Chapter 6: Thermochemistry

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73 Terms

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Thermochemistry

Study of energy absorbed of released in chemical reactions

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Energy

the capacity to do work or transfer heat

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Mechanical energy

energy that is due to an object’s motion, position, or both

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Kinetic energy

energy of motion

KE = 1/2mv²

dont have to memorize equation

m = mass (kg)

v = velocity (x/s)

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Potential Energy

energy related to position

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Table 6.1: Common Units of Energy

memorize: 1 calorie (cal) = 4.184 J

<p><strong>memorize: 1 calorie (cal) = 4.184 J</strong></p>
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The System and the Surroundings

  • It is important to be able to define where this energy is coming from and where it is going

  • System: the source of the energy, such as a container with chemical reactants and products

  • Surroundings: the rest of the universe

universe = system + surroundings

Energy can move between a system and its surroundings

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Open System

both matter and energy can move between the system and the surroundings (i.e. fireplace burning wood)

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Closed System

Energy but not matter can move between the system and the surroundings (i.e. Pressure cooker)

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Isolated System

Neither matter nor energy can leave or enter the system

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First Law of Thermodynamics

  • Energy can move between a system and its surroundings

  • states that energy cannot be created or destroyed, just transferred from one form to another and that the energy of the universe is constant

<ul><li><p>Energy can move between a system and its surroundings</p></li><li><p>states that <u>energy cannot be created or destroyed, just transferred from one form to another and that the energy of the universe is constant</u></p></li></ul><p></p>
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Work and Heat Flow

  • work, w, is the energy resulting froma force acting on an objec over a distance

  • The flow of energy that causes a temperature change in an object or its surroundings is known as heat, q

  • Both work and heat can be exchanged between the system and surroundings

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Work Done on the System

  • w is positive (w > 0)

  • If the system is a car, an example is a person pushing the car

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Work done by the system

  • w is negative (w < 0)

  • If the system is a car, an example is a moving car hitting a person

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Heat Can be absorbed or released by the system: Heat added to the system

  • q is positive (q > 0)

  • An example is heat transferred from a person’s hand to an ice cube (the system)

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Heat Can be absorbed or released by the system: Heat released by the system

  • q is negative (q < 0)

  • An examle is a bowl of cold water (the system) left in the freezer. Heat is transferred from the bowl into the freezer as the water cools and then freezes

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<p>Example 6.2.</p>

Example 6.2.

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Internal Energy, U or E

  • The internal energy of a system is defined as the sum of all kinetic and potential energies of the particles within a system

  • This includes the motions of fthe molecules inside the system

  • Absolute values of U or E are difficut to determine and usually we determine delta U (delta E)

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Changes in Internal Energy, DeltaE

  • Changes in internal energy, deltaU, are more commonly measured

deltaE = Efinal - Einitial

deltaE = q + w

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<p>Ex. 6.3</p>

Ex. 6.3

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State Function

describes the current state of a system and is independent of the path taken to achieve its value

  • Internal energy is a state function; delta H

<p>describes the current state of a system and is <em>independent</em> of the <em>path taken to achieve its value</em></p><ul><li><p>Internal energy is a state function; delta H</p></li></ul><p></p>
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Path Functions

  • The value of path functions depends on the path taken

  • They depend on the sequence of steps taken between initial and final states

  • Work and heat are path functions

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<p>E.x. 6.4</p>

E.x. 6.4

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Pressure-Volume Work

work done when there is a volume change as measured against an external pressure

  • Work; pressure, P; and the change in volume, deltaV are related

    • w = -PdeltaV

  • Where DeltaV = Vfinal - Vinitial

and 1 L atm = 101.325J

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Example: Work, heat and internal energy: work has sign

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<p>Example2: Work, heat and internal energy</p>

Example2: Work, heat and internal energy

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<p>Example3: Work, heat and internal energy</p>

Example3: Work, heat and internal energy

  1. Is heat positive or negative? (absorb or release)

  2. Is system or surrounding doing the work? (on or by)x

<ol><li><p>Is heat positive or negative? (absorb or release)</p></li><li><p>Is system or surrounding doing the work? (on or by)x</p></li></ol><p></p>
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Enthalpy (Heat at constant P)

  • Most chemical reactions are perofmed in open systems with exchange of both heat and work between system and surroundings

  • A state function that relates internal energy, pressure, and volume

H = U + PV

delta h = q subscript p

a change in enthalpy is equal to the flow of heat at constant pressure

When you heat water at a stove, you do not seal the container, means you are heating water at a constant pressure

Most reactions that you do inside chemistry lab (in test tube, spot plate and beaker, are done at constant pressure)

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Endothermic Process

reactions that absorb heat from surroundings

deltaH > 0 (positive) for endothermic reactions

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Exothermic Process

  • reactions that transfer heat to the surroundings

DeltaH < 0 (negative) for exothermic reactions

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<p>E.x. 6.6</p>

E.x. 6.6

Soln a: Heat is transferred from the system to the surroundings, so the reaction is exothermic and deltaH < 0

Soln b: Since this is an open system, the pressure is constant; deltaH = q = -250kJ

Soln c: deltaU = q + w = (-250kJ) + (-15 kJ) = -265 kJ

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Specific Heat

the number of calories required to raise exactly 1g of the substance by exactly 1ºC Typical units are J/(g x ºC)

In the absence of a phase change, the amount of heat required to change the temperature of a substance is given by q = mcdeltaT

m is the mass of the substance, c is the specific heat of the substance, and deltaT is the change in temperature in either degrees Celsius or Kelvins

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Table 6.3: Specific Heats of Selected Substances

memorize specific heat of water

<p>memorize specific heat of water</p>
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<p>E.x. 6.7</p>

E.x. 6.7

q = mcdeltaT

q/m x deltaT = c

cant cancel anything, so all units will be part of the answer

<p>q = mcdeltaT</p><p>q/m x deltaT = c</p><p>cant cancel anything, so all units will be part of the answer</p>
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<p>E.x. 6.9</p>

E.x. 6.9

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DeltaT is the ____ in temperature; what is its formula?

change

delta T (tfinal - tinitial)

OR

delta T = q / m x c

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What is the enthalpy of the solution

the amount of heat absorbed or given off during the process

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Calorimetry

the amount of heat lost by the metal is equal to the heat gained by the water

-qmetal = qwater OR 0 = qwater + qmetal

qmetal = (mmetal)(cmetal)(deltaTmetal)

qwater= (mwater)(cwater)(deltaTwater)

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<p>Ex. 6.10</p>

Ex. 6.10

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<p>E.x. 6.11</p>

E.x. 6.11

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Constant-Pressure Calorimetry

The system is not sealed, so the calorimeter pressure is the atmospheric pressure

qp - deltaH, the enthalpy change for the reaction

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Heat of Solution

  • For processes happening in solution, qsoln = -qp

  • Heat of soln / enthalpy soln (DeltaH soln): the release or absorption of heat that occurs when a solid compound dissolves in a solvent at a constant pressure to infinite dilution

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<p>E.x. 6.13</p>

E.x. 6.13

qsoln = -qrxn

Delta H = qsoln

<p>qsoln = -qrxn</p><p>Delta H = qsoln</p>
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<p>Practice Problem 6.13: Enthalpy of Reaction</p>

Practice Problem 6.13: Enthalpy of Reaction

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<p>Example: Enthalpy of Reaction</p>

Example: Enthalpy of Reaction

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Constant-Volume Calorimetry

carried out in a bomb calorimeter

designed to withstand high temperature and pessure changes, and to be sealed and insulated from the surroundings

They are close to a true isolated system

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Heat flow in a bomb

A measure of deltaU or deltaE

deltaU = deltaE = q + w

w = 0 inside a rigid container with a constant volume

Therefore, deltaU = qv

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Bomb Calorimetry Calculations

These calculations use the total heat capacity of their calorimeter components and the water, Ccal, not just the specific heat of water, cwater

deltaUrxn = -CcaldeltaT

or (remember): qrxn = -CcaldeltaT

q=cdeltaT

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<p>E.x&gt; 6.14</p>

E.x> 6.14

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<p>E.x. 6.15</p>

E.x. 6.15

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<p>Practice Problem 6.15</p>

Practice Problem 6.15

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Constant V Equation

qrxn = -CcaldeltaT/qsoln

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Constant P Equation

qrxn = -qsoln

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<p>E.x. 6.17</p>

E.x. 6.17

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E.x. 6.18

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<p>Your Turn </p>

Your Turn

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Hess’ Law

a method to calculate deltaH values for rxns by manipulating the equations for other reactions with known deltaH values

Advantage: we don’t need to use actual calorimeter to collect any data needed for the calculations of deltaH values

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Hess’ Law: Manipulating Chem Equations: Rule 1

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Hess’ Law: Manipulating Chem Equations: Rule 2

same equation, diff quantities

<p>same equation, diff quantities</p>
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Hess’ Law: Manipulating Chem Equations: Rule 3

C(diamond) + O2(g) + CO2(g) → CO2(gr) + O2(g) add the two rxns

C(diamond) → C(graphite)

delta H = ?

<p>C(diamond) + O2(g) + CO2(g) → CO2(gr) + O2(g) <strong>add the two rxns</strong></p><p>C(diamond) → C(graphite) </p><p>delta H = ?</p>
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<p>Another Example$</p>

Another Example$

Reactants: C(s) + O2(g) + CO2(g)

Products: CO2(g) + CO(g)+1/2O2(g)

See what you can cancel out; the ½ O2 multiplies wit the other O2

What’s Left: C(s) + 1/2O2(g) → CO(g) (Target Equation)

DeltaH: Add the -393 and new DeltaH2 +283

<p>Reactants: C(s) + O2(g) + CO2(g)</p><p>Products: CO2(g) + CO(g)+1/2O2(g)</p><p>See what you can cancel out; the ½ O2 multiplies wit the other O2</p><p>What’s Left: C(s) + 1/2O2(g) → CO(g) (Target Equation)</p><p>DeltaH: Add the -393 and new DeltaH2 +283</p>
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<p>E.x. 6.19</p>

E.x. 6.19

  1. Try to see similarities between the I & II reaction and the Target reaction.

  2. II reaction has a different product, this means it has to be reversed. (I rxn has the same product) This changes the sign of the deltaH

  3. Determine Reactants and Products

  4. Cancel Out any similarities

  5. Target Equation is found

<ol><li><p>Try to see similarities between the I &amp; II reaction and the Target reaction.</p></li><li><p>II reaction has a different product, this means it has to be reversed. (I rxn has the same product) This changes the sign of the deltaH</p></li><li><p>Determine Reactants and Products</p></li><li><p>Cancel Out any similarities</p></li><li><p>Target Equation is found</p></li></ol><p></p>
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Standard State

standard state of a substance is the state in which the substance is most stable at 25ºC (298K) and 1 atm pressure

The standard state of a substance is the natural state of the substance

Standard state of Na (s) not Na (l)

Standard state of H2 (g) not H (g) or H (l)

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Enthalpy of Formation + Enthalpy Change

The enthalpy change of the rxn below is called Enthalpy of formation, deltaHf

2H2(g) + O2 (g) → 2H2O (l)

The enthalpy change of the rxn below is called Standard Enthalpy of formation, delta H^0 subscript f

H2 (g) +1/2 O2(g) → H2O (l)

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What is the Standard Enthalpy of Formation deltaHºsubscriptf

the enthalpy of formation of 1 mol of a compound from its constituent elements in their standard states (25ºC and 1 atm)

Note that the deltaHºf of any element in its standard state is 0

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Standard Enthalpies of Formation

Standard Enthalpy formation deltaHºf of water (l) is -285.8 kJ/mole

It means that H2(g) + 1/2O2 → H2O (l)

Since this rxn forms one mole of the product in the standard state, for this rxn deltaHºf = deltaHrxn

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Table 6.4: Some Standard Enthalpies of Formation at 298K

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Molar deltaH

find the deltaH and divide by the numbero f moles

dissolution proess; opposite of given (exo/endo) change in enthalpy changes

Use molar mass of given formula and the given mass to find the mole

<p>find the deltaH and divide by the numbero f moles</p><p>dissolution proess; opposite of given (exo/endo) change in enthalpy changes</p><p>Use molar mass of given formula and the given mass to find the mole</p>
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Using deltaHºf to Calculate deltaHºrxn

deltaHºrxn = [sum of deltaHºf (products)] - [sum of deltaHºf (reactants)]

when the reaction involves multiple moles of products and reactants

deltaHºrxn = summation m [deltaHºf (products)] - summation n [deltaHºf(reactants)]

m = number of moles of each product

n = number of moles of each reactant

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<p>E.x.  6.21</p>

E.x. 6.21

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<p>E.x. 6.22</p>

E.x. 6.22

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<p>E.x 6.23</p>

E.x 6.23

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Standard Enthalpies of Formation

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