Rate-determining step Knowledge-Based Practice

Questions

1. 1.

   A reaction proceeds via a mechanism with an initial fast, reversible step followed by a slow, rate-determining second step. How does increasing the concentration of a product of the first step affect the overall reaction rate?

   • A) It increases the overall rate by shifting the initial equilibrium toward the formation of more intermediate.
   • B) It decreases the overall rate by shifting the initial equilibrium in reverse, thereby reducing the intermediate concentration available for the rate-determining step.
   • C) It does not affect the overall rate because the rate-determining step depends solely on the initial reactants.
   • D) It increases the overall rate by lowering the activation energy barrier of the second elementary step.
2. 2.

   A multi-step reaction mechanism consists of a slow rate-determining step followed by a fast second step. How does increasing the concentration of a reactant that participates exclusively in the second step affect the overall reaction rate?

   • A) The overall rate doubles because the second elementary step proceeds at a higher speed.
   • B) The overall rate increases because the activation energy barrier of the second step is lowered.
   • C) The overall rate decreases because the accumulation of intermediate inhibits the first step.
   • D) The overall rate remains unchanged because the rate is governed by the rate-determining step and any steps preceding it.
3. 3.

   A reaction proceeds via a three-step mechanism with activation energies of $E_{\text{a,1}} = 25\text{ kJ/mol}$, $E_{\text{a,2}} = 85\text{ kJ/mol}$, and $E_{\text{a,3}} = 30\text{ kJ/mol}$. If a modifier selectively decreases the activation energy of Step 1 to $15\text{ kJ/mol}$ without altering the other steps, how will the overall reaction rate be affected?

   • A) The overall reaction rate will increase significantly because the initial step initiates the entire pathway.
   • B) The overall reaction rate will remain essentially unchanged because Step 2 remains the rate-determining step.
   • C) The overall reaction rate will decrease because the energy difference between Step 1 and Step 2 increases.
   • D) The overall reaction rate will double because the activation energy of Step 1 is lowered by nearly half.
4. 4.

   A reaction occurs according to the proposed two-step mechanism: Step 1: $\text{NO}_2\text{(g)} + \text{F}_2\text{(g)} \rightarrow \text{NO}_2\text{F}\text{(g)} + \text{F}\text{(g)}$ (slow) Step 2: $\text{NO}_2\text{(g)} + \text{F}\text{(g)} \rightarrow \text{NO}_2\text{F}\text{(g)}$ (fast) Which rate law is predicted by this reaction mechanism?

   • A) $\text{Rate} = k[\text{NO}_2][\text{F}_2]$
   • B) $\text{Rate} = k[\text{NO}_2]^2[\text{F}_2]$
   • C) $\text{Rate} = k[\text{NO}_2][\text{F}]$
   • D) $\text{Rate} = k[\text{NO}_2]^2$
5. 5.

   Which statement correctly describes the rate-determining step in a multi-step chemical reaction mechanism?

   • A) It is the elementary step with the lowest activation energy that initiates the reaction sequence.
   • B) It is always the final elementary step that converts unstable reaction intermediates into stable products.
   • C) It is the slowest elementary step in the proposed mechanism, and its rate determines the overall reaction rate law.
   • D) It is an elementary step whose rate constant is vastly greater than the rate constants of all other steps.

Answers and step-by-step solutions appear after you finish the test.