Investigation 12: Reaction Rates and Equilibrium Study Guide

Reaction Chemistry and Reaction Rates

• Definition of Reaction Rate: The rate of change in the concentration of a reactant or product over a specific time interval.

• Investigation of $H_2(g) + I_2(g) \rightarrow 2 HI(g)$:     * Time Period: $0.0$ to $10.0$ seconds.     * Initial Concentration of $H_2$: $0.210\,M$.     * Final Concentration of $H_2$: $0.185\,M$.     * Rate of Change of $H_2$ Calculation:         * $\text{Rate} = \frac{0.185\,M - 0.210\,M}{10.0\,s - 0.0\,s}$         * $\text{Rate} = \frac{-0.025\,M}{10.0\,s}$         * $\text{Rate} = -0.0025\,M/s$     * Rate of Change of $HI$:         * Because the stoichiometric coefficient of $HI$ is $2$ while $H_2$ is $1$, the rate of $HI$ production is twice the rate of $H_2$ consumption (ignoring the negative sign for production).         * $\text{Rate} = 0.0025\,M/s \times 2 = 0.0050\,M/s$     * Overall Reaction Rate: Calculated as the rate of the species with a coefficient of $1$ (or dividing any species rate by its coefficient).         * $\text{Overall Rate} = 0.0025\,M/s$

• Investigation of $N_2(g) + 3 H_2(g) \rightarrow 2 NH_3(g)$:     * Time Period: $0.0$ to $10.0$ seconds.     * Initial Concentration of $NH_3$: $0\,M$.     * Final Concentration of $NH_3$: $2.50\,M$.     * Rate of Change of $NH_3$ Calculation:         * $\text{Rate} = \frac{2.50\,M - 0\,M}{10.0\,s - 0.0\,s} = 0.250\,M/s$     * Rate of Change of $H_2$:         * Determined by the ratio of coefficients ($3:2$ ratio between $H_2$ and $NH_3$).         * $\text{Rate} = 0.25\,M/s \times \frac{3}{2} = -0.375\,M/s$     * Overall Reaction Rate:         * $\text{Overall Rate} = \frac{0.250\,M/s}{2} = 0.125\,M/s$

Activation Energy and Energy Diagrams

• Activation Energy ($E_a$): The minimum amount of energy required for a chemical reaction to occur.     * Comparison of Curves: In a comparison between two energy graphs (Reaction A and Reaction B), the reaction with the higher peak relative to the reactants level has the higher activation energy.     * Reaction Speed: At the same temperature, a reaction with a lower activation energy will be faster because more molecules will have enough kinetic energy to overcome the smaller barrier.     * Quantifying $E_a$: In a provided example diagram, the activation energy needed was identified as $100\,kJ$.

• Role of a Catalyst:     * A catalyst provides an alternative reaction pathway with a lower activation energy.     * On an energy diagram, this is represented by a lower "hump" or peak between the reactants and products.

• Energy Changes in Reactions:     * Bond Breaking: Requires energy input.     * Bond Formation: Releases energy.     * Exothermic Reactions: Release heat to the surroundings; the potential energy of the products is lower than the potential energy of the reactants.     * Endothermic Reactions: Absorb heat; the potential energy of the products is higher than the potential energy of the reactants.

Collision Theory and Reaction Rate Factors

• Fundamental Requirements: For a reaction to occur, molecules must collide. However, a collision only results in a reaction if the molecules collide with enough energy (to overcome activation energy) and the correct orientation.

• The Effect of Temperature:     * Example: $4 Al(s) + 3 O_2(g) \rightarrow 2 Al_2O_3(s)$.     * A reaction is faster at higher temperatures (e.g., $75^{\circ}C$ vs. $25^{\circ}C$).     * Reasoning: Higher temperature increases the kinetic energy of the reactants. This leads to a higher frequency of collisions and ensures a greater proportion of collisions have sufficient energy to overcome the activation energy barrier.

• The Effect of Concentration:     * Example: Aluminum reacting in pure oxygen gas ($100\%$) vs. air ($21\%$).     * The rate increases with higher concentration (pure oxygen).     * Reasoning: A higher concentration of reactant molecules increases the frequency of collisions, thus increasing the reaction rate.

• The Effect of Surface Area:     * Example: A large chunk of aluminum metal vs. fine aluminum powder.     * The reaction is faster with the fine powder.     * Reasoning: Grinding a solid into powder exposes significantly more surface area. This provides more sites for collisions to occur, leading to a higher rate of successful collisions.

• The Effect of Catalysts:     * Adding a catalyst increases the reaction rate.     * Reasoning: By lowering the activation energy, it allows a larger fraction of total collisions to be successful.

Chemical Equilibrium Fundamentals

• Characteristics of Equilibrium:     * The concentrations of the reactants and products remain constant over time.     * The rates of the forward reaction and the reverse reaction are equal.

• Equilibrium Expressions ($K$):     * The expression is written as the ratio of the product concentrations to the reactant concentrations, each raised to the power of their stoichiometric coefficients.     * Example 1: $2 SO_2(g) + O_2(g) \rightleftharpoons 2 SO_3(g)$         * Expression: $K = \frac{[SO_3]^2}{[SO_2]^2[O_2]}$     * Example 2: $N_2(g) + 3 Cl_2(g) \rightleftharpoons 2 NCl_3(g)$         * Expression: $K = \frac{[NCl_3]^2}{[N_2][Cl_2]^3}$     * Example 3: $3 O_2 \rightleftharpoons 2 O_3$         * Reverse Derived Expression: If $K = \frac{[O_3]^2}{[O_2]^3}$, the balanced equation is $3 O_2 \rightleftharpoons 2 O_3$.

• Special Rules for Expressions:     * Only gases $(g)$ and aqueous solutions $(aq)$ are included in the equilibrium expression.     * Pure solids $(s)$ and pure liquids $(l)$ are excluded because their concentrations do not change.     * Example Expression for $Fe_2O_3(s) + 3 H_2(g) \rightleftharpoons 2 Fe(s) + 3 H_2O(g)$:         * $K = \frac{[H_2O]^3}{[H_2]^3}$

Calculating and Interpreting Equilibrium Constants ($K$)

• Calculations:     * For the $SO_2/O_2/SO_3$ system: If equilibrium concentrations are $[SO_2] = 0.44\,M$, $[O_2] = 0.06\,M$, and $[SO_3] = 0.03\,M$:         * $K = 0.08$         * Interpretation: The value is small ($K < 1$), meaning the left side (reactants) is favored.     * For the $N_2/Cl_2/NCl_3$ system: If equilibrium concentrations are $[N_2] = 0.42\,M$, $[Cl_2] = 0.42\,M$, and $[NCl_3] = 3.16\,M$:         * $K = 320$         * Interpretation: The value is large ($K > 1$), meaning the right side (products) is favored.

• Interpreting $K$ at Different Temperatures:     * For the reaction $NH_4Cl(s) \rightleftharpoons NH_3(g) + HCl(g)$, experimental data shows:         * At $25^{\circ}C$, $K = 0.000041$ (Left side/reactants favored).         * At $400^{\circ}C$, $K = 1$ (Neither side favored).         * At $600^{\circ}C$, $K = 1200$ (Right side/products favored).

Le Chatelier’s Principle and Reaction Shifts

• General Principle: If a stress is applied to a system at equilibrium, the system will shift in the direction that counteracts the stress.

• Concentration Changes:     * Adding Reactant: Shifts the reaction to the Right (towards products) to consume the excess.     * Removing Product: Shifts the reaction to the Right to replace what was lost (e.g., removing $O_2$ in an $SO_3$ system shifts right).

• Temperature Changes (Le Chatelier):     * Endothermic Reaction ($2 SO_3 + Energy \rightleftharpoons 2 SO_2 + O_2$):         * Increasing Temperature: Shifts Right (uses up added heat).         * Result: $K$ increases because more products are formed.     * Exothermic Reaction ($3 H_2 + N_2 \rightleftharpoons 2 NH_3 + Energy$):         * Decreasing Temperature: Shifts Right (produces more heat stable).         * Result: $K$ increases because it shifts toward products.         * Increasing Temperature: Shifts Left.

• Volume and Pressure Changes:     * Increasing Volume (Decreasing Pressure): The system shifts to the side with the more molecules of gas to counteract the drop in pressure.         * Example: In $3 H_2(g) + N_2(g) \rightleftharpoons 2 NH_3(g)$, there are 4 moles of gas on the left and 2 on the right. Increasing volume shifts the reaction Left.     * Decreasing Volume (Increasing Pressure): The system shifts to the side with the fewer molecules of gas.

• Effect of Catalysts on Equilibrium:     * Adding a catalyst results in no shift in the equilibrium position.     * Reasoning: A catalyst speeds up both the forward and the reverse reactions equally, so the ratio of products to reactants (equilibrium) remains unchanged.

Case Studies and Applications

• The Hydrogen-Oxygen Balloon:     * Scenario: A balloon filled with $2\,mol\,H_2$ and $1\,mol\,O_2$ gas sits without reacting at room temperature. When touched by a match, it explodes.     * Observation Explanation: At room temperature, the molecules collide, but they do not have sufficient energy to overcome the high activation energy required for the reaction. The flame from the match provides the necessary activation energy to start the reaction, which then proceeds rapidly.

• Ammonia Synthesis Temperature Data:     * Reaction: $N_2(g) + 3 H_2(g) \rightleftharpoons 2 NH_3(g)$.     * Data: $K = 0.50$ at $400^{\circ}C$ and $K = 0.012$ at $600^{\circ}C$.     * Conclusions:         * Since $K$ decreases as temperature increases, the reaction must be exothermic (releasing heat).         * At $600^{\circ}C$, the $K$ value is much less than 1, meaning the reaction favors the left side (reactants).