Advanced Chemistry Kinetics and Equilibrium Study Guide

Student Information and Assessment Overview

• Student Name: Lottie Davis

• Date: 4/30/26

• Hour: 5th

• Assessment Title: Advanced Chemistry Kinetics Test B

• Total Score Indicated: $-22$ (13/35)

Reaction Kinetics Data and Analysis

• Concentration Change Over Time:   - Raw data for reaction concentration as it relates to time (in seconds):     - Time $0\,s$: Concentration $9.5$     - Time $10\,s$: Concentration $8.3$     - Time $20\,s$: Concentration $7.3$     - Time $30\,s$: Concentration $6.5$     - Time $40\,s$: Concentration $5.9$     - Time $50\,s$: Concentration [Not explicitly provided but chart trends downward]

• Average Rate Calculations (as recorded in the transcript):   - Average rate over the first 10 seconds: Calculated as $17.8 = 1.78 \text{ (1.8 rounded)}$ over 10 seconds.   - Average rate of change from 10 to 20 seconds: Calculated as $15.6 = 1.56 \text{ (1.6 rounded)}$ over 10 seconds.   - Average rate of change from 30 to 40 seconds: Calculated as $12.4 = 1.24$ over 10 seconds.

• Observation of Reaction Dynamics:   - Question: Why does the rate of reaction slow down over time?   - Response: The rate slows down because the concentration of reactants decreases as the reaction proceeds.

Surface Area and Collision Theory

• Comparative Study of Reaction States (Chunks vs. Powder):   - A student performed two sets of reactions using solid reactants in chunks and in powder form ($KI$ and $Pb(NO_3)_2$).   - Reaction Time Data:     - Reaction Set A:       - Chunks: $45\,sec$       - Powder: $25\,sec$     - Reaction Set B:       - Chunks: $75\,sec$       - Powder: $55\,sec$

• Data Analysis and Collision Theory:   - Concentration Comparison: Based on the recorded data and student response, Reaction Set B was considered the more concentrated set.   - Powder Form Acceleration: Reactions involving reactants in powder form took less time than those in chunks.   - Explanation through Collision Theory: The reactions in powder form proceeded faster due to the increased frequency/rate in which collisions occurred, attributed to the higher surface area available for interaction.

Fundamental Principles of Chemical Equilibrium

• Definition of Equilibrium Solution: In a solution at equilibrium, certain conditions regarding concentration and reaction rates apply.

• Multiple Choice Question Analysis:   - Question: In an equilibrium solution, which of the following is true?   - Options:     - A. Concentration of Reactants = Concentration of Products     - B. Products will be in greater concentration than reactants     - C. Reactants will be in greater concentration than products     - D. The rate of the forward reaction will equal the rate of the reverse   - Student Choice: A (Note: Technically, at equilibrium, the rates of the forward and reverse reactions are equal, which is option D; however, the transcript notes an 'A' next to this question).

Le Chatelier’s Principle and System Perturbations

• Effect of Pressure Changes on Equilibrium Equilibrium:   - Increasing pressure in the container causes shifts based on the number of gas moles on either side of the equation.   - Reaction (a): $2H_2O_{(g)} + N_{2(g)} \rightarrow 2H_{2(g)} + 2NO_{(g)}$     - Predicted Shift: Left   - Reaction (b): $SiO_{2(g)} + 4HF_{(g)} \rightarrow SiF_{4(g)} + 2H_2O_{(g)}$     - Predicted Shift: Left   - Reaction (c): $CO_{(g)} + H_{2(g)} \rightarrow C_{(g)} + H_2O_{(g)}$     - Predicted Shift: Right

• Effect of Concentration and Temperature Changes:   - Equilibrium Reaction: $Heat + 2SO_{3(g)} \rightleftharpoons 2SO_{2(g)} + O_{2(g)}$   - Change: Sulfur dioxide ($SO_2$) is added to the system.     - Predicted Result: Student wrote "Remains the same."   - Change: Sulfur trioxide ($SO_3$) is removed from the system.     - Predicted Result: Shift Right.   - Change: Oxygen ($O_2$) is added to the system.     - Predicted Result: Shift Left.   - Change: Pressure is increased.     - Predicted Result: Shift Left.   - Change: Increased Heat.     - Predicted Result: Shift Right.

• Endothermic Shift Optimization:   - Reaction: $Heat + CH_{4(g)} + 2H_2S_{(g)} \rightleftharpoons CS_{2(g)} + 4H_{2(g)}$   - Question: Which change would cause this endothermic reaction in equilibrium to shift right?   - Options:     - (a) Decrease the concentration of dihydrogen monosulfide     - (b) Increase the pressure on the system     - (c) Increase the temperature of the system     - (d) Increase the concentration of carbon disulfide     - (e) Decrease the concentration of methane   - Correct Answer: (c) Increase the temperature of the system (endothermic reactions shift right with added heat).

Quantitative Equilibrium Calculations ($K_{eq}$)

• Reaction 1: Combustion-style Equilibrium:   - Chemical Equation: $C_3H_{8(g)} + 5O_{2(g)} \rightarrow 3CO_{2(g)} + 4H_2O_{(g)}$   - Equilibrium Concentrations:     - $[C_3H_8] = 1.25$     - $[O_2] = 4.67$     - $[CO_2] = 3.25$     - $[H_2O] = 1.67$   - Reaction Favoring: The reaction was identified as Product Favored.

• Reaction 2: General Gaseous Equilibrium:   - Chemical Equation: $A_{2(g)} + B_{2(g)} \rightarrow 2AB_{(g)}$   - Equilibrium Concentrations:     - $[A_2] = 3.45\,M$     - $[B_2] = 5.67\,M$     - $[AB] = 0.67\,M$   - Equilibrium Constant Expression ($K_{eq}$):     - $K_{eq} = \frac{[AB]^2}{[A_2][B_2]}$     - Substitution as recorded: $\frac{0.67}{(3.45)(5.67)}$   - Calculated Value of $K_{eq}$: 1.10 (Based on student data recorded).

• Reaction 3: Hydrogen Iodide Formation:   - Chemical Equation: $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$   - Given Values:     - $[H_2] = 0.46\,M$     - $[I_2] = 0.39\,M$     - $K_{eq} = 50.16$   - Task: Find the concentration of $[HI]$.