Reaction Mechanisms & Catalysts practice questions

Question 1

The decomposition of ozone, $\text{O}_3\text{(g)}$, is proposed to occur through the following two-step mechanism:

Step 1: $\text{O}_3\text{(g)} \rightleftharpoons \text{O}_2\text{(g)} + \text{O(g)}$
Step 2: $\text{O}_3\text{(g)} + \text{O(g)} \rightarrow 2\,\text{O}_2\text{(g)}$

Which of the following represents the overall balanced chemical equation for the reaction?

Accepted Answer

$2\,\text{O}_3\text{(g)} \rightarrow 3\,\text{O}_2\text{(g)}$

Answer

$2\,\text{O}_3\text{(g)} + \text{O(g)} \rightarrow 3\,\text{O}_2\text{(g)} + \text{O(g)}$

Answer

$\text{O}_3\text{(g)} + \text{O(g)} \rightarrow 2\,\text{O}_2\text{(g)}$

Answer

$\text{O}_3\text{(g)} \rightarrow \text{O}_2\text{(g)} + \text{O(g)}$

Question 2

Nitric oxide gas reacts with ozone gas in the atmosphere in an elementary reaction step represented below.

$\text{NO(g)} + \text{O}_3\text{(g)} \rightarrow \text{NO}_2\text{(g)} + \text{O}_2\text{(g)}$

Which of the following correctly identifies the molecularity and the rate law for this elementary step?

Accepted Answer

Bimolecular, with a rate law of $\text{Rate} = k[\text{NO}][\text{O}_3]$

Answer

Unimolecular, with a rate law of $\text{Rate} = k[\text{NO}]$

Answer

Bimolecular, with a rate law of $\text{Rate} = k[\text{NO}_2][\text{O}_2]$

Answer

Termolecular, with a rate law of $\text{Rate} = k[\text{NO}][\text{O}_3][\text{NO}_2]$

Question 3

A student analyzes the reaction energy profile for the gas-phase reaction 
$\text{NO}_2\text{(g)} + \text{CO(g)} \rightarrow \text{NO(g)} + \text{CO}_2\text{(g)}$, as shown in the graph below. Based on the profile, what are the activation energy ($E_a$) and the enthalpy change ($\Delta H$) for the forward reaction?

Accepted Answer

$E_a = 50 \text{ kJ/mol}$ and $\Delta H = -30 \text{ kJ/mol}$

Answer

$E_a = 90 \text{ kJ/mol}$ and $\Delta H = -30 \text{ kJ/mol}$

Answer

$E_a = 50 \text{ kJ/mol}$ and $\Delta H = +30 \text{ kJ/mol}$

Answer

$E_a = 80 \text{ kJ/mol}$ and $\Delta H = +30 \text{ kJ/mol}$

Question 4

A student investigates the thermal decomposition of nitryl chloride, $\text{NO}_2\text{Cl(g)}$. The first step of the proposed reaction mechanism is a unimolecular elementary step:

$\text{NO}_2\text{Cl(g)} \rightarrow \text{NO}_2\text{(g)} + \text{Cl(g)}$

Which of the following is the correct rate law expression for this elementary step?

Accepted Answer

$\text{Rate} = k [\text{NO}_2\text{Cl}]$

Answer

$\text{Rate} = k [\text{NO}_2\text{Cl}]^2$

Answer

$\text{Rate} = k [\text{NO}_2][\text{Cl}]$

Answer

$\text{Rate} = k \dfrac{[\text{NO}_2][\text{Cl}]}{[\text{NO}_2\text{Cl}]}$

Question 5

A student investigates a reaction that may be used in a thermal-energy storage system. The reaction-energy profile is shown in the diagram.

Which choice correctly compares the forward and reverse activation energies and classifies the forward reaction?

Accepted Answer

$E_{a,\text{forward}} > E_{a,\text{reverse}}$; the forward reaction is endothermic.

Answer

$E_{a,\text{forward}} > E_{a,\text{reverse}}$; the forward reaction is exothermic.

Answer

$E_{a,\text{forward}} < E_{a,\text{reverse}}$; the forward reaction is endothermic.

Answer

$E_{a,\text{forward}} < E_{a,\text{reverse}}$; the forward reaction is exothermic.

Question 6

A student adds a small amount of potassium iodide to accelerate the decomposition of residual hydrogen peroxide in a water sample. The proposed mechanism is shown below.

Step 1: $\text{H}_2\text{O}_2\text{(aq)}+\text{I}^-\text{(aq)}\rightarrow\text{IO}^-\text{(aq)}+\text{H}_2\text{O(l)}$

Step 2: $\text{H}_2\text{O}_2\text{(aq)}+\text{IO}^-\text{(aq)}\rightarrow\text{I}^-\text{(aq)}+\text{H}_2\text{O(l)}+\text{O}_2\text{(g)}$

Which species acts as a catalyst in the proposed mechanism?

Accepted Answer

$\text{I}^-\text{(aq)}$

Answer

$\text{H}_2\text{O}_2\text{(aq)}$

Answer

$\text{IO}^-\text{(aq)}$

Answer

$\text{O}_2\text{(g)}$

Question 7

A proposed mechanism for a elementary two-step reaction is analyzed using the potential energy profile shown in the graph. Based on the reaction energy profile, which step is the rate-determining step and what is its activation energy?

Accepted Answer

Step 1, because its activation energy is $60 \text{ kJ/mol}$, which is greater than the activation energy of Step 2.

Answer

Step 2, because its transition state reaches the highest overall potential energy at $90 \text{ kJ/mol}$.

Answer

Step 1, because its activation energy is $80 \text{ kJ/mol}$, which is greater than the activation energy of Step 2.

Answer

Step 2, because its activation energy is $70 \text{ kJ/mol}$, which is greater than the activation energy of Step 1.

Question 8

A student proposes the following two-step mechanism for the reaction $\text{NO}_2\text{(g)} + \text{CO(g)} \rightarrow \text{NO(g)} + \text{CO}_2\text{(g)}$:

Step 1: $\text{NO}_2\text{(g)} + \text{NO}_2\text{(g)} \rightarrow \text{NO}_3\text{(g)} + \text{NO(g)}$ (slow)
Step 2: $\text{NO}_3\text{(g)} + \text{CO(g)} \rightarrow \text{NO}_2\text{(g)} + \text{CO}_2\text{(g)}$ (fast)

Which of the following rate laws is consistent with the proposed mechanism, and why?

Accepted Answer

$\text{Rate} = k [\text{NO}_2]^2$, because Step 1 is the slow rate-determining step and its rate depends on the bimolecular collision of two $\text{NO}_2$ reactant molecules.

Answer

$\text{Rate} = k [\text{NO}_2][\text{CO}]$, because the reaction orders in the rate law must match the stoichiometric coefficients of the reactants in the overall balanced equation.

Answer

$\text{Rate} = k [\text{NO}_2][\text{CO}]$, because $\text{CO}$ is a reactant in Step 2, which rapidly converts the intermediate $\text{NO}_3$ into products.

Answer

$\text{Rate} = k [\text{NO}_2]^2$, because $\text{NO}_2$ acts as a catalyst that is consumed in Step 1 and regenerated in Step 2.

Question 9

A student investigates the kinetics of the reaction between solid calcium carbonate and hydrochloric acid, represented by the balanced equation below:

$$\text{CaCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}$$

In Trial 1, the student adds a single $2.0 \text{ g}$ chip of $\text{CaCO}_3\text{(s)}$ to $50.0 \text{ mL}$ of $1.0 \text{ M HCl(aq)}$ at $25^\circ\text{C}$ and measures the initial rate of $\text{CO}_2\text{(g)}$ production. In Trial 2, the student repeats the experiment using $2.0 \text{ g}$ of finely powdered $\text{CaCO}_3\text{(s)}$ with $50.0 \text{ mL}$ of $1.0 \text{ M HCl(aq)}$ at $25^\circ\text{C}$. Which of the following best predicts and explains the effect of using powdered $\text{CaCO}_3\text{(s)}$ on the initial reaction rate?

Accepted Answer

The initial rate increases because crushing the solid exposes a greater surface area, increasing the frequency of collisions per unit time between $\text{H}^+\text{(aq)}$ ions and $\text{CaCO}_3\text{(s)}$.

Answer

The initial rate increases because crushing the solid increases the average kinetic energy of the $\text{CaCO}_3\text{(s)}$ particles, increasing the fraction of collisions with energy equal to or greater than $E_a$.

Answer

The initial rate remains unchanged because the mass of $\text{CaCO}_3\text{(s)}$, temperature, and initial concentration of $\text{HCl(aq)}$ are identical in both trials.

Answer

The initial rate decreases because smaller particles have less mass per particle, which reduces the force of collision upon impact with $\text{H}^+\text{(aq)}$ ions.

Question 10

A student investigates the gas-phase formation of nitrosyl bromide, represented by the overall reaction

$$2\text{NO(g)}+\text{Br}_2\text{(g)}\rightarrow 2\text{NOBr(g)}$$

The following two-step mechanism is proposed.

$$\text{NO(g)}+\text{Br}_2\text{(g)}\rightleftharpoons\text{NOBr}_2\text{(g)}\qquad \text{fast}\;(k_1,k_{-1})$$

$$\text{NOBr}_2\text{(g)}+\text{NO(g)}\rightarrow 2\text{NOBr(g)}\qquad \text{slow}\;(k_2)$$

Which of the following is the rate law for the overall reaction expressed only in terms of reactant concentrations and elementary-step rate constants?

Accepted Answer

$\text{rate}=\dfrac{k_1k_2}{k_{-1}}[\text{NO}]^2[\text{Br}_2]$

Answer

$\text{rate}=k_2[\text{NOBr}_2][\text{NO}]$

Answer

$\text{rate}=\dfrac{k_1k_2}{k_{-1}}[\text{NO}][\text{Br}_2]$

Answer

$\text{rate}=\dfrac{k_{-1}k_2}{k_1}[\text{NO}]^2[\text{Br}_2]$

Question 11

A proposed mechanism for the gas-phase reaction $2\,\text{NO}_2\text{(g)} + \text{F}_2\text{(g)} \rightarrow 2\,\text{NO}_2\text{F(g)}$ consists of the following two elementary steps:

Step 1: $\text{NO}_2\text{(g)} + \text{F}_2\text{(g)} \rightarrow \text{NO}_2\text{F(g)} + \text{F(g)}$ (slow)
Step 2: $\text{NO}_2\text{(g)} + \text{F(g)} \rightarrow \text{NO}_2\text{F(g)}$ (fast)

Which of the following claims and justifications best explains why the experimentally observed rate law is $\text{Rate} = k[\text{NO}_2][\text{F}_2]$?

Accepted Answer

Step 1 determines the overall rate law because it has a higher activation energy than Step 2, making it the rate-determining step that limits the overall reaction rate.

Answer

Step 1 determines the overall rate law because $\text{F}_2\text{(g)}$ is completely consumed in Step 1, making it the limiting reactant of the reaction mechanism.

Answer

Step 2 determines the overall rate law because its lower activation energy enables molecules to overcome the energy barrier more rapidly.

Answer

Step 2 determines the overall rate law because the fast consumption of the intermediate $\text{F(g)}$ drives the overall reaction forward.

Question 12

A student investigates the elementary gas-phase reaction between carbon monoxide and nitrogen dioxide, represented by the equation below.

$\text{CO(g)} + \text{NO}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + \text{NO(g)}$

The reaction occurs via a single bimolecular collision in which a carbon-oxygen bond is formed and a nitrogen-oxygen bond is broken. Which of the following particulate representations best shows a collision orientation that can lead to the formation of products?

Accepted Answer

Diagram C

Answer

Diagram A

Answer

Diagram B

Answer

Diagram D

Question 13

At a particular temperature, the reaction

$2\text{SO}_2\text{(g)}+\text{O}_2\text{(g)}\rightleftharpoons2\text{SO}_3\text{(g)}$

is at equilibrium in a sealed, rigid vessel. A small amount of solid $\text{V}_2\text{O}_5$, which acts as a catalyst and has negligible volume, is added while the temperature remains constant. Which choice correctly predicts both the forward and reverse reaction rates immediately after the catalyst is added and the partial pressure of $\text{SO}_3\text{(g)}$ after sufficient time?

Accepted Answer

Forward and reverse rates: both increase and remain equal; final $P_{\text{SO}_3}$: unchanged

Answer

Forward and reverse rates: both decrease and remain equal; final $P_{\text{SO}_3}$: increases

Answer

Forward and reverse rates: both decrease and remain equal; final $P_{\text{SO}_3}$: unchanged

Answer

Forward and reverse rates: both increase and remain equal; final $P_{\text{SO}_3}$: increases

Question 14

A student investigates an enzyme that catalyzes the reversible reaction

$\text{X(aq)} \rightleftharpoons \text{Y(aq)}$

The student measures the forward and reverse rate constants over a temperature range in which the enzyme remains active. For each direction, the student constructs a plot of $\ln k$ versus $1/T$, for which the slope is $-E_a/R$. The student also measures the equilibrium concentration ratio after starting separate trials with only $\text{X(aq)}$ or only $\text{Y(aq)}$.

Which set of results provides the strongest evidence that the enzyme lowers the activation energy without altering the thermodynamic equilibrium?

Accepted Answer

The enzyme gives Arrhenius plots with slopes of smaller magnitude for both directions, and $[\text{Y}]_{\text{eq}}/[\text{X}]_{\text{eq}}$ is unchanged at every temperature tested.

Answer

The enzyme gives Arrhenius plots with the same slopes but larger intercepts for both directions, and $[\text{Y}]_{\text{eq}}/[\text{X}]_{\text{eq}}$ is unchanged at every temperature tested.

Answer

The enzyme gives Arrhenius plots with slopes of smaller magnitude for both directions, and $[\text{Y}]_{\text{eq}}/[\text{X}]_{\text{eq}}$ is larger at every temperature tested.

Answer

The enzyme and control give the same equilibrium ratio from either initial substance at every temperature tested, but no temperature-dependent rate constants are measured before equilibrium.

Question 15

A materials scientist studies the hypothetical gas-phase conversion $\text{R(g)} \rightarrow \text{P(g)}$ in an exhaust-treatment prototype. The reaction-coordinate diagram for the uncatalyzed pathway is shown.

The scientist adds a solid catalyst that provides an alternative pathway with a lower activation energy. The chemical identities and states of the reactant and product remain unchanged. Which set of relative potential energies could represent the catalyzed pathway?

Accepted Answer

Reactants: $60$ units; highest point: $120$ units; products: $20$ units

Answer

Reactants: $60$ units; highest point: $120$ units; products: $60$ units

Answer

Reactants: $60$ units; highest point: $160$ units; products: $20$ units

Answer

Reactants: $60$ units; highest point: $200$ units; products: $20$ units

Question 16

A student investigates two elementary bimolecular gas-phase reactions for removing a hypothetical contaminant, $\text{X(g)}$. The collision partner is either structural isomer $\text{Y}_1\text{(g)}$ or structural isomer $\text{Y}_2\text{(g)}$. Both experiments are performed at $400\text{ K}$, with an initial concentration of $0.010\text{ M}$ for each reactant.

| Property | Reaction with $\text{Y}_1\text{(g)}$ | Reaction with $\text{Y}_2\text{(g)}$ |
|---|---:|---:|
| Molar mass of collision partner | $72\text{ g/mol}$ | $72\text{ g/mol}$ |
| Effective collision diameter | $0.50\text{ nm}$ | $0.50\text{ nm}$ |
| Measured $E_a$ | $40\text{ kJ/mol}$ | $40\text{ kJ/mol}$ |

Structural modeling indicates that the reactive atom of $\text{Y}_2\text{(g)}$ is accessible over a smaller fraction of the molecular surface than the reactive atom of $\text{Y}_1\text{(g)}$. Let $A_1$ and $A_2$ be the respective pre-exponential factors in the Arrhenius equation. Which prediction correctly compares reaction 2 with reaction 1?

Accepted Answer

\(A_2

Answer

$A_2>A_1$, and reaction 2 has a greater initial rate, because steric shielding increases the frequency of energetic collisions.

Answer

$A_2=A_1$, but reaction 2 has a lower initial rate, because steric shielding increases the activation energy.

Question 17

A student investigates the oxidation of carbon monoxide in two sealed vessels at $298\ \text{K}$. The vessels have the same volume and initially contain the same partial pressures of $\text{CO(g)}$ and $\text{O}_2\text{(g)}$. The reaction and its standard Gibbs free-energy change are shown below.

$2\text{CO(g)}+\text{O}_2\text{(g)}\rightarrow 2\text{CO}_2\text{(g)}\qquad \Delta G^\circ_{\mathrm{rxn}}=-514\ \text{kJ/mol}_{\mathrm{rxn}}$

| Surface in vessel | Percentage of $\text{CO(g)}$ converted after $30\ \text{min}$ |
|---|---:|
| Uncoated ceramic | Less than $1\%$ |
| Palladium-coated ceramic | Approximately $70\%$ |

Which statement best explains both the sign of $\Delta G^\circ_{\mathrm{rxn}}$ and the difference between the results?

Accepted Answer

The reaction is thermodynamically favorable under standard conditions but kinetically slow without palladium, because \(\Delta G^\circ_{\mathrm{rxn}}

Answer

The reaction is thermodynamically favorable under standard conditions but kinetically slow without palladium, because the large magnitude of $\Delta G^\circ_{\mathrm{rxn}}$ causes reactants to cross the activation-energy barrier slowly.

Answer

The reaction is thermodynamically unfavorable under standard conditions and kinetically slow without palladium, because the small conversion after $30\ \text{min}$ shows that the equilibrium mixture contains mostly reactants.

Answer

The reaction is thermodynamically unfavorable under standard conditions and kinetically slow without palladium, because a high activation-energy barrier can limit reaction rate independently of thermodynamic favorability.

Question 18

A materials scientist considers the conversion of diamond to graphite at standard conditions.

$\text{C(s, diamond)} \rightarrow \text{C(s, graphite)}$

| Substance | $\Delta_{\mathrm f}G^\circ$ at $298\text{ K}$ |
|---|---:|
| $\text{C(s, diamond)}$ | $+2.9\text{ kJ/mol}$ |
| $\text{C(s, graphite)}$ | $0.0\text{ kJ/mol}$ |

After several years at $298\text{ K}$, no measurable graphite is detected in a sample of diamond. Which statement best reconciles the thermodynamic data with the observation?

Accepted Answer

The conversion is thermodynamically favorable but effectively unobservable, because graphite has a lower standard Gibbs free energy, whereas the extensive bond rearrangement requires crossing a large activation-energy barrier.

Answer

The conversion is thermodynamically favorable but effectively unobservable, because the product-favored equilibrium constant makes the forward reaction rate extremely small.

Answer

The conversion is thermodynamically unfavorable and effectively unobservable, because the large activation energy raises $\Delta G^\circ$ above zero.

Answer

The conversion is thermodynamically unfavorable and effectively unobservable, because subtracting the product value from the reactant value gives $\Delta G^\circ=+2.9\text{ kJ/mol}$.

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