is positive, <em>w</em> &gt; 0</p></li><li><p>Work done by the system on the surroundings is negative, <em>w</em> &lt; 0</p></li></ul><p></p>
18
New cards

Exchange of Heat Between System and Surroundings

When heat is released by the system into the surroundings, the process is exothermic. An exothermic reaction feels hot.

When heat is absorbed by the system from the surroundings, the process is endothermic. An endothermic reaction feels cold.

<p>When heat is released by the system into the surroundings, the process is exothermic. An exothermic reaction feels hot.</p><p>When heat is absorbed by the system from the surroundings, the process is endothermic. An endothermic reaction feels cold.</p>
19
New cards

State Functions (1)

  • Usually, we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem

  • However, we do know that the internal energy of a system is independent of the path by which the system achieved that state. A state function depends only on the initial and final states of a system.

  • Internal energy is a state function

  • It depends only on the present state of the system, not on the path by which the system arrived at that state

  • Therefore, △E depends only on Efinal and Einitial

  • Example: In the system below, the water could have reached room temperature from either direction. The internal energy of 50 g of H2O (l) at 25℃ does not depend on whether we cool 50 g of H2O (l) from 100℃ to 25℃ or heat 50 g of H2O (l) at 0℃ to 25℃


<ul><li><p>Usually, we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem</p></li><li><p>However, we do know that the internal energy of a system is independent of the path by which the system achieved that state. A state function depends only on the initial and final states of a system.</p></li><li><p>Internal energy is a state function</p></li><li><p>It depends only on the present state of the system, not on the path by which the system arrived at that state</p></li><li><p>Therefore, △<em>E</em> depends only on <em>E</em><sub>final</sub> and <em>E</em><sub>initial</sub></p></li><li><p>Example: In the system below, the water could have reached room temperature from either direction. The internal energy of 50 g of H<sub>2</sub>O (<em>l</em>) at 25℃ does not depend on whether we cool 50 g of H<sub>2</sub>O (<em>l</em>) from 100℃ to 25℃ or heat 50 g of H<sub>2</sub>O (<em>l</em>) at 0℃ to 25℃</p></li></ul><p></p>
20
New cards

State Functions (2)

  • However, q and w are not state functions

  • Whether the battery is shorted out or is discharged by running the fan, its △E is the same, but q and w are different in the two cases


<ul><li><p>However, <em>q</em> and <em>w</em> are not state functions</p></li><li><p>Whether the battery is shorted out or is discharged by running the fan, its △<em>E </em>is the same, but <em>q</em> and <em>w</em> are different in the two cases</p></li></ul><p></p>
21
New cards

Enthalpy of Reaction

  • The change in enthalpy, △H, is the enthalpy of the products minus the enthalpy of the reactants:

    • ∆Hrxn = Eproducts – Ereactants          

  • This quantity, ∆Hrxn is called the enthalpy of reaction, or the heat of reaction

  • Consider the thermochemical equation for the combustion of hydrogen:

    • 2H2(g) + O2(g) → 2H2O(g)            ∆H = – 483.6 kJ

    • The equation tells us that 483.6 kJ of energy is released to the surroundings when water is formed

    • ∆H noted at the end of the balanced equation depends on the number of moles of reactants and products associated with the ∆H value

    • These equations are called thermochemical equations


<ul><li><p>The change in enthalpy, △<em>H</em>, is the enthalpy of the products minus the enthalpy of the reactants:</p><ul><li><p><span>∆<em>H</em><sub>rxn</sub> = E<sub>products</sub> – E<sub>reactants</sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</span></p></li></ul></li><li><p>This quantity, ∆<em>H</em><sub>rxn</sub> is called the enthalpy of reaction, or the heat of reaction</p></li><li><p>Consider the thermochemical equation for the combustion of hydrogen:</p><ul><li><p><span>2H<sub>2</sub>(<em>g</em>) + O<sub>2</sub>(<em>g</em>) → 2H<sub>2</sub>O(<em>g)</em>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; ∆<em>H</em> = – 483.6 kJ</span></p></li><li><p>The equation tells us that 483.6 kJ of energy is released to the surroundings when water is formed</p></li><li><p>∆<em>H</em> noted at the end of the balanced equation depends on the number of moles of reactants and products associated with the ∆<em>H</em> value</p></li><li><p>These equations are called thermochemical equations</p></li></ul></li></ul><p></p>
22
New cards

Enthalpy Diagram

  • are used to represent enthalpy changes associated with a reaction

  • in the enthalpy diagram for the combustion of hydrogen the reactants, H2(g) and O2(g), have a higher enthalpy than the product, H2O

  • Thus the reaction is exothermic


<ul><li><p>are used to represent enthalpy changes associated with a reaction</p></li><li><p>in the enthalpy diagram for the combustion of hydrogen the reactants, H<sub>2</sub>(<em>g</em>) and O<sub>2</sub>(<em>g</em>), have a higher enthalpy than the product, H<sub>2</sub>O</p></li><li><p>Thus the reaction is exothermic</p></li></ul><p></p>
23
New cards

Endothermic and Exothermic

  • Chemists use a property known as enthalpy (H) to describe the thermodynamics of chemical and physical processes

  • Heat transferred from the surroundings to the system has a positive enthalpy (i.e., ∆H > 0 for an endothermic reaction)

  • Heat transferred from the system to the surroundings has a negative enthalpy (i.e., ∆H < 0 for an exothermic reaction


<ul><li><p>Chemists use a property known as enthalpy (<em>H</em>) to describe the thermodynamics of chemical and physical processes</p></li><li><p>Heat transferred from the surroundings to the system has a positive enthalpy (i.e., ∆<em>H</em> &gt; 0 for an endothermic reaction)</p></li><li><p>Heat transferred from the system to the surroundings has a negative enthalpy (i.e., ∆<em>H</em> &lt; 0 for an exothermic reaction</p></li></ul><p></p>
24
New cards

The Truth About Enthalpy

  • Enthalpy is an extensive property

    • Therefore, the magnitude of enthalpy id directly proportional to the amount of reactant consumed

      • Example: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)        ∆H = –890 kJ

      • If one mol of CH4 is burned in oxygen to produce CO2 and water, then 890 kJ of heat is released to the surroundings. If two mol of CH4 is burned, then 1780 kJ of heat is released.

  • The sign ∆H depends on the direction of the reaction

    • The enthalpy change for a reaction is equal in magnitude to, but opposite in sign from, ∆H for the reverse reaction

    • Example: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)  ∆H = -890 kJ,

      But  CO2(g) + 2H2O(l) → CH4(g) + 2O2(g)  ∆H = +890 kJ.

  • Enthalpy change depends on state of the products and reactants

    • 2H2O(g) → 2H2O(l)  ∆H = –88 kJ


25
New cards

Heat Capacity and Specific Heat

The amount of energy required to raise the temperature of a substance by 1℃ (1 degree Celsius) is its heat capacity, C. If the amount of the substance heated is one gram, it is the specific heat, s. If the amount is one mole, it is the molar heat capacity.

<p><span>The amount of energy required to raise the temperature of a substance by 1℃ (1 degree Celsius) is its <strong>heat capacity, C</strong>. If the amount of the substance heated is one gram, it is the <strong>specific heat, s</strong>. If the amount is one mole, it is the <strong>molar heat capacity</strong>.</span></p>
26
New cards

Calorimetry

  • Since we cannot know the exact enthalpy of the reactants and products, we measure ∆H through calorimetry, the measurement of heat flow.

  • The instrument used to measure heat flow is called a calorimeter.

  • By carrying out a reaction in aqueous solution in a simple calorimeter, the heat change for the system can be found by measuring the heat change for the water in the calorimeter.

  • The specific heat for water is 4.184 J/g℃. We use this value for dilute solutions.

  • We can calculate q for the reaction with this equation: q = ms∆T.

  • Be careful with the sign of q.


<ul><li><p><span>Since we cannot know the exact enthalpy of the reactants and products, we measure ∆<em>H</em> through <strong>calorimetry</strong>, the measurement of heat flow.</span></p></li><li><p><span>The instrument used to measure heat flow is called a <strong>calorimeter</strong>.</span></p></li><li><p><span>By carrying out a reaction in aqueous solution in a simple calorimeter, the heat change for the system can be found by measuring the heat change for the water in the calorimeter.</span></p></li><li><p><span>The specific heat for water is 4.184 J/g℃. We use this value for dilute solutions.</span></p></li><li><p><span>We can calculate q for the reaction with this equation: q = ms∆T.</span></p></li><li><p><span>Be careful with the sign of <em>q</em>.</span></p></li></ul><p></p>
27
New cards

Heat Capacity and Specific Heat (2)

  • We can determine the amount of heat, q, entering or leaving the substance by measuring the temperature change before and after the heat is gained or lost.

  • We can calculate q with this equation:

    • q = (mass of substance) x (specific heat) x (temperature change)

    • q = m x s x (Tfinal –Tinitial)

    • q = m x s x ∆ T

  • Be careful with the sign of q.

  • Heat Capacity is defined mathematically as: C = m x s


28
New cards
<p>Hess’s Law</p>

Hess’s Law

  • ΔH is well known for many reactions, and it is inconvenient to measure ΔH for every reaction in which we are interested.

  • However, we can calculate ΔH using published ΔH values and the properties of enthalpy.

  • Hess’s law: If a reaction is carried out in a series of steps, ΔH for the overall reaction equals the sum of the enthalpy changes for the individual steps.

  • Because H is a state function, for a particular set of reactants and products, ΔH is the same whether the reaction takes place in one step or in a series of steps.  

  • The total change in enthalpy is independent of the number of steps.

  • This is true because enthalpy is a state function!


<ul><li><p><span>Δ<em>H</em> is well known for many reactions, and it is inconvenient to measure Δ<em>H</em> for every reaction in which we are interested.</span></p></li><li><p><span>However, we can calculate Δ<em>H</em> using published Δ<em>H</em> values and the properties of enthalpy.</span></p></li><li><p><span>Hess’s law: If a reaction is carried out in a series of steps, Δ<em>H</em> for the overall reaction equals the sum of the enthalpy changes for the individual steps.</span></p></li><li><p><span>Because <em>H</em> is a state function, for a particular set of reactants and products, Δ<em>H</em> is the same whether the reaction takes place in one step or in a series of steps. &nbsp;</span></p></li><li><p><span>The total change in enthalpy is independent of the number of steps.</span></p></li><li><p><span>This is true because enthalpy is a state function!</span></p></li></ul><p></p>
29
New cards

Standard Enthalpies of Formation

  • A standard enthalpy of formation, ΔHf°, is defined as the enthalpy change for the reaction in which 1 mol of compound is made from its constituent elements in their elemental forms and are measured under standard conditions (25 °C and 1.00 atm pressure).

  • ∆H°f values for selected substances are tabulated in Table 6.2.

  • A large majority of ∆H°f values tabulated in Table 6.2 are negative, meaning that most formation reactions are exothermic.

  • The standard enthalpy of formation of the most stable form of an element is zero.


<ul><li><p><span>A <strong>standard</strong> <strong>enthalpy of formation</strong>, ΔH<sub>f</sub>°, is defined as the enthalpy change for the reaction in which 1 mol of compound is made from its constituent elements in their elemental forms and are measured under standard conditions (25 °C and 1.00 atm pressure).</span></p></li><li><p><span>∆<em>H</em>°<sub>f</sub> values for selected substances are tabulated in Table 6.2.</span></p></li><li><p><span>A large majority of ∆<em>H</em>°<sub>f</sub> values tabulated in Table 6.2 are negative, meaning that most formation reactions are exothermic.</span></p></li><li><p><span>The standard enthalpy of formation of the most stable form of an element is zero.</span></p></li></ul><p></p>
30
New cards

Hess’s law! Calculation of ΔHrxn

  • We can use Hess’s law in this way: the standard enthalpy of reaction is the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants.

  • where Σ means “sum of” and n and m are the stoichiometric coefficients from the balanced chemical equation


<ul><li><p><span>We can use Hess’s law in this way:<em> the standard enthalpy of reaction is the sum of the standard enthalpies of formation of the</em>&nbsp;<strong><em>products</em></strong>&nbsp;<em>minus the sum of the standard enthalpies of formation of the</em>&nbsp;<strong><em>reactants.</em></strong></span></p></li><li><p><span>where Σ means “sum of” and <em>n</em> and <em>m</em> are the stoichiometric coefficients from the balanced chemical equation</span></p></li></ul><p></p>
31
New cards

The Quantum Mechanical Model of the Atom

Light is a form of electromagnetic radiations, a type of energy

that travels through space at a constant speed of C = 3.00 × 108

m/s.

32
New cards

Electron Energy Levels

A rainbow forms when light passes through water droplets.

<p><span style="font-family: &quot;Times New Roman&quot;;">A rainbow forms when light passes through water droplets.</span></p>
33
New cards

Electromagnetic Radiation

We experience electromagnetic radiation in different forms, such as light, the colors of a rainbow, or when we have an X-ray taken.

Electromagnetic radiation consists of energy particles that move as waves of energy.

Low-energy radiation has longer wavelengths.

High energy radiation has shorter wavelength.

<p><span>We experience electromagnetic radiation in different forms, such as light, the colors of a rainbow, or when we have an X-ray taken.</span></p><p style="text-align: left;"><span><strong>Electromagnetic radiation </strong>consists of energy particles that move as waves of energy.</span></p><p style="text-align: left;"><span>Low-energy radiation has longer wavelengths.</span></p><p style="text-align: left;"><span>High energy radiation has shorter wavelength.</span></p>
34
New cards
<p>Electromagnetic Radiation (2)</p>

Electromagnetic Radiation (2)

  • All waves have a characteristic wavelength, λ (lambda), which is the  distance between two peaks or troughs in a wave. The unit of λ is usually nm.

  • The frequency, ν (nu), of a wave is the number of cycles that pass a point in one second. The unit of ν is hertz (1 Hz = 1 s–1).

  • The amplitude of a wave is the height of the crest (or depth of the trough). For an electromagnetic wave, the amplitude is related to the intensity of the radiation, or its brightness in the case of visible light


<ul><li><p><span style="font-family: &quot;Times New Roman&quot;;">All waves have a characteristic </span><strong>wavelength</strong><span style="font-family: &quot;Times New Roman&quot;;">, λ (lambda), which is the&nbsp; distance between two peaks or troughs in a wave. The unit of λ is usually <em>nm.</em></span></p></li><li><p><span style="font-family: &quot;Times New Roman&quot;;">The <strong>frequency</strong>, ν (nu), of a wave is the number of cycles that pass a point in one second. The unit of ν is <em>hertz </em>(1 Hz = 1 s<sup>–1</sup>).</span></p></li><li><p><span>The&nbsp;<strong>amplitude</strong>&nbsp;of a wave is the height of the crest (or depth of the trough). For an electromagnetic wave, the amplitude is related to the&nbsp;<em>intensity</em>&nbsp;of the radiation, or its brightness in the case of visible light</span></p></li></ul><p></p>
35
New cards
<p>Light is Electromagnetic Radiation</p>

Light is Electromagnetic Radiation

Emission and transmission of energy in the form of electromagnetic waves. Red light with the wavelength of 750nm has the longest wavelength where as Violet light shorter wavelength 400 nm

The presence of color in white light is responsible for the colors we see in our everyday vision. A red color shirt is red because it reflects red light.

Our eyes see only the reflected light

<p><span>Emission and transmission of energy in the form of electromagnetic waves. Red light with the wavelength of 750nm has the longest wavelength where as Violet light shorter wavelength 400 nm</span></p><p><span>The presence of color in white light is responsible for the colors we see in our everyday vision. A red color shirt is red because it reflects red light.</span></p><p><span>Our eyes see only the reflected light</span></p>
36
New cards

Light is Electromagnetic Radiation (2)

  • Visible light makes up a tiny portion of the entire electromagnetic spectrum which includes all wavelengths of electromagnetic radiation.

  • Main region of electro magnetic spectrum , ranging in wavelength

  • 10-15 m gamma rays to 10-5 m radio waves.

  • Shorter wavelength : higher frequency and high energy ( right side)

  • Longer wavelength : lower  frequency and lower energy ( left side)

  • Shortest : wave : Gamma rays:  can damage biological molecules.

  • X rays: medical use, can pass through many substances that block visible light and therefore used to take images of bones and internal organs.

  • UV light : main component of sun and responsible for skin cancer and sunburn.

  • visible light (violet : shorter wavelength – red: longer wavelength)

  • infrared light:  All warm objects including human body emit IR light.

  • Microwave: used  for radar and micro ovens. Low energy waves

  • Radio wave:( longest wavelength): to transmit the signals responsible for AM and FM radio, cellular phones and other forms of communications.


<ul><li><p><span>Visible light makes up a tiny portion of the entire electromagnetic spectrum which includes all wavelengths of electromagnetic radiation.</span></p></li><li><p><span>Main region of electro magnetic spectrum , ranging in wavelength</span></p></li><li><p><span>10<sup>-15</sup> m gamma rays to 10<sup>-5</sup> m radio waves.</span></p></li><li><p><span>Shorter wavelength : higher frequency and high energy ( right side)</span></p></li><li><p><span>Longer wavelength : lower&nbsp; frequency and lower energy ( left side)</span></p></li><li><p><span>Shortest : wave : Gamma rays:&nbsp; can damage biological molecules.</span></p></li><li><p><span>X rays: medical use, can pass through many substances that block visible light and therefore used to take images of bones and internal organs.</span></p></li><li><p><span>UV light : main component of sun and responsible for skin cancer and sunburn.</span></p></li><li><p><span>visible light (violet : shorter wavelength – red: longer wavelength)</span></p></li><li><p><span>infrared light:&nbsp; All warm objects including human body emit IR light.</span></p></li><li><p><span>Microwave: used&nbsp; for radar and micro ovens. Low energy waves</span></p></li><li><p><span>Radio wave:( longest wavelength): to transmit the signals responsible for AM and FM radio, cellular phones and other forms of communications.</span></p></li></ul><p></p>
37
New cards

Amplitude and Wavelength

 We characterize a wave by its amplitude and wavelength. Amplitude of wave is the vertical height of a crest(or depth of the trough).The amplitude of the wave in light determines the light intensity or brightness- the greater the amplitude, the greater the brightness.

<p><span>&nbsp;We characterize a wave by its amplitude and wavelength. Amplitude of wave is the vertical height of a crest(or depth of the trough).The amplitude of the wave in light determines the light intensity or brightness- the greater the amplitude, the greater the brightness.</span></p>
38
New cards

Photon

  • A particle of light is called a Photon

  • Photon is a single packet of light energy and the amount of energy carried in a packet depends on the wavelength of  the light.

  • Shorter the wave length= greater the energy

  • Violet light (shorter wavelength) carries more energy per photon than red light energy( longer wavelength)

  • The frequency and energy of electromagnetic radiations are  inversely related to its wavelength.


39
New cards
<p>Electromagnetic Radiation (3)</p>

Electromagnetic Radiation (3)

Einstein assumed that light traveled in energy packets called photons.

  •  The energy of one photon is: image

h = 6.626 ´ 10–34 Js is (Planck’s constant )

c = 3.00 x 108 m/s  (speed of light )

Electromagnetic Radiation Exhibits Wave and Particulate Properties

<p><span style="font-family: &quot;Times New Roman&quot;;">Einstein assumed that light traveled in energy packets called <strong>photons</strong>.</span></p><p style="text-align: left;"><span style="font-family: &quot;Times New Roman&quot;;">&nbsp; •&nbsp; The energy of one photon is: image</span></p><p style="text-align: left;"><span style="font-family: &quot;Times New Roman&quot;;"><em>h = </em>6.626 </span><span>´</span><span style="font-family: &quot;Times New Roman&quot;;"> 10<sup>–34</sup> Js is (Planck’s constant )</span></p><p style="text-align: left;"><span style="font-family: &quot;Times New Roman&quot;;"><em>c = </em>3.00 x 10<sup>8</sup> m/s <em>&nbsp;</em>(<em>speed of light </em>)</span></p><p style="text-align: left;"><span style="font-family: &quot;Times New Roman&quot;;">Electromagnetic Radiation Exhibits Wave and Particulate Properties</span></p>
40
New cards

Each Element Has Its Own Atomic Emission Spectrum of Light

  • Neon atoms inside a glass tube absorb energy in the form of heat, light or electricity, then re-emit the energy as light.  For  example:

  • Neon sign composed of two or more glass tubes filled with neon gas. When electric current is passed through the tube, the neon gas absorb some of the electrical energy and reemit it as red color. If the atoms in the tube are different, they  emit light of different color.

  • Mercury atoms emit light that appears blue.


<ul><li><p><span>Neon atoms inside a glass tube absorb energy in the form of heat, light or electricity, then re-emit the energy as light.&nbsp; For&nbsp; example:</span></p></li><li><p><span>Neon sign composed of two or more glass tubes filled with neon gas. When electric current is passed through the tube, the neon gas absorb some of the electrical energy and reemit it as red color. If the atoms in the tube are different, they&nbsp; emit light of different color.</span></p></li><li><p><span>Mercury atoms emit light that appears blue.</span></p></li></ul><p></p>
41
New cards

Exciting Gas Atoms to Emit Light with Electrical Energy

knowt flashcard image
42
New cards

Emission Spectra

A white-light spectrum is continuous, with some radiation emitted at every wavelength.
The emission spectrum of an individual element includes only certain specific wavelength and produces its own, unique and distinctive emission spectrum. They consists of bright spots or lines at specific wavelengths with complete darkness in between.

<p><span style="font-family: &quot;Aptos Display&quot;;">A white-light spectrum is continuous, with some radiation emitted at every wavelength. <br>The emission spectrum of an individual element includes only certain specific wavelength and produces its own, unique and distinctive emission spectrum. They consists of bright spots or lines at specific wavelengths with complete darkness in between.</span></p>
43
New cards

Continuous Spectrum

  • A spectrum is produced when radiation from a polychromatic source is separated into its component wavelengths. This rainbow of colors, containing light of all wavelengths, is called a  continuous spectrum.

  • For atoms and molecules, one does not observe a continuous spectrum, as one gets from a white light source. Only a line spectrum of discrete wavelengths is observed.

  • (a) A continuous spectrum containing all  wavelengths of visible light.

    (b) The hydrogen line spectrum contains only a few discrete wavelengths.


<ul><li><p><span style="font-family: &quot;Times New Roman&quot;;">A spectrum is produced when radiation from a polychromatic source is separated into its component wavelengths. This rainbow of colors, containing light of all wavelengths, is called a&nbsp; <strong>continuous spectrum</strong>.</span></p></li><li><p><span style="font-family: &quot;Times New Roman&quot;;">For atoms and molecules, one does not observe a continuous spectrum, as one gets from a white light source. Only a <strong>line spectrum</strong> of discrete wavelengths is observed.</span></p></li><li><p><span style="font-family: &quot;Times New Roman&quot;;"><strong>(a) </strong>A continuous spectrum containing all&nbsp; wavelengths of visible light.</span></p><p style="text-align: left;"><span style="font-family: &quot;Times New Roman&quot;;"><strong>(b)</strong> The hydrogen line spectrum contains only a few discrete wavelengths.</span></p></li></ul><p></p>
44
New cards
<p>Line Spectra</p>

Line Spectra

One of the biggest scientific mysteries in the early twentieth century involved the emission spectra observed from energy emitted by atoms and molecules. A line spectrum shows only certain colors or specific wavelengths of light. When atoms are heated, they emit light. This process produces a line spectrum that is specific to that atom. The emission spectra of three elements are shown on the left of this slide.


<p><span style="font-family: &quot;Times New Roman&quot;;">One of the biggest scientific mysteries in the early twentieth century involved the emission spectra observed from energy emitted by atoms and molecules. A <strong>line spectrum</strong> shows only certain colors or specific wavelengths of light. When atoms are heated, they emit light. This process produces a line spectrum that is specific to that atom. The emission spectra of three elements are shown on the left of this slide.</span></p><p></p>
45
New cards
<p>The Bohr Model</p>

The Bohr Model

  • To explain the line spectrum of hydrogen, Bohr assumed that electrons in hydrogen atoms move in circular orbits around the nucleus.

  • Bohr noted the line spectra of certain elements and assumed that electrons were confined to specific energy states. These were called orbits.

  • Bohr’s model is based on three postulates:

  • In his model, Bohr used Planck’s and Einstein’s ideas about quantized energy and proposed three postulates:

    • The H atom has only certain energy levels, which Bohr called stationary states. Each state is associated with a fixed, circular orbit of the electron around the nucleus. The higher the energy level, the farther the orbit is from the nucleus.

    • The atom does not radiate energy while in one of its stationary states. Even though it violates principles of classical physics, the atom does not change its energy while the electron moves within an orbit.

    • The atom changes to another stationary state (the electron moves to another orbit) only by absorbing or emitting a photon. The energy of the photon (hv) equals the difference in the energies of the two states:

               Ephoton =∆Eatom = Efinal -  Einitial = hv      

  • In 1913, Neils Bohr, a Danish scientist, set down postulates to account for:

  • The stability of the hydrogen atom.

  • The line spectrum of the atom.


<ul><li><p><span style="font-family: &quot;Times New Roman&quot;;">To explain the line spectrum of hydrogen, Bohr assumed that electrons in hydrogen atoms move in circular orbits around the nucleus.</span></p></li><li><p><span style="font-family: &quot;Times New Roman&quot;;">Bohr noted the line spectra of certain elements and assumed that electrons were confined to specific energy states. These were called orbits.</span></p></li><li><p><span style="font-family: &quot;Times New Roman&quot;;">Bohr’s model is based on three postulates:</span></p></li><li><p><span>In his model, Bohr used Planck’s and Einstein’s ideas about quantized energy and proposed three postulates:</span></p><ul><li><p><span>The H atom has only certain energy levels,&nbsp;which Bohr called&nbsp;<strong>stationary states.</strong>&nbsp;Each state is associated with a fixed, circular orbit of the electron around the nucleus. The higher the energy level, the farther the orbit is from the nucleus.</span></p></li><li><p><span>The atom does&nbsp;<strong>not</strong>&nbsp;radiate energy while in one of its stationary states.&nbsp;Even though it violates principles of classical physics, the atom does not change its energy while the electron moves&nbsp;within&nbsp;an orbit.</span></p></li><li><p><span>The atom changes to another stationary state&nbsp;(the electron moves to another orbit)&nbsp;only by absorbing or emitting a photon. The energy of the photon (hv) equals the difference in the energies of the two states:</span></p></li></ul></li></ul><p><span style="font-family: &quot;Times New Roman&quot;;">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; E<sub>photon</sub> =</span><span style="font-family: &quot;Cambria Math&quot;;">∆</span><span style="font-family: &quot;Times New Roman&quot;;">E<sub>atom</sub> = E<sub>final</sub> -&nbsp; E<sub>initial</sub> = hv&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</span></p><ul><li><p style="text-align: left;"><span style="font-family: &quot;Times New Roman&quot;;">In 1913, Neils Bohr, a Danish scientist, set down postulates to account for:</span></p></li><li><p style="text-align: left;"><span style="font-family: &quot;Times New Roman&quot;;">The stability of the hydrogen atom.</span></p></li><li><p style="text-align: left;"><span style="font-family: &quot;Times New Roman&quot;;">The line spectrum of the atom.</span></p></li></ul><p></p>
46
New cards
<p>The Bohr Model (2)</p>

The Bohr Model (2)

When an atom absorbs energy, the electrons can be promoted to higher energy levels

When an atom emits energy, the electrons can be decayed to a lower energy level.

  • When an atom absorbs energy, an e- in one of these fixed orbits is excited and promoted to the higher energy level farther away from the nucleus.

  • However, in this new configuration, the atom is less stable and the electron quickly fall back to the lower energy level,

  • As it does so, It releases a photon of light containing precise amount of energy- the Quantum of energy- that correspond to the energy difference between the two orbits.


<p><span style="font-family: &quot;Times New Roman&quot;;"><strong>When an atom absorbs energy, the electrons can be promoted to higher energy levels</strong></span></p><p><span style="font-family: &quot;Times New Roman&quot;;"><strong>When an atom emits energy, the electrons can be decayed to a lower energy level.</strong></span></p><ul><li><p>When an atom absorbs energy, an e- in one of these fixed orbits is excited and promoted to the higher energy level farther away from the nucleus.</p></li><li><p>However, in this new configuration, the atom is less stable and the  electron quickly fall back to the lower energy level, </p></li><li><p>As it does so, It releases a photon of light containing precise amount of energy- the Quantum of energy- that correspond to the energy difference between the two orbits.</p></li></ul><p></p>
47
New cards

Hydrogen Emission Lines

When an atom absorbs energy, an electron in one of these orbits is excited and promoted to the orbit that is farther away from the nucleus and therefore higher in energy. In this configuration, an atom is less stable and the electron quickly falls back to a lower energy orbit and emits  photon of light containing precise amount of energy- quantum of energy equal to the difference between the two levels.

<p><span>When an atom absorbs energy, an electron in one of these orbits is excited and promoted to the orbit that is farther away from the nucleus and therefore higher in energy. In this configuration, an atom is less stable and the electron quickly falls back to a lower energy orbit and emits&nbsp; photon of light containing precise amount of energy- quantum of energy equal to the difference between the two levels.</span></p>
48
New cards

Hydrogen Emission Lines (2)

The amount of energy in a photon is directly related to its wavelength.

The light emitted by an excited electron in an atom consist of

  • Specific lines at specific wavelengths, each corresponding to a specific transition between two orbits.


<p><span>The amount of energy in a photon is directly related to its wavelength.</span></p><p style="text-align: left;"><span>The light emitted by an excited electron in an atom consist of</span></p><ul><li><p style="text-align: left;"><span>Specific lines at specific wavelengths, each corresponding to a specific transition between two orbits.</span></p></li></ul><p></p>
49
New cards

The Bohr Model (3)

Limitations of the Bohr Model

  • The Bohr Model has several limitations:

    • It cannot explain the spectra of atoms other than hydrogen.

    • Electrons do not move about the nucleus in circular orbits.

  • However, the model introduces two important ideas:

    • The energy of an electron is quantized: electrons exist only in certain energy levels described by quantum numbers.

    • Energy gain or loss is involved in moving an electron from one energy level to another.


50
New cards
<p><span style="font-family: &quot;Times New Roman&quot;;">Line Spectra and the Rydberg Equation</span></p>

Line Spectra and the Rydberg Equation

Spectroscopists studying atomic hydrogen identified several series of spectral lines in different regions of the electromagnetic spectrum. Figure: shows the spectral lines in the ultraviolet, visible, and infrared regions

Using data, not theory, the Swedish physicist Johannes Rydberg (1854–1919) developed a relationship, called the Rydberg equation, that predicts the position and wavelength of any line in a given series: (image)

where λ is the wavelength of the line, n1 and n2 are positive integers with n2>n1, and R is the Rydberg constant 1.096776×107m−1.

<p><span>Spectroscopists studying atomic hydrogen identified several series of spectral lines in different regions of the electromagnetic spectrum.&nbsp;<strong>Figure:</strong>&nbsp;shows the spectral lines in the ultraviolet, visible, and infrared regions</span></p><p><span>Using data, not theory, the Swedish physicist Johannes Rydberg (1854–1919) developed a relationship, called the&nbsp;<em>Rydberg equation,</em>&nbsp;that predicts the position and wavelength of any line in a given series: (image)</span></p><p><span>where λ is the wavelength of the line,&nbsp;n<sub>1</sub>&nbsp;and&nbsp;n<sub>2</sub>&nbsp;are positive integers with&nbsp;n<sub>2</sub>&gt;n<sub>1</sub>, and&nbsp;<em>R</em>&nbsp;is the Rydberg constant&nbsp;1.096776×10<sup>7</sup>m<sup>−1</sup>.</span></p>
51
New cards

Quantum Mechanics

  • In 1924, the French physicist Louis de Broglie suggested a surprising hypothesis: All matter exhibits both particulate and wave properties.

  • The wavelength of a particle of mass, m (kg), and velocity, v (m/s), is given by the de Broglie relation:  λ = h/mν

  • In 1926, Erwin Schrödinger, an Austrian physicist, building on de Broglie’s work developed an equation, now know as Schrödinger’s wave equation, that incorporates both the wave-like behavior of the electron and its particle like behavior.  Solving the wave equation leads to wave functions, or orbitals, that describe the electron in an atom and are designated with a lowercase Greek letter  psi ψ. Psi-squared, ψ 2, gives us the probability of finding the particle within a region of space. Wave functions ψ are also called orbitals. Electron density is another way of expressing probability. A region of high electron density is one where there is a high probability of finding an electron.

  • This theory, known as quantum mechanics, allows one mathematically describes the wave properties of subatomic particles, and to calculate the probability of finding an electron in space


<ul><li><p><span style="font-family: &quot;Times New Roman&quot;;">In 1924, the French physicist Louis de Broglie suggested a surprising hypothesis: <em>All matter exhibits both particulate and wave properties</em></span><span>.</span></p></li><li><p><span style="font-family: &quot;Times New Roman&quot;;">The wavelength of a particle of mass, <em>m</em> (kg), and velocity, <em>v</em> (m/s), is given by the <strong>de</strong> <strong>Broglie</strong> <strong>relation</strong>: <strong><em>&nbsp;λ = h/mν</em></strong></span></p></li><li><p><span style="font-family: &quot;Times New Roman&quot;;">In 1926, Erwin Schrödinger, an Austrian physicist, building on de Broglie’s work developed an equation, now know as Schrödinger’s wave equation, that incorporates both the wave-like behavior of the electron and its particle like behavior.&nbsp; Solving the wave equation leads to <strong>wave functions</strong>, or <strong>orbitals</strong>, that describe the electron in an atom and are designated with a lowercase Greek letter&nbsp; psi <em>ψ</em>. Psi-squared, <em>ψ</em> <sup>2</sup>, gives us the probability of finding the particle within a region of space. Wave functions<em> ψ</em> are also called <strong>orbitals</strong>. <strong>Electron density</strong> is another way of expressing probability. A region of high electron density is one where there is a high probability of finding an electron.</span></p></li><li><p><span style="font-family: &quot;Times New Roman&quot;;">This theory, known as quantum mechanics, allows one mathematically describes the wave properties of subatomic particles, and to calculate the probability of finding an electron in space</span></p></li></ul><p></p>
52
New cards

Quantum-Mechanical Orbitals

  • In the quantum-mechanical model, a number and a letter specify an orbital (or orbitals).

  • The lowest-energy orbital (ground state) in the quantum-mechanical model is called the 1s orbital. 

  • Quantum numbers: (image)


<ul><li><p><span>In the quantum-mechanical model, a number and a letter specify an orbital (or orbitals).</span></p></li><li><p><span>The lowest-energy orbital (ground state) in the quantum-mechanical model is called the <em>1s orbital</em>.&nbsp;</span></p></li><li><p><span>Quantum numbers:&nbsp;(image)</span></p></li></ul><p></p>
53
New cards

Quantum-Mechanical Orbitals (2)

  • In the quantum-mechanical model, a number and a letter specify an orbital (or orbitals).

  • The lowest-energy orbital (ground state) in the quantum-mechanical model is called the 1s orbital.

  • It is specified by the number 1 and the letter s.

  • The number is called the principal quantum number (n) and specifies the principal shell of the orbital.

  • The letter indicates the subshell of the orbital and specifies its shape.

  • The possible letters are s, p, d, and f, each with a different shape

  • The number of subshells in a given principal shell is equal to the value of n.


<ul><li><p><span>In the quantum-mechanical model, a number and a letter specify an orbital (or orbitals).</span></p></li><li><p><span>The lowest-energy orbital (ground state) in the quantum-mechanical model is called the <em>1s orbital</em>.</span></p></li><li><p><span>It is specified by the number 1 and the letter <em>s</em>.</span></p></li><li><p><span>The number is called the <strong>principal quantum number</strong> (<em>n</em>) and specifies the <strong>principal shell</strong> of the orbital.</span></p></li><li><p><span>The letter indicates the <strong>subshell</strong> of the orbital and specifies its shape.</span></p></li><li><p><span>The possible letters are <em>s</em>, <em>p</em>, <em>d</em>, and <em>f</em>, each with a different shape</span></p></li><li><p><span>The number of subshells in a given principal shell is equal to the value of <em>n</em>.</span></p></li></ul><p></p>
54
New cards

Sublevels

With in each principal shell, are subshells. The number of subshells in a principal shell is equal to the value of its quantum number, n.

  • Energy levels are subdivided into sublevels (l).  Each subshell is identified with a one-letter label:

  • s, p, d, f,

  • n = 1 : the subshell 1. The orbital is 1S

  • n = 2 : the subshell 2: The orbitals are 2s, 2p.

  • n = 3 : the subshell 3:  the orbitals are 3s, 3p, 3d.

  • n = 4 : the subshell 4:  the orbitals are 4s, 4p, 4d and 4f.


<p><span>With in each principal shell, are subshells. The number of subshells in a principal shell is equal to the value of its quantum number, <strong>n.</strong></span></p><ul><li><p><span>Energy levels are subdivided into <strong>sublevels (<em>l</em>)</strong>.&nbsp; Each subshell is identified with a one-letter label:</span></p></li><li><p><span><strong><em> s, p, d, f,</em></strong></span></p></li><li><p><span><strong><em>n</em> = 1</strong> : the subshell <strong>1</strong>. The orbital is <strong>1<em>S</em></strong></span></p></li><li><p><span><strong><em>n</em> = 2</strong> : the subshell <strong>2</strong>: The orbitals are <strong>2<em>s</em></strong>, <strong>2<em>p.</em></strong></span></p></li><li><p><span><strong><em>n</em> = 3</strong> : the subshell <strong>3</strong>:&nbsp; the orbitals are <strong>3<em>s</em></strong>, <strong>3<em>p</em></strong>, <strong>3<em>d.</em></strong></span></p></li><li><p><span><strong><em>n</em> = 4</strong> : the subshell <strong>4</strong>:&nbsp; the orbitals are <strong>4<em>s</em></strong>, <strong>4<em>p</em></strong>, <strong>4<em>d and 4f.</em></strong></span></p></li></ul><p></p>
55
New cards

Quantum Numbers, Their Properties, and Significance

knowt flashcard image
56
New cards

Orbitals and Quantum Numbers

  • The Bohr model was a one-dimensional model that used one quantum number to describe the distribution of electrons in the atom. The only information that was important was the size of the orbit, which was described by the n quantum number.

  • Schrödinger's model allowed the electron to occupy three-dimensional space. It therefore required three coordinates, or three quantum numbers, to describe the orbitals in which electrons can be found.

  • The three coordinates that come from Schrödinger's wave equations are:

  • the principal quantum number (n=1,2,3,4,…),

  • the angular momentum quantum number (l=0,1,…(n-1)), and

  • the magnetic quantum number (ml= -l,…0,…+l).

  • These quantum numbers are related to the size (energy), shape, and orientation in space respectively of the orbitals on an atom.


<ul><li><p><span style="font-family: &quot;Times New Roman&quot;;">The Bohr model was a one-dimensional model that used one quantum number to describe the distribution of electrons in the atom. The only information that was important was the size of the orbit, which was described by the n quantum number.</span></p></li><li><p><span style="font-family: &quot;Times New Roman&quot;;">Schrödinger's model allowed the electron to occupy three-dimensional space. It therefore required three coordinates, or three quantum numbers, to describe the orbitals in which electrons can be found.</span></p></li><li><p><span style="font-family: &quot;Times New Roman&quot;;">The three coordinates that come from Schrödinger's wave equations are:</span></p></li><li><p><span style="font-family: &quot;Times New Roman&quot;;">the principal quantum number (</span><span><em>n=1,2,3,4,…)</em></span><span style="font-family: &quot;Times New Roman&quot;;">,</span></p></li><li><p><span style="font-family: &quot;Times New Roman&quot;;">the angular momentum quantum number (<em>l=0,1,…(n-1)</em>), and</span></p></li><li><p><span style="font-family: &quot;Times New Roman&quot;;">the magnetic quantum number (<em>m<sub>l</sub>= -l,…0,…+l</em>).</span></p></li><li><p><span style="font-family: &quot;Times New Roman&quot;;">These quantum numbers are related to the size (energy), shape, and orientation in space respectively of the orbitals on an atom.</span></p></li></ul><p></p>
57
New cards
<p><span style="font-family: &quot;Times New Roman&quot;;">The s Orbitals</span></p>

The s Orbitals

• All s orbitals are spherical.

• As n increases, the s orbitals get larger.

•n orbitals hold a maximum of two electrons

<p><span style="font-family: &quot;Times New Roman&quot;;">• All <em>s </em>orbitals are spherical.</span></p><p style="text-align: left;"><span style="font-family: &quot;Times New Roman&quot;;">• As <em>n </em>increases, the <em>s </em>orbitals get larger.</span></p><p><span style="font-family: &quot;Times New Roman&quot;;">•<strong><em>n</em> orbitals hold a maximum of two electrons</strong></span></p>
58
New cards

The p Orbitals

  • There are three p orbitals: px, py and pz.

    • The three p orbitals lie along the x-, y-, and z-axes of a Cartesian system.

    • The letters correspond to allowed values of ml of -1, 0, and +1.

  • The orbitals are dumbbell shaped; each has two lobes.

  • As n increases, the p orbitals get larger.

  • All p orbitals have a node at the nucleus.

  • 6 total electrons can occupy the three p orbitals of a subshell.


<ul><li><p><span style="font-family: &quot;Times New Roman&quot;;">There are three <em>p </em>orbitals: <em>p<sub>x</sub></em>, <em>p<sub>y</sub> </em>and <em>p<sub>z</sub></em>.</span></p><ul><li><p><span style="font-family: &quot;Times New Roman&quot;;">The three <em>p </em>orbitals lie along the <em>x</em>-, <em>y</em>-, and <em>z</em>-axes of a Cartesian system.</span></p></li><li><p><span style="font-family: &quot;Times New Roman&quot;;">The letters correspond to allowed values of <em>m<sub>l </sub></em>of -1, 0, and +1.</span></p></li></ul></li><li><p><span style="font-family: &quot;Times New Roman&quot;;">The orbitals are dumbbell shaped; each has two <em>lobes</em>.</span></p></li><li><p><span style="font-family: &quot;Times New Roman&quot;;">As <em>n </em>increases, the <em>p </em>orbitals get larger.</span></p></li><li><p><span style="font-family: &quot;Times New Roman&quot;;">All <em>p </em>orbitals have a node at the nucleus.</span></p></li><li><p><span style="font-family: &quot;Times New Roman&quot;;"><strong>6 total electrons can occupy the three <em>p</em> orbitals of a subshell.</strong></span></p></li></ul><p></p>
59
New cards

The d Orbitals

  • There are five d orbitals.

  • The letters correspond to allowed values of ml of -2, -1, 0, +1 and +2.

  • Three of the d orbitals lie in a plane bisecting the x-, y-, and z-axes.

  • Two of the d orbitals lie in a plane aligned along the x-, y-, and z-axes.

  • Four of the d orbitals have four lobes each.

  • One d orbital has two lobes and a collar.

  • 10 total electrons can occupy the five d orbitals of a subshell.


<ul><li><p><span style="font-family: &quot;Times New Roman&quot;;">There are five <em>d </em>orbitals<em>.</em></span></p></li><li><p><span style="font-family: &quot;Times New Roman&quot;;">The letters correspond to allowed values of <em>m<sub>l</sub> </em>of -2, -1, 0, +1 and +2.</span></p></li><li><p><span style="font-family: &quot;Times New Roman&quot;;">Three of the <em>d </em>orbitals lie in a plane bisecting the </span><span><em>x</em></span><span style="font-family: &quot;Times New Roman&quot;;">-, <em>y</em>-, and <em>z</em>-axes.</span></p></li><li><p><span style="font-family: &quot;Times New Roman&quot;;">Two of the <em>d </em>orbitals lie in a plane aligned along the <em>x</em>-, <em>y</em>-, and <em>z</em>-axes.</span></p></li><li><p><span style="font-family: &quot;Times New Roman&quot;;">Four of the <em>d </em>orbitals have four lobes each.</span></p></li><li><p><span style="font-family: &quot;Times New Roman&quot;;">One <em>d </em>orbital has two lobes and a collar.</span></p></li><li><p><span style="font-family: &quot;Times New Roman&quot;;"><strong>10 total electrons can occupy the five <em>d</em> orbitals of a subshell.</strong></span></p></li></ul><p></p>
60
New cards

The f Orbitals

  • There are seven f orbitals.

  • The letters correspond to allowed values of ml of -3,-2, -1, 0, +1, +2 and +3.

  • 14 total electrons can occupy the seven f orbitals of a subshell.


<ul><li><p><span style="font-family: &quot;Times New Roman&quot;;">There are seven <em>f </em>orbitals.</span></p></li><li><p><span style="font-family: &quot;Times New Roman&quot;;">The letters correspond to allowed values of <em>m<sub>l</sub> </em>of -3,-2, -1, 0, +1, +2 and +3.</span></p></li><li><p><span style="font-family: &quot;Times New Roman&quot;;"><strong>14 total electrons can occupy the seven <em>f</em> orbitals of a subshell.</strong></span></p></li></ul><p></p>
61
New cards
<p><span style="font-family: &quot;Times New Roman&quot;;">Spin Quantum Number, <em>m</em></span><span><em><sub>s</sub></em></span></p>

Spin Quantum Number, ms

The Stern- Gerlach experiment

  • In 1921, it was discovered that two electrons in the same orbital do not have exactly the same energy.

  • The “spin” of an electron describes its magnetic field, which affects its energy.

  • This led to a fourth quantum number, the spin quantum number, ms.

  • The spin quantum number has only 2 allowed values:  +1/2 and −1/2.


<p><span style="font-family: &quot;Times New Roman&quot;;"><strong><em>The Stern- Gerlach experiment</em></strong></span></p><ul><li><p><span style="font-family: &quot;Times New Roman&quot;;">In 1921, it was discovered that two electrons in the same orbital do not have exactly the same energy.</span></p></li><li><p><span style="font-family: &quot;Times New Roman&quot;;">The “spin” of an electron describes its magnetic field, which affects its energy.</span></p></li><li><p><span style="font-family: &quot;Times New Roman&quot;;">This led to a fourth quantum number, the spin quantum number, <em>m<sub>s</sub>.</em></span></p></li><li><p><span style="font-family: &quot;Times New Roman&quot;;">The spin quantum number has only 2 allowed values:&nbsp; +1/2 and −1/2.</span></p></li></ul><p></p>
62
New cards

Energy Levels of Electrons

Quantum Shell Summary

  • With each new principal quantum shell, a new subshell type is introduced

  • n:  gives the shell

    l:  gives the subshell

    ml:  gives the orbital


<p><span style="font-family: &quot;Times New Roman&quot;;">Quantum Shell Summary</span></p><ul><li><p><span style="font-family: &quot;Times New Roman&quot;;">With each new principal quantum <strong>shell</strong>, a new <strong>subshell</strong> type is introduced</span></p></li><li><p><span><em>n: </em>&nbsp;gives the shell</span></p><p style="text-align: left;"><span><em>l:&nbsp; </em>gives the subshell</span></p><p><span><em>m<sub>l</sub>: </em>&nbsp;gives the orbital</span></p></li></ul><p></p>
63
New cards

Atomic Structure - Electron Configuration

There are several methods of depicting atomic structures of atoms:

1) An Electron Configuration shows the distribution of all electrons in an atom.

  • Each component consists of nl#

    • A number denoting the energy level,

    • A letter denoting the type of orbital,

    • A superscript denoting the number of electrons in those orbitals.

  • Maximum Electrons in Subshells

  • An s subshell, with one orbital, can hold a maximum of 2 electrons.

  • A p subshell, with three orbitals, can hold a maximum of 6 electrons.

  • A d subshell, with five orbitals, can hold a maximum of 10 electrons.

  • An f subshell, with seven orbitals, can hold a maximum of 14 electrons


<p><span style="font-family: &quot;Times New Roman&quot;;">There are several methods of depicting atomic structures of atoms:</span></p><p><span style="font-family: &quot;Times New Roman&quot;;">1) An <strong><em>Electron Configuration </em></strong>shows<em> </em>the distribution of all electrons in an atom.</span></p><ul><li><p><span style="font-family: &quot;Times New Roman&quot;;">Each component consists of <em>nl<sup>#</sup></em></span></p><ul><li><p><span style="font-family: &quot;Times New Roman&quot;;">A number denoting the energy level,</span></p></li><li><p><span style="font-family: &quot;Times New Roman&quot;;">A letter denoting the type of orbital,</span></p></li><li><p><span style="font-family: &quot;Times New Roman&quot;;">A superscript denoting the number of electrons in those orbitals.</span></p></li></ul></li><li><p style="text-align: left;"><span style="font-family: &quot;Times New Roman&quot;;">Maximum Electrons in Subshells</span></p></li><li><p style="text-align: left;"><span style="font-family: &quot;Times New Roman&quot;;">An <em>s</em> subshell, with one orbital, can hold a maximum of 2 electrons.</span></p></li><li><p style="text-align: left;"><span style="font-family: &quot;Times New Roman&quot;;">A <em>p</em> subshell, with three orbitals, can hold a maximum of 6 electrons.</span></p></li><li><p style="text-align: left;"><span style="font-family: &quot;Times New Roman&quot;;">A <em>d</em> subshell, with five orbitals, can hold a maximum of 10 electrons.</span></p></li><li><p style="text-align: left;"><span style="font-family: &quot;Times New Roman&quot;;">An <em>f</em> subshell, with seven orbitals, can hold a maximum of 14 electrons</span></p></li></ul><p></p>
64
New cards

Electron Configuration

Distribution of electrons among the various atomic orbitals in an atom (help us understand and predict properties of elements)

Aufbau principle: “Fill up” electrons in lowest energy orbitals

<p><span>Distribution of electrons among the various atomic orbitals in an atom<em> </em>(help us understand and predict properties of elements)</span></p><p><span><strong>Aufbau principle: </strong>“Fill up” electrons in lowest energy orbitals</span></p>
65
New cards

Electron Filling

The colored areas of elements show the different subshells that are filling with those elements.

  • We fill orbitals in increasing order of energy.

  • Different blocks on the periodic table (shaded in different colors in this chart) correspond to different types of orbitals


<p><span>The colored areas of elements show the different subshells that are filling with those elements.</span></p><ul><li><p><span style="font-family: &quot;Times New Roman&quot;;">We fill orbitals in increasing order of energy.</span></p></li><li><p><span style="font-family: &quot;Times New Roman&quot;;">Different blocks on the periodic table (shaded in different colors in this chart) correspond to different types of orbitals</span></p></li></ul><p></p>
66
New cards
<p><span>Electron Configuration (2)</span></p>

Electron Configuration (2)

Energy order of atomic subshells:

1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p

Arrange all electrons in an atom from the lowest energy level to the higher energy levels. 

(Ground State Electron Configuration)

<p><span><strong>Energy order of atomic subshells:</strong></span></p><p><span>1s &lt; 2s &lt; 2p &lt; 3s &lt; 3p &lt; 4s &lt; 3d &lt; 4p</span></p><p><span>Arrange all electrons in an atom from the lowest energy level to the higher energy levels.&nbsp;</span></p><p><span>(<strong>Ground State Electron Configuration)</strong></span></p>
67
New cards

General Properties of gases

You have learned that matter can exist in three phases; solid, liquid, and gas.

In the gas phase, gas molecules:

  • Have an indefinite volume

  • Have an indefinite shape

  • Have low densities

  • Have high kinetic energies

Four factors affect the properties of gases:

1)P = pressure

2)T = Kelvin temperature

3)V = volume

4)Number of gas molecules

when the gas is released, the pressure of the air escaping from the neck propels the balloon in a rapid irregular flight. If the balloon is inflated until it bursts, the gas escaping all at once causes an explosive noise.

<p><span>You have learned that matter can exist in three phases; solid, liquid, and gas.</span></p><p><span>In the gas phase, gas molecules:</span></p><ul><li><p><span>Have an indefinite volume</span></p></li><li><p><span>Have an indefinite shape</span></p></li><li><p><span>Have low densities</span></p></li><li><p><span>Have high kinetic energies</span></p></li></ul><p><span>Four factors affect the properties of gases:</span></p><p><span>1)P = pressure</span></p><p><span>2)T = Kelvin temperature</span></p><p><span>3)V = volume</span></p><p><span>4)Number of gas molecules</span></p><p><span>when the gas is released, the pressure of the air escaping from the neck propels the balloon in a rapid irregular flight. If the balloon is inflated until it bursts, the gas escaping all at once causes an explosive noise.</span></p>
68
New cards

Pressure dependence on the number of molecules and the temperature:

Gas particles are extremely small and move rapidly. When they hit the walls of the container, they exert a pressure. At specific temperature and volume, the number of collision depends on the number of  gas molecules present. The number of collision can be  increased by increasing the number of molecules.


Pressure is directly proportional to the number of molecules of gas present.


The pressure of a gas in a fixed volume also varies with temperature.

When the temperature is increased, the kinetic energy of the molecules increases, causing more frequent and more energetic  collisions of the molecules with the walls of the container. This results in the increase in pressure of the gas.

69
New cards

Kinetic Molecular Theory: A Model For Gases

  • Gas particles move in straight lines in all directions, colliding frequently with one another or with the walls of the container. As a result they fill the entire container.

  • Gases consist of tiny particles and they have no attraction for one another.  Particles are very far apart from each other.

  • There is a lot of space between gas particles compared with the size of the particles themselves. As a result gases have low density.

  • Gas particles have kinetic energy that increases with an increase in temperature.


<ul><li><p><span>Gas particles move in straight lines in all directions, colliding frequently with one another or with the walls of the container. As a result they fill the entire container.</span></p></li><li><p><span>Gases consist of tiny particles and they have no&nbsp;attraction for one another.&nbsp; Particles are very far apart from each other.</span></p></li><li><p><span>There is a lot of space between gas particles compared with the size of the particles themselves. As a result gases have low density.</span></p></li><li><p><span>Gas particles have kinetic energy that increases with an increase in temperature.</span></p></li></ul><p></p>
70
New cards

Atmospheric Pressure

Atmospheric pressure:  

It is the pressure exerted by a column of air from the top of the atmosphere to the surface of  Earth produces pressure on each of us of about 1 atm at sea level.

It decreases as altitude increases because there are fewer particles of gases in the air.

<p><span><strong>Atmospheric pressure:&nbsp;&nbsp;</strong></span></p><p style="text-align: left;"><span><strong>It </strong>is the pressure exerted by a column of air from the top of the atmosphere to the surface of &nbsp;Earth produces pressure on each of us of about 1 atm at sea level.</span></p><p style="text-align: left;"><span>It decreases as altitude increases because there are fewer particles of gases in the air.</span></p>
71
New cards

Pressure

Pressure: The pressure exerted by a gas

Sample is defined as the force per unit area that results from the collision of air particles with surrounding surfaces.

 P= force /  Area

At a pressure of exactly 1 atmosphere (atm), A mercury column in a inverted tube would be exactly 760 mm high, at a temperature of  0°C In the barometer, the pressure rises or falls with the changes in pressure.

Air pressure is measured in many units.

1 atm   =   760 mmHg

760 mmHg = 760 torr (exact)

 1atm = 101,325 Pa or 101.3KPa = 1.013 bar = 14.69 psi.  

The mercury barometer was invented by the physicist E. Torricelli for whom the unit of torr  was named.

<p><span>Pressure: The pressure exerted by a gas</span></p><p style="text-align: left;"><span>Sample is defined as the force per unit area that results from the collision of air particles with surrounding surfaces.</span></p><p style="text-align: left;"><span>&nbsp;P= force /&nbsp; Area</span></p><p style="text-align: left;"><span>At a pressure of exactly 1 atmosphere (atm), A mercury column in a inverted tube would be exactly 760 mm high, at a temperature of&nbsp; 0°C<sup> </sup>In the barometer, the pressure rises or falls with the changes in pressure.</span></p><p style="text-align: left;"><span>Air pressure is measured in many units.</span></p><p style="text-align: left;"><span>1 atm&nbsp;&nbsp; =&nbsp;&nbsp; 760 mmHg</span></p><p style="text-align: left;"><span>760 mmHg = 760 torr (exact)</span></p><p style="text-align: left;"><span>&nbsp;1atm = 101,325 Pa or 101.3KPa = 1.013 bar = 14.69 psi.&nbsp;&nbsp;</span></p><p style="text-align: left;"><span>The mercury barometer was invented by the physicist E. Torricelli for whom the unit of torr&nbsp; was named.</span></p>
72
New cards

Barometer

A mercury barometer can be prepared by completely filling a long tube with pure dry mercury and inverting the open end into a open dish of mercury.

It  measures the pressure exerted by the gases in the atmosphere which is equal to the downward pressure of the mercury column in a closed glass tube.

Convert

740. mmHg to

1) torr

2) atmospheres.

<p><span>A mercury barometer can be prepared by completely filling a long tube with pure dry mercury and inverting the open end into a open dish of mercury.</span></p><p style="text-align: left;"><span>It&nbsp; measures the pressure exerted by the gases in the atmosphere which is equal to the downward pressure of the mercury column in a closed glass tube.</span></p><p style="text-align: left;"><span>Convert</span></p><p style="text-align: left;"><span>740. mmHg to</span></p><p style="text-align: left;"><span>1) torr</span></p><p style="text-align: left;"><span>2) atmospheres.</span></p>
73
New cards

Gas Laws

By studying the gas laws, we can understand the behavior of a gas with respect to its volume, temperature and pressure as well as the number of moles.

74
New cards
<p>Boyle’s Law</p>

Boyle’s Law

At constant temperature (T), the volume (V) of a fixed mass of a gas is inversely proportional to the pressure (P)

where V is the volume, P is the pressure and k is a proportionality constant that would depend on the number of moles of gas and the temperature.

<p><span>At constant temperature (T), the volume (<strong>V</strong>) of a fixed mass of a gas is inversely proportional to the pressure (<strong>P</strong>)</span></p><p><span>where V is the volume, P is the pressure and k is a proportionality constant that would depend on the number of moles of gas and the temperature.</span></p>
75
New cards

Boyle's Law (2)

As the volume decreases, there is a decrease in the surface area of the container. The air particles are crowded together, more collision occurs, and the pressure increases with in the container.

<p><span>As the volume decreases, there is a decrease in the surface area of the container. The air particles are crowded together, more collision occurs, and the pressure increases with in the container.</span></p>
76
New cards
<p>Calculations using Boyle’s Law</p>

Calculations using Boyle’s Law

Freon-12, CCl2F2, was used in refrigeration systems. What is
the new volume of an 8.0L sample of Freon gas initially at

550 mmHg after its pressure is changed to 2200 mmHg at
constant temperature and moles?

SOLUTION:

STEP 1  Organize the data in a table of initial and final

               conditions.

     Temperature and moles remain constant.

STEP 2  Rearrange the gas law equation to solve for the          unknown quantity.   

      P1V1  =   P2V2              Boyle’s law

     To solve for V2, divide both sides by P2.

STEP 3  Substitute values into the gas law equation and calculate.

<p><span>Freon-12, CCl<sub>2</sub>F<sub>2</sub>, was used in refrigeration systems. What is <br>the new volume of an 8.0L sample of Freon gas initially at</span></p><p style="text-align: left;"><span>550 mmHg after its pressure is changed to 2200 mmHg at <br>constant temperature and moles?</span></p><p style="text-align: left;"><span><strong>SOLUTION:</strong></span></p><p style="text-align: left;"><span><strong>STEP 1&nbsp; Organize the data in a table of initial and final</strong></span></p><p style="text-align: left;"><span><strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; conditions.</strong></span></p><p style="text-align: left;"><span><strong>&nbsp; &nbsp; </strong>&nbsp;Temperature and moles remain constant.</span></p><p style="text-align: left;"><span><strong>STEP 2 &nbsp;Rearrange the gas law equation to solve for the &nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;unknown quantity.&nbsp; </strong>&nbsp;</span></p><p style="text-align: left;"><span>&nbsp; &nbsp; &nbsp; <em>P</em><sub>1</sub><em>V</em><sub>1</sub>&nbsp; =&nbsp;&nbsp; <em>P</em><sub>2</sub><em>V</em><sub>2&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </sub><strong><sub>&nbsp;</sub>Boyle’s law</strong></span></p><p><span>&nbsp; &nbsp;&nbsp; To solve for <em>V</em><sub>2</sub>, divide both sides by <em>P</em><sub>2</sub>.</span></p><p><span><strong>STEP 3&nbsp; Substitute values into the gas law equation and calculate.</strong></span></p>
77
New cards

Charles’ Law

Charles’ Law states that, at constant pressure, the volume of a gas is directly       proportional to the absolute temperature. 

Mathematically,   we would write: V ∝ T

                                                or   V =  kT     or   V  = k ( at constant pressure)

                                                                              T

where V is the volume, T is the temperature and k is a proportionality constant that would depend on the number of moles of gas and the pressure. If the pressure and amount of gas is constant, and the gas experiences a change in temperature, we can use Charles’ Law to predict the new volume.

<p><span>Charles’ Law states that, at constant pressure, the volume of a gas is directly&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; proportional to the absolute temperature.&nbsp;</span></p><p style="text-align: left;"><span>Mathematically,&nbsp;&nbsp; we would write: V ∝ T</span></p><p style="text-align: left;"><span>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; or&nbsp;&nbsp; V =&nbsp; kT&nbsp;&nbsp;&nbsp;&nbsp; or&nbsp;&nbsp; <u>V </u>&nbsp;= k ( at constant pressure)</span></p><p style="text-align: left;"><span>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; T</span></p><p style="text-align: justify;"><span>where V is the volume, T is the temperature and k is a proportionality constant that would depend on the number of moles of gas and the pressure. If the pressure and amount of gas is constant, and the gas experiences a change in temperature, we can use Charles’ Law to predict the new volume.</span></p>
78
New cards
<p>Charles’ Law (2)</p>

Charles’ Law (2)

When the gas is heated, the molecules move faster, and their K.E. energy increases, causing increase in the number of collision .This action  should increase the pressure. With the external pressure constant , The net result is in increase in volume due to an increase in temperature.

If we increase the temperature of a gas sample, the volume of the gas container must increase. If the temperature of the gas is decreased ,  the volume of the container must also decrease when the pressure and amount of gas are not changed.

All temperatures used in gas law calculations must be converted to their corresponding Kelvin (K) temperatures.

<p><span>When the gas is heated, the molecules move faster, and their K.E. energy increases, causing increase in the number of collision .This action&nbsp; should increase the pressure. With the external pressure constant , The net result is in increase in volume due to an increase in temperature.</span></p><p><span>If we increase the temperature of a gas sample, the volume of the gas container must increase. If the temperature of the gas is decreased ,&nbsp; the volume of the container must also decrease when the pressure and amount of gas are not changed.</span></p><p style="text-align: left;"><span>All temperatures used in gas law calculations must be converted to their corresponding Kelvin (K) temperatures.</span></p>
79
New cards
<p><span>Calculations using the Charles’s Law</span></p>

Calculations using the Charles’s Law

A balloon has a volume of 785 mL at 21o C. If the temperature drops to 0°C , what is the new volume of the balloon at constant pressure and moles?


SOLUTION:

STEP 1  Organize the data into a table of initial and final      conditions.

Pressure and moles remain constant

All temperatures used in gas law calculations must be converted to their corresponding Kelvin (K) temperatures.

STEP 2  Rearrange to solve for unknown quantity, V2.

STEP  3  Substitute the values into the gas law
            equation and calculate.

<p><span>A balloon has a volume of 785 mL at 21</span><span style="font-family: &quot;Times New Roman&quot;;"><sup>o </sup></span><span>C. If the temperature drops to 0°C , what is the new volume of the balloon at constant pressure and moles?</span></p><p style="text-align: left;"></p><p style="text-align: left;"><span><strong>SOLUTION:</strong></span></p><p style="text-align: left;"><span><strong>STEP 1&nbsp; Organize the data into a table of initial and final &nbsp; &nbsp;&nbsp;&nbsp;conditions.</strong></span></p><p style="text-align: left;"><span>Pressure and moles remain constant</span></p><p style="text-align: left;"><span>All temperatures used in gas law calculations must be converted to their corresponding Kelvin (K) temperatures.</span></p><p style="text-align: left;"><span><strong>STEP 2 </strong>&nbsp;<strong>Rearrange to solve for unknown quantity, <em>V</em><sub>2</sub>.</strong></span></p><p style="text-align: left;"><span><strong>STEP&nbsp; 3&nbsp; Substitute the values into the gas law<br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; equation and calculate.</strong></span></p>
80
New cards

Charles’ Law (3)

As the temperature rises , the particles of the gas move faster and  hit the walls of the container more often and with greater force. If volume and amount of gas  do not change, the pressure would increase.

<p><span>As the temperature rises , the particles of the gas move faster and&nbsp; hit the walls of the container more often and with greater force. If volume and amount of gas&nbsp; do not change, the pressure would increase.</span></p>
81
New cards

 Avogadro’s Law

Avogadro’s law, which states that the volume of a gas is directly related  to the  number of moles of a  gas when temperature and pressure do not change.

Equal volumes of different gases at constant T  and P contain the same number of molecules. 

Mathematically, this means V = kn   or   V ∝ n

<p><span>Avogadro’s law, which states that the volume of a gas is directly related&nbsp; to the&nbsp; number of moles of a&nbsp; gas when temperature and pressure do not change.</span></p><p><span>Equal volumes of different gases at constant <em>T</em>&nbsp; and <em>P </em>contain the same number of molecules.&nbsp;</span></p><p><span>Mathematically, this means <em>V</em> = <em>kn&nbsp;&nbsp; or&nbsp;&nbsp; V</em> ∝ n</span></p>
82
New cards
<p>Calculations using Avogadro’s Law</p>

Calculations using Avogadro’s Law

If 0.75 mole of helium gas occupies a volume of 1.5 L, what volume (in liters) will 1.2 moles of helium occupy at the same temperature and pressure?

A.  0.94 L 

B.  1.8 L 

C    2.4 L

If 0.75 mole of helium gas occupies a volume of 1.5 L, what volume (in liters) will 1.2 moles of helium occupy at the same temperature and pressure?

STEP 2  Rearrange to solve for unknown quantity, V2.


STEP 3  Substitute the values into the gas law equation and calculate.

The answer is C.

<p><span style="font-family: &quot;Times New Roman&quot;;">If 0.75 mole of helium gas occupies a volume of 1.5 L, what volume (in liters) will 1.2 moles of helium occupy at the same temperature and pressure?</span></p><p style="text-align: left;"><span style="font-family: &quot;Times New Roman&quot;;">A.&nbsp; 0.94 L&nbsp;</span></p><p style="text-align: left;"><span style="font-family: &quot;Times New Roman&quot;;">B.&nbsp; 1.8 L&nbsp;</span></p><p style="text-align: left;"><span style="font-family: &quot;Times New Roman&quot;;">C&nbsp;&nbsp;&nbsp; 2.4 L</span></p><p style="text-align: left;"><span style="font-family: &quot;Times New Roman&quot;;">If 0.75 mole of helium gas occupies a volume of 1.5 L, what volume (in liters) will 1.2 moles of helium occupy at the same temperature and pressure?</span></p><p style="text-align: left;"><span><strong>STEP 2&nbsp; Rearrange to solve for unknown quantity, <em>V</em><sub>2</sub>.</strong></span></p><p style="text-align: left;"></p><p style="text-align: left;"><span><strong>STEP 3&nbsp; Substitute the values into the gas law equation and calculate.</strong></span></p><p style="text-align: left;"><span>The answer is C.</span></p>
83
New cards

VOLUME AND MOLES :Avogadro’s Law

Using Avogadro’s  law,  we can say that any two gases will have equal volumes if they contain same number of moles of gas at the same temperature and pressure (STP)

STP is Standard Temperature and Pressure conditions. Standard temperature (T )  is 0° C (or 273 K); and standard pressure (P ) is 1 atm (760 mmHg).

At STP,  one mole of any gas occupies a volume of  22.4 L which is same as about the volume of three basket balls. This volume,


       22.4 L of any gas is called the molar volume.


 1 mole of gas = 22.4 L.(STP)

<p><span style="font-family: &quot;Times New Roman&quot;;">Using Avogadro’s&nbsp; law,&nbsp; we can say that any two gases will have equal volumes if they contain same number of moles of gas at the same temperature and pressure (STP)</span></p><p style="text-align: left;"><span style="font-family: &quot;Times New Roman&quot;;"><strong>STP </strong>is <strong>S</strong>tandard <strong>T</strong>emperature and <strong>P</strong>ressure conditions. <strong>Standard temperature (<em>T </em>)</strong>&nbsp; is 0</span><span>° </span><span style="font-family: &quot;Times New Roman&quot;;">C (or 273 K); and <strong>standard pressure (<em>P </em>)</strong> is 1 atm (760 mmHg).</span></p><p style="text-align: left;"><span style="font-family: &quot;Times New Roman&quot;;"><u>At STP</u>,&nbsp; one mole of any gas occupies a volume of&nbsp; <strong>22.4 L </strong>which is same as about the volume of three basket balls.<strong> This volume,</strong></span></p><p style="text-align: left;"></p><p style="text-align: left;"><span style="font-family: &quot;Times New Roman&quot;;"><strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 22.4 L of any gas is called the molar volume.</strong></span></p><p style="text-align: left;"></p><p style="text-align: left;"><span style="font-family: &quot;Times New Roman&quot;;">&nbsp;<strong>1 mole </strong>of gas = <strong>22.4 L</strong>.(STP)</span></p>
84
New cards

Mole/Mass/Volume Relationships

Standard Molar Volume: 1 mole of any gas occupies at STP a molar volume of 22.4 L. Standard temperature and pressure (STP) for reporting properties of gases:

273.15 K and 1 atm.

Molar volume can be used as a conversion factor if the

mass and volume occupied by a gas are known.

<p><span>Standard Molar Volume: 1 mole of any gas occupies at STP a molar volume of 22.4 L. Standard temperature and pressure <strong>(STP) </strong>for reporting properties of gases:</span></p><p style="text-align: center;"><span>273.15 K and 1 atm.</span></p><p style="text-align: center;"><span>Molar volume can be used as a conversion factor if the</span></p><p style="text-align: center;"><span>mass and volume occupied by a gas are known.</span></p>
85
New cards
<p><span><u>Gay-Lussac’s Law</u></span></p>

Gay-Lussac’s Law

At constant volume (V), the pressure (P) of a fixed mass of a gas is directly proportional to the absolute temperature (T)

No change in volume and amount of gas

All temperatures used in gas law calculations must be converted to their corresponding Kelvin (K) temperatures.


<p><span>At constant volume (V), the pressure (<strong>P</strong>) of a fixed mass of a gas is directly proportional to the absolute temperature (<strong>T</strong>)</span></p><p><span>No change in volume and amount of gas</span></p><p><span>All temperatures used in gas law calculations must be converted to their corresponding Kelvin (K) temperatures.</span></p><p></p>
86
New cards
<p><span>Calculations using Gay- Lussac’s law</span></p>

Calculations using Gay- Lussac’s law

A gas has a pressure at 2.0 atm at 18°C  . What is the new pressure when the temperature is 62o C ? (Volume and moles remain constant.)


SOLUTION:

STEP 1  Organize the data in a table of initial and final

               conditions.

     Volume and moles remain constant.

A gas has a pressure at 2.0  atm at 18°C . What is the new pressure when the temperature is 62o C ? (Volume and moles remain constant.)


STEP 2  Rearrange to solve for unknown quantity, P2.

Solve Gay-Lussac’s law for P2:

 


 

STEP 3  Substitute the values into the gas law equation and      calculate.

<p><span>A gas has a pressure at 2.0 atm at 18°C<sup> </sup>&nbsp;. What is the new pressure when the temperature is 62</span><span style="font-family: &quot;Times New Roman&quot;;"><sup>o</sup> </span><span>C ? (Volume and moles remain constant.)</span></p><p style="text-align: left;"></p><p style="text-align: left;"><span><strong>SOLUTION:</strong></span></p><p style="text-align: left;"><span><strong>STEP 1&nbsp; Organize the data in a table of initial and final</strong></span></p><p style="text-align: left;"><span><strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; conditions.</strong></span></p><p style="text-align: left;"><span>&nbsp; &nbsp;&nbsp; Volume and moles remain constant.</span></p><p style="text-align: left;"><span>A gas has a pressure at 2.0&nbsp; atm at 18°C . What is the new pressure when the temperature is 62</span><span style="font-family: &quot;Times New Roman&quot;;"><sup>o</sup> </span><span>C ? (Volume and moles remain constant.)</span></p><p style="text-align: left;"></p><p style="text-align: left;"><span><strong>STEP 2</strong>&nbsp; <strong>Rearrange to solve for unknown quantity, <em>P</em><sub>2</sub>.</strong></span></p><p style="text-align: left;"><span>Solve Gay-Lussac’s law for <em>P</em><sub>2</sub>:</span></p><p style="text-align: left;"><span>&nbsp;</span></p><p style="text-align: left;"></p><p style="text-align: left;"><span>&nbsp;</span></p><p style="text-align: left;"><span><strong>STEP 3&nbsp; Substitute the values into the gas law equation and &nbsp; &nbsp;&nbsp;&nbsp;calculate.</strong></span></p>
87
New cards

Combined Gas Laws

A combination of Boyle’s, Charles’s, and Gay-Lussac’s Laws.

Used in problems involving changes in P, T, and V with a constant amount of gas.

The volume of a fixed quantity of gas depends on the temperature and pressure.

It is not possible to state the volume of gas without stating the temperature and pressure.

Standard Temperature and Pressure (STP):

0.00 °C (273.15 K) and 1 atm (760 torr)

<p><span>A combination of Boyle’s, Charles’s, and Gay-Lussac’s Laws.</span></p><p><span>Used in problems involving changes in <em>P</em>, <em>T</em>, and <em>V </em>with a constant amount of gas.</span></p><p><span>The volume of a fixed quantity of gas depends on the temperature and pressure.</span></p><p><span>It is not possible to state the volume of gas without stating the temperature and pressure.</span></p><p><span>Standard Temperature and Pressure (STP):</span></p><p><span>0.00 °C (273.15 K) and 1 atm (760 torr)</span></p>
88
New cards
<p><span>Calculations using Combined gas laws</span></p>

Calculations using Combined gas laws

A gas has a volume of 675 mL at 35°C and 646 mmHg pressure. What is the volume (in milliliters) of the gas at −95o C and a pressure of 802 mmHg? (n is constant.)


SOLUTION:

STEP 1  Organize the data into a table of initial and final      conditions.

A gas has a volume of 675 mL at 35° C and 646 mmHg pressure. What is the volume (in milliliters) of the gas at −95o C  and a pressure of 802 mmHg? (n is constant.)

STEP 2  Rearrange to solve for unknown quantity, V2.


 



STEP 3  Substitute the values into the gas law equation
     and calculate.


<p><span>A gas has a volume of 675 mL at 35°C and 646 mmHg pressure. What is the volume (in milliliters) of the gas at −95</span><span style="font-family: &quot;Times New Roman&quot;;"><sup>o</sup> </span><span>C and a pressure of 802 mmHg? (<em>n</em> is constant.)</span></p><p style="text-align: left;"></p><p style="text-align: left;"><span><strong>SOLUTION:</strong></span></p><p style="text-align: left;"><span><strong>STEP 1</strong>&nbsp; <strong>Organize the data into a table of initial and final &nbsp; &nbsp;&nbsp;&nbsp;conditions.</strong></span></p><p style="text-align: left;"><span>A gas has a volume of 675 mL at 35° C and 646 mmHg pressure. What is the volume (in milliliters) of the gas at −95</span><span style="font-family: &quot;Times New Roman&quot;;"><sup>o</sup> </span><span>C &nbsp;and a pressure of 802 mmHg? (<em>n</em> is constant.)</span></p><p style="text-align: left;"><span><strong>STEP 2</strong>&nbsp; <strong>Rearrange to solve for unknown quantity, <em>V</em><sub>2</sub>.</strong></span></p><p style="text-align: left;"></p><p style="text-align: left;"><span><em>&nbsp;</em></span></p><p style="text-align: left;"></p><p style="text-align: left;"></p><p style="text-align: left;"><span><strong>STEP 3&nbsp; Substitute the values into the gas law equation <br>&nbsp; &nbsp;&nbsp; and calculate.</strong></span></p><p style="text-align: left;"></p>
89
New cards

Ideal-Gas Equation

So far, we’ve seen that

V ∝ 1/P, constant at constant T and n (Boyle’s law)

V ∝ T  , constant at constant P and n (Charles’s law)

V ∝ n , constant at constant P and T (Avogadro’s law)

Combining these, we get (image)

<p><span>So far, we’ve seen that</span></p><p><span><em>V</em> ∝ 1/<em>P</em>, constant at constant <em>T </em>and <em>n </em>(Boyle’s law)</span></p><p style="text-align: left;"><span><em>V</em> </span>∝<span> <em>T</em>&nbsp; , constant at constant <em>P </em>and <em>n </em>(Charles’s law)</span></p><p style="text-align: left;"><span><em>V</em>  </span>∝<span>&nbsp;n , constant at constant <em>P </em>and <em>T </em>(Avogadro’s law)</span></p><p style="text-align: left;"><span>Combining these, we get (image)</span></p>
90
New cards
<p>Ideal-Gas Equation (2)</p>

Ideal-Gas Equation (2)

The constant of proportionality is known as R, the gas constant.

<p><span>The constant of proportionality is known as <em>R</em>, the gas constant.</span></p>
91
New cards

Ideal Gas Law (3)

A single equation relating all properties of a gas.

<p><span>A single equation relating all properties of a gas.</span></p>
92
New cards

Ideal Gas Law and Molar Mass

The ideal gas law can also be written in terms of molar mass of a gas.

then substitute the “n”  into the ideal gas law:  PV = nRT

<p><span>The ideal gas law can also be written in terms of molar mass of a gas.</span></p><p><span>then substitute the “<em>n”</em>&nbsp; into the ideal gas law:&nbsp; <em>PV </em>= <em>nRT</em></span></p>
93
New cards

Density of gases:

The density d, of a gas is its mass per unit volume.

<p><span>The density d, of a gas is its mass per unit volume.</span></p>
94
New cards

Dalton’s Law of Partial Pressures

The total pressure of a mixture is the sum of the partial pressures of the different gases in the mixture.

Ptotal = P1 + P2 + P3 …

In the equation Ptotal is the total pressure of a mixture of gases, P1 is the partial pressure of gas 1; P2 is the partial pressure of gas 2; P3 is the partial pressure of gas 3; and so on.

Each gas behaves independently in the mixture.

Each gas obeys the Ideal Gas law.

<p><span>The total pressure of a mixture is the sum of the partial&nbsp;pressures of the different gases in the mixture.</span></p><p><span><em>P</em><sub>total</sub> = <em>P</em><sub>1</sub> + <em>P</em><sub>2</sub> + <em>P</em><sub>3 </sub>…</span></p><p style="text-align: center;"><span>In the equation <em>P<sub>total </sub></em>is the total pressure of a mixture of gases, P<sub>1</sub> is the partial pressure of gas 1; <em>P<sub>2</sub> </em>is the partial pressure of gas 2; <em>P<sub>3</sub> </em>is the partial pressure of gas 3; and so on.</span></p><p style="text-align: center;"><span>Each gas behaves independently in the mixture.</span></p><p><span>Each gas obeys the Ideal Gas law.</span></p>
95
New cards

Dalton’s Law of Partial Pressures (2)

Using the Ideal Gas Law:

<p><span>Using the Ideal Gas Law:</span></p>
96
New cards

Gas Stoichiometry Solution Map

At STP: the molar volume can be used as a conversion factor to convert between moles and volume.

Non STP Conditions: use the ideal gas law to convert between moles and volume.

<p><span>At STP: the molar volume can be used as a conversion factor to convert between moles and volume.</span></p><p><span>Non STP Conditions: use the ideal gas law to convert between moles and volume.</span></p>