Chem 132 Exam 2 Review: Chemical Kinetics and Equilibrium

Chemical Kinetics

Rate Laws and Reaction Rates

Chemical kinetics studies the speed at which chemical reactions occur and the factors affecting these rates. The rate law expresses the relationship between the rate of a reaction and the concentration of reactants.

• Rate Law: For a general reaction $aA + bB \rightarrow cC + dD$, the rate is given by $\text{Rate} = k[A]^m[B]^n$, where k is the rate constant, and m, n are reaction orders.

• Integrated Rate Laws: These equations relate reactant concentration to time for zero, first, and second order reactions.

• Half-life ($t_{1/2}$): The time required for half of the reactant to be consumed.

Order	Rate Law	Integrated Rate Law	Plot for Straight Line	Slope	Half-life
Zero	Rate = k	$[A] = -kt + [A]_0$	$[A]$ vs. $t$	-k	$t_{1/2} = \frac{[A]_0}{2k}$
First	Rate = k[A]	$\ln[A] = -kt + \ln[A]_0$	$\ln[A]$ vs. $t$	-k	$t_{1/2} = \frac{0.693}{k}$
Second	Rate = k[A]^2	$\frac{1}{[A]} = kt + \frac{1}{[A]_0}$	$1/[A]$ vs. $t$	k	$t_{1/2} = \frac{1}{k[A]_0}$

• Arrhenius Equation: Describes how rate constants depend on temperature and activation energy: $k = Ae^{-E_a/RT}$

• Linearized Arrhenius Equation: $\ln(k) = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln(A)$

Experimental Data Analysis

To determine reaction order, plot concentration data as follows:

• Zero order: $[\text{Reactant}]$ vs. time

• First order: $\ln[\text{Reactant}]$ vs. time

• Second order: $1/[\text{Reactant}]$ vs. time

Example: Half-life Calculation

• For first-order decomposition of $N_2O$ at 1000 K with $t_{1/2} = 0.915$ s:

• $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{0.915} = 0.758\ \text{s}^{-1}$

Reaction Mechanisms

Elementary Steps and Rate-Determining Step

Reaction mechanisms describe the sequence of elementary steps by which a reaction occurs. The slowest step is the rate-determining step.

• Elementary Step: A single molecular event in a reaction mechanism.

• Rate Law for Elementary Step: Directly derived from the stoichiometry of the step.

• Reaction Intermediate: Species formed in one step and consumed in another; does not appear in the overall equation.

• Requirements for Valid Mechanism: (1) The sum of elementary steps must give the overall reaction; (2) The mechanism must agree with the experimentally determined rate law.

Example Mechanism

• For $2\ NO\ (g) + O_2\ (g) \rightarrow 2\ NO_2\ (g)$:

• Step 1 (fast): $2\ NO \xrightarrow{k_1} N_2O_2$

• Step 2 (slow): $N_2O_2 + O_2 \xrightarrow{k_2} 2\ NO_2$

• The slow step is rate-determining.

• Predicted rate law: Rate depends on $[NO]^2[O_2]$ if the slow step involves both $NO$ and $O_2$.

Chemical Equilibrium

Dynamic Equilibrium

At equilibrium, the rates of the forward and reverse reactions are equal, and concentrations of reactants and products remain constant.

• Law of Mass Action: For $aA + bB \rightleftharpoons cC + dD$, $K = \frac{[C]^c[D]^d}{[A]^a[B]^b}$

• Equilibrium Constant ($K$): Quantifies the ratio of product to reactant concentrations at equilibrium.

Relationships Between $K$ and Chemical Equations

• If the equation is reversed, $K$ is inverted: $K_{\text{reverse}} = 1/K_{\text{forward}}$

• If coefficients are multiplied by $n$, $K$ is raised to the $n$th power: $K' = K^n$

• If equations are added, $K_{\text{overall}} = K_1 \times K_2$

Example Table: Relating Equilibrium Constants

Reaction	Equilibrium Constant
$2\ PCl_3 (g) \rightarrow 2\ PCl_5 (g)$	$K_1$
$2\ PCl_5 (g) \rightarrow 2\ PCl_3 (g)$	$K_2$

For $PCl_3 (g) + Cl_2 (g) \rightleftharpoons PCl_5 (g)$, $K$ can be expressed in terms of $K_1$ and $K_2$.

Calculating Equilibrium Constant ($K_c$)

Given initial and equilibrium concentrations, $K_c$ can be calculated using the law of mass action.

• Example: $N_2O_4 (g) \rightleftharpoons 2\ NO_2 (g)$

• Initial $[N_2O_4] = 0.0400$ mol/L, equilibrium $[N_2O_4] = 0.0055$ mol/L

• $K_c = \frac{[NO_2]^2}{[N_2O_4]}$

Calculating Equilibrium Concentrations

Equilibrium concentrations can be found using ICE tables (Initial, Change, Equilibrium) and the equilibrium constant.

• For $H_2 (g) + I_2 (g) \rightleftharpoons 2\ HI (g)$, with initial $[H_2] = 2.0$ M, $[I_2] = 2.0$ M:

• Let $x$ be the change in concentration; solve for $x$ using $K_c$.

• No quadratic equations required if $K$ is very small (approximation method).

Le Châtelier’s Principle

Response to Changes in Equilibrium

Le Châtelier’s Principle states that if a system at equilibrium is disturbed, it will shift to counteract the disturbance and restore equilibrium.

• Change in Concentration: Adding/removing reactants or products shifts equilibrium to oppose the change.

• Change in Temperature: For endothermic reactions ($\Delta H^\circ > 0$), increasing temperature shifts equilibrium toward products.

• Change in Volume/Pressure: Decreasing volume (increasing pressure) shifts equilibrium toward the side with fewer moles of gas.

• Addition of Catalyst: Speeds up both forward and reverse reactions; does not affect equilibrium position.

• Addition of Inert Gas: No effect if volume is constant.

Example Reaction

• $CO_2 (g) + 2\ H_2O (l) \rightleftharpoons CH_4 (g) + 2\ O_2 (g)$, $\Delta H^\circ = +890$ kJ

• Increasing $O_2$ shifts equilibrium left (toward reactants).

• Increasing temperature shifts equilibrium right (toward products).

Acid Strength Comparison

Comparing Acids

Acid strength can be compared based on molecular structure and electronegativity.

• Example: Monochloroacetic acid ($ClCH_2COOH$) vs. acetic acid ($CH_3COOH$)

• Electron-withdrawing groups (like Cl) increase acid strength by stabilizing the conjugate base.