Comprehensive Study Notes on Chemical Reactions, Stoichiometry, Solutions, and Gases

Chemical Reactions and Chemical Quantities

Chapter 4

Chemical Reactions

• Reactions must be balanced.

  • Reactants lead to products.

  • The number of each atom must be equal on each side of the equation to be considered balanced.

  • To balance reactions, add coefficients in front of the reactants/products. These coefficients change the number of molecules to achieve balance.

• Example of chemical reaction:

  • Unbalanced: $CH<em>4(g) + O</em>2(g) \rightarrow CO<em>2(g) + H</em>2O(g)$

  • Balanced: $CH<em>4(g) + 2O</em>2(g) \rightarrow CO<em>2(g) + 2H</em>2O(g)$

Stoichiometry - How many to Balance

• Continuing with the balanced reaction:

  • $CH<em>4(g) + 2O</em>2(g) \rightarrow CO<em>2(g) + 2H</em>2O(g)$

  • The coefficients (the numbers in front) specify the number of molecules required to carry out the reaction:

  • 1 mol of methane (CH₄) reacts with 2 mol of oxygen (O₂) to generate:

    • 1 mol of carbon dioxide (CO₂)

    • 2 mol of water (H₂O)

Stoichiometry II - Mass to Moles and Back

• Example:

  • Given the balanced reaction: $CH<em>4(g) + 2O</em>2(g) \rightarrow CO<em>2(g) + 2H</em>2O(g)$

  • If we have 50.0 g of methane, how do we calculate the mass of water (H₂O) produced?

  • Convert mass of methane to moles of methane.

  • Then convert moles of methane to moles of water.

  • Finally, convert moles of water back to mass of water.

Stoichiometry - One Reactant in Excess

• Reaction example:

  • $N<em>2(g) + 3H</em>2(g) \rightarrow 2NH_3(g)$

  • Situation: Given 50.0 g of hydrogen gas (H₂) with nitrogen (N₂) in excess.

  • Question: How many grams of ammonia (NH₃) will be produced?

  • Reverse situation:

  • If 50.0 g of ammonia is formed, how many grams of nitrogen are required when hydrogen is in excess?

Stoichiometry III - One Reagent will be Limiting

• Definition of limiting reagent: one reactant is completely consumed, causing the reaction to stop, while the other reactant remains in excess.

• Example: From the reaction $N<em>2(g) + 3H</em>2(g) \rightarrow 2NH_3(g)$

  • If we start with 100.0 g of nitrogen gas and 50.0 g of hydrogen gas, calculate grams of ammonia produced.

Stoichiometry - Yields

• Theoretical yield: calculated amount of product expected based on limiting reagent.

• Actual yield: the amount of product obtained from an experiment.

• Percent yield formula:

  • $ext{Percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%$

Limiting Reagent and Theoretical Yield

• Given the reaction: $2NO(g) + 5H<em>2(g) \rightarrow 2NH</em>3(g) + 2H_2O(g)$

  • Starting with 86.30 g of NO and 25.60 g of H₂, determine:

  • Theoretical (maximum) yield of ammonia (NH₃) in grams.

  • Amount (in grams) of excess reactant remaining.

BCA Tables: Before, Change, After a Reaction

• Use BCA (Before, Change, After) tables to track the amounts of reactants and products:

  • Example: $N<em>2(g) + 3H</em>2(g) \rightarrow 2NH_3(g)$

  • For starting amounts: 100.0 g of nitrogen and 50.0 g of hydrogen, use to calculate:

    • Grams of ammonia produced.

    • Grams of excess reagent left.

BCA Tables: Problem 2

• Given the equation from above, repeat process with the initial amounts of 86.30 g NO and 25.60 g H₂.

  • Determine both theoretical yield of NH₃ and the remaining amount of excess reactant.

Introduction to Solutions and Aqueous Reactions

Chapter 5

A Solution: Solute in Solvent

• Definition of solution: A solution consists of a solute dissolved in a solvent.

  • Most common in general chemistry: aqueous solutions, where water is the solvent in greatest proportion.

  • Dissociation of ions occurs when ionic compounds dissolve in water.

Dilute and Concentrated Solutions

• Concentration affects the amount of solute.

  • To create a concentrated solution, add more solute without changing the volume.

  • A saturated solution occurs when additional solute can no longer dissolve, indicating maximum solubility.

How to Make a 1 Molar Solution: 1 mol/L

• Steps to prepare a 1M solution:

  • Weigh out 1 mol of the substance.

  • Dissolve the solute in a volumetric flask with approximately 80% of the desired volume of water.

  • After thorough dissolution, add water to reach a total volume of 1 L.

Molar Solutions

• Molar solutions allow for concentration calculations (mol/L) enabling determination of dilution or concentration.

  • Molarity: $M = \frac{\text{moles}}{\text{L}}$

  • 1M corresponds to 1 mol of solute per liter of solution.

Making Molar Solutions with Dry Chemicals

• Example problem for making a specific concentration:

  • Need to prepare 0.200 L of 5.00M KOH:

  • Calculate required KOH: $5.00 \text{ mol/L} \times 0.200 \, \text{L} = 1.00 \, \text{mol KOH}$

  • KOH molar mass = 56.1 g, thus:

    • $1.00 \text{ mol} \times 56.1 \, \text{g/mol} = 56.1 \, \text{g}$

  • Dissolve in approximately 180 mL of DI water and bring to a final volume of 200 mL.

Concentrated Acids – Use with Care

• Common concentrated acids and their respective molarity:

  • HCl: 12.4M

  • HNO₃: 16.2M

  • Glacial acetic acid: 17.5M

• Note: It's commonplace to work with these acids in a diluted form; typically a stock solution concentration of 6.00M is prepared for regular lab work.

Making a 6M Stock Solution of HCl

• Process to dilute concentrated HCl:

  • Given concentrated HCl: 12.4M

  • Objective: Prepare 2.00L of a 6.00M stock solution.

• Calculate target volume:

  • It is crucial to pour acid into water, never the reverse, to mitigate safety hazards.

Solution Dilution: Using M₁V₁ = M₂V₂

• This equation describes the relationship before and after solution dilution.

  • M₁: initial molarity

  • V₁: initial volume

  • M₂: final molarity

  • V₂: final volume

Serial Dilutions - Another Use of M₁V₁ = M₂V₂

• Serial dilution is a technique where dilutions of known concentrations are sequentially produced to achieve desired lower concentrations, often starting with 10.0M solutions, for example.

Aqueous Solutions: Electrolytes (Ions)

• Electrolyte solutions contain ionic compounds that dissociate in water into solvated ions.

  • These ions can move freely which allows solutions to conduct electricity.

Aqueous Solutions: Nonelectrolytes

• Definition: Polar molecules that do not dissociate into ions when in aqueous solution.

  • Electricity cannot flow through these solutions because they lack charged elements necessary for conductivity.

Aqueous Solutions: Strong Acids

• Arrhenius definition: Strong acids fully dissociate in aqueous solution to produce H⁺ ions.

  • Examples include:

  • $HCl(aq) \rightarrow H^{+}(aq) + Cl^{-}(aq)$

  • $H<em>2SO</em>4(aq) \rightarrow H^{+}(aq) + HSO_4^{-}(aq)$

  • Such solutions exhibit conductivity with significant amounts of H⁺.

Aqueous Solutions: Weak Acids

• Weak acids partially dissociate in aqueous solutions, maintaining some intact molecules and forming conjugate bases.

  • Examples include:

  • $CH<em>3COOH \leftrightarrow H^{+}(aq) + CH</em>3COO^{-}(aq)$ (weak)

  • $HF \leftrightarrow H^{+}(aq) + F^{-}(aq)$ (weak)

  • Thus, conductivity correlates with the concentration of H⁺ ions.

Aqueous Solutions: Strong Bases

• Arrhenius definition: Strong bases dissociate completely in solution to generate OH⁻ ions.

  • Examples include:

  • $NaOH(aq) \rightarrow Na^{+}(aq) + OH^{-}(aq)$

  • $KOH(aq) \rightarrow K^{+}(aq) + OH^{-}(aq)$

Aqueous Solutions: Weak Bases

• Weak bases partially form OH⁻ ions in solution.

  • Bronsted bases accept H⁺ and often result in lower H⁺ concentrations.

  • An example:

  • $NH<em>3(g) + H</em>2O(l) \leftrightarrow OH^{-}(aq) + NH_4^{+}(aq)$ (weak, where ammonia is the most common weak base).

Conjugate Acids/Bases: Add/Subtract H⁺

• In acid-base reactions, H⁺ transfer occurs:

  • Acids provide H⁺ and convert into their conjugate base upon losing the proton.

  • Bases accept H⁺ and transform into their conjugate acids after accepting the proton.

Is Water an Acid or a Base?

• Determining whether water acts as an acid or a base depends on what it is reacting with:

  • Consider the H⁺ transfer event to assess its role.

General Reaction Types

• Types of chemical reactions:

  • Synthesis: $3H<em>2 + N</em>2 \rightarrow 2NH_3$

  • Decomposition: $H<em>2CO</em>3 \rightarrow H<em>2O + CO</em>2$

  • Single Displacement: $Zn + 2HCl \rightarrow ZnCl<em>2 + H</em>2$

  • Double Displacement: $AgNO<em>3 + NaBr \rightarrow AgBr + NaNO</em>3$

Solubility Rules: Generally Soluble Ionic Compounds in Water

• Common soluble ions include:

  • $Li^{+}, Na^{+}, K^{+}, NH<em>4^{+}, NO</em>3^{-}, CH_3COO^{-}$

  • Halides $Cl^{-}, Br^{-}, I^{-}$ are soluble except when paired with $Ag^{+}, Hg_2^{2+}, Pb^{2+}$

  • Sulfates $SO_{4}^{2-}$ are soluble except with $Ca^{2+}, Ba^{2+}, Ag^{+}, Sr^{2+}, Pb^{2+}$

Solubility Rules: Generally Insoluble Ionic Compounds in Water

• Common insoluble ions and their exceptions:

  • $OH^{-}$ and $S^{2-}$ except when with $Li^{+}, Na^{+}, K^{+}, NH_4^{+}$

  • $CO<em>{3}^{2-}$ and $PO</em>{4}^{3-}$ are generally insoluble except with $Li^{+}, Na^{+}, K^{+}, NH_4^{+}$

Precipitation Reactions

• A precipitate forms when two ionic solutions react:

  • Example: $2KI(aq) + Pb(NO<em>3)</em>{2}(aq) \rightarrow PbI<em>2(s) + 2KNO</em>3(aq)$

  • The insoluble product, $PbI_2$, is a solid precipitate.

Writing Reaction Equations (Balance)

• Example equations to balance:

  • $CuSO_4(aq) + NaOH(aq) \rightarrow$

  • $AgNO_3(aq) + NaCl(aq) \rightarrow$

  • $Ba(NO<em>3)</em>{2}(aq) + ZnSO_4(aq) \rightarrow$

  • $K<em>3PO</em>{4}(aq) + Ca(NO<em>3)</em>{2}(aq) \rightarrow$

Aqueous Reactions: Soluble Ions Separate

• While writing molecular equations, soluble ions dissociate:

  • Example:

  • $2KOH(aq) + Mg(NO<em>3)</em>{2}(aq) \rightarrow 2KNO<em>3(aq) + Mg(OH)</em>{2}(s)$

  • Reality in solution involves separate ions:

    • $2K^{+}(aq) + 2OH^{-}(aq) + Mg^{2+}(aq) + 2NO<em>{3}^{-}(aq) \rightarrow 2K^{+}(aq) + 2NO</em>{3}^{-}(aq) + Mg(OH)_{2}(s)$

Net Ionic Equation: Remove Spectator Ions

• The overall ionic equation shows all ions, but the net ionic equation excludes the spectator ions:

  • From earlier example:

  • $2OH^{-}(aq) + Mg^{2+}(aq) \rightarrow Mg(OH)_{2}(s)$

• Spectator ions remain unchanged and do not influence the precipitation outcome.

Acid-Base Reactions

• Basic definitions:

  • Acid: substance that produces H⁺ in aqueous solution.

  • Base: substance that produces OH⁻ in aqueous solution (or accepts H⁺).

• Reaction example:

  • $H^{+}(aq) + H<em>2O(l) \rightarrow H</em>{3}O^{+}(aq)$ suggests that H⁺ ions cannot exist alone in solution; they exist as hydronium ions.

• Polyprotic acids can donate more than one H⁺, like:

  • $H<em>2SO</em>4 \rightarrow H^{+} + HSO<em>{4}^{-} \rightarrow H^{+} + SO</em>{4}^{2-}$

Writing Acid-Base Reactions

• Format for writing balanced acid-base reactions:

  • Acid + Base = Water + Salt

  • Example with a strong acid:

    • $HCl(aq) + NaOH(aq) \rightarrow H_2O(l) + NaCl(aq)$

  • Example with a weak acid:

    • $HC<em>2H</em>3O<em>2(aq) + KOH(aq) \leftrightarrow H</em>2O(l) + KC<em>2H</em>3O_2(aq)$

Acid-Base Titrations

• Titration definition:

  • The process of adding a solution of known concentration to one of unknown concentration until reaching the equivalence point.

  • Equivalence point marks where mol H⁺ = mol OH⁻ at neutral pH and produces water.

  • Color change at certain pHs is often monitored using indicators.

Titration Problem: Find Moles of Known

• Example problem:

  • With 0.100 L of 0.500M HBr titrated with 0.200M NaOH, determine the volume of NaOH needed to reach equivalence point.

  • Utilize BCA table as part of calculations.

Titration Problem: Find Moles of Known

• Changing titrant:

  • For a situation with 0.100 L of 0.500M HBr titrated with 0.200M Ba(OH)₂, calculate the necessary volume of Ba(OH)₂ for equivalence.

  • Maintaining principles of stoichiometry, ensure all mol H⁺ = mol OH⁻ are equilibrated during reaction.

Gas-Evolution Reactions

• Example equation: $H<em>2SO</em>4(aq) + Li<em>2S(aq) \rightarrow H</em>2S(g) + Li<em>2SO</em>4(aq)$

• Types of compounds that undergo gas-evolution reactions:

  • Sulfides, carbonates, bicarbonates, sulfites, and ammonium compounds.

• Additional examples:

  • $2HCl(aq) + K<em>2S(aq) \rightarrow H</em>2S(g) + 2KCl(aq)$

  • $2HCl(aq) + K<em>2CO</em>3(aq) \rightarrow H<em>2O(l) + CO</em>2(g) + 2KCl(aq)$

Oxidation-Reduction (Redox) Reactions

• Definition: reactions that involve the transfer of electrons between atoms or molecules.

  • Oxidation: loss of electrons (OIL - Oxidation Is Loss).

  • Reduction: gain of electrons (RIG - Reduction Is Gain).

• Example: $4Fe(s) + 3O<em>2(g) \rightarrow 2Fe</em>2O_3(s)$ (rust formation).

• Another example: $2H<em>2(g) + O</em>2(g) \rightarrow 2H_2O(g)$ (combustion).

Oxidation States and Seeing Redox

• Oxidation states help determine the electron transfer actions in compounds, helping reveal which atoms are oxidized or reduced. Oxidation states can differ from ionic charges in covalent bonds, approximating ionic behavior to highlight redox activity.

Oxidation State Rules

• Basic rules for determining oxidation states:

  • All elements in their elemental form have an oxidation state of 0.

  • Nonmetals can be treated similarly to metals in ionic compounds to evaluate their oxidation states in covalent bonding situations.

Redox Reactions

• Characteristics of redox reactions:

  • Both oxidation and reduction occur simultaneously; products undergo oxidation/reduction while reactants act as oxidants or reductants.

Redox Examples

• Classifying reactions:

  • Example 1: $Ni + Pb(NO<em>3)</em>{2} \rightarrow Pb + Ni(NO<em>3)</em>{2}$

  • Example 2: $Cu + Zn(NO<em>3)</em>{2} \rightarrow Zn + Cu(NO<em>3)</em>{2}$

  • Example 3: $H<em>2 + Br</em>{2} \rightarrow 2HBr$

  • Example 4: $CH<em>4 + 2O</em>2 \rightarrow CO<em>2 + 2H</em>2O$

• Evaluating products/reactants based on oxidation and reduction states and identify the oxidizing/reducing agents.

Gases

Chapter 6

The Barometer: Atmospheric Pressure

• Definition: Average atmospheric pressure at sea level is approximately 760 mm Hg, equivalently 1.00 atm or 760 Torr.

Simple Gas Laws

• Boyle's Law: volume is inversely proportional to gas pressure.

  • If one value decreases, the other increases.

• Charles' Law: volume is directly proportional to gas temperature (in Kelvin).

  • Conversion to Kelvin indicated by adding 273.15 to the Celsius temperature.

  • Absolute zero is defined as 0K, which marks a theoretical state where gases would condense to liquids.

• Avogadro's Law: volume of gas is directly proportional to the number of moles (n) of the gas.

• Gay-Lussac’s Law: pressure is directly proportional to gas temperature (in Kelvin).

Combined Gas Laws

• The combined gas law enables one to relate pressure, volume, and temperature in a single calculation:

  • Example:

  • Starting with 2.00L of O₂ gas at 793 Torr and 25°C, calculate the final temperature upon expansion to 4.00L and pressure reduction to 511 Torr.

Ideal Gas Law

• Formulated as: $PV = nRT$

  • Where R is the gas constant $0.08206 \, \text{L} \cdot \text{atm} / (mol \cdot K)$.

  • Allows for calculating pressure, volume, temperature, or number of moles of a gas, given the other three variables.

Ideal Gas Law Example Problems

• Problem 1: Calculate the volume of 2.00 mol of N₂ at 1.20 atm and 300K:

  • Using $PV = nRT$ to solve for volume (V).

• Problem 2: Determine the pressure of 1.50 mol of O₂ at 1.40L and 330K.

• Problem 3: Find the moles of Cl₂ at 4.00L, 1.50 atm, and 350K.

Standard Conditions: Temperature and Pressure

• Standard temperature is defined as 273.15K (0°C) and standard pressure is 1 atm.

• Example calculation from PV=nRT gives a volume of 22.40L at standard temperature and pressure for 1.00 mol of an ideal gas (6.022 x 10²³ molecules).

Gas Density at STP (Standard Temperature & Pressure)

• Gas density can be described using the formula:

  • $\text{Density} = \frac{\text{mass}}{\text{volume}}$

  • Gas density can additionally be expressed as molar mass/molar volume, increasing with molecular mass.

  • Example: the density of H₂ is $0.09000 \, \text{g/L}$ versus $dCl₂ = 3.1654 \, \text{g/L}$ both calculated at 22.4L/mol.

Partial Pressures of Gas Mixtures

• Total pressure within a gas mixture is the sum of the partial pressures of each component gas:

  • Mole fraction can be calculated as the percentage of total volume occupied by a specific gas.

  • For example, nitrogen gas contributes a partial pressure of approximately 0.7808 atm in a total 1.00 atm.

Collecting Gases over Water

• When collecting gases over water, the vapor pressure of water contributes to the total pressure measured:

  • Formulation: $P<em>{total} = P</em>O<em>2 + P</em>{H_2O}$

  • Reference water vapor tables at specific temperatures to evaluate and remove water vapor from calculations to find the pure gas pressure.

Gases in Chemical Reactions: Stoichiometry

• Example reaction: $CO(g) + 2H<em>2(g) \rightarrow CH</em>3OH(g)$; find required volume of H₂ for reactions under given conditions: 50.00g CH₃OH at 350.0K and 1.20 atm.

More PV=nRT Problems

• Additional problems using the ideal gas law to evaluate volumes, pressures, and temperatures relating to reactions producing methanol, including specific calculations for each component's required quantities.