The Question is from neet-chemistry chemical-kinetics , and it was created by the author A.RAMA NAGESWARARAO

Question

A reaction proceeds via a two-step mechanism. The first step is a fast, reversible equilibrium with a small equilibrium constant. The second step is slow.  Which of the following is MOST likely true about the overall reaction rate?

A.RAMA NAGESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: C
## CONCEPT FRAMEWORK
 **Primary Concept**: Reaction Mechanisms and Rate Determining Steps
 **Related Topics**: Equilibrium Constants, Rate Laws, Steady-State Approximation
 **Prerequisites**: Understanding of reaction mechanisms, rate laws, and equilibrium constants.
## SOLUTION PATHWAY
**Given Information**:
  * Two-step mechanism
  * Step 1: Fast, reversible equilibrium, small $K_{eq}$
  * Step 2: Slow step
**Method & Steps**:
 1. Identify the rate-determining step.
  Reasoning: The overall reaction rate is determined by the slowest step in the mechanism. In this case, step 2 is the slow step.
 2. Express the rate law for the slow step.
  Working: Let's assume the slow step involves an intermediate, I.  The rate law will be in the form: Rate = $k_2$[I], where $k_2$ is the rate constant for the slow step.
 3. Express the concentration of the intermediate in terms of reactants using the equilibrium of the fast step.
  Reasoning: Since step 1 is a fast equilibrium, we can use the equilibrium expression to relate the concentration of the intermediate to the reactants.  Let's say the reactants are A and B, and the intermediate is I. The equilibrium for step 1 can be written as: $A + B ightleftharpoons I$, with $K_{eq} = \frac{[I]}{[A][B]}$.  Therefore, $[I] = K_{eq}[A][B]$.
 4. Substitute the expression for the intermediate concentration into the rate law for the slow step.
  Working: Substituting $[I] = K_{eq}[A][B]$ into Rate = $k_2$[I], we get: Rate = $k_2 K_{eq}[A][B]$.
 5. Analyze the overall rate law.
  Result: The overall rate law is Rate = $k_{overall}[A][B]$, where $k_{overall} = k_2 K_{eq}$. This shows that the overall rate depends on the concentrations of both reactants A and B.
  Verification: The rate depends on the concentrations of reactants in both steps, and the equilibrium constant of the fast step affects the overall rate constant.
 ## OPTION ANALYSIS
A :x: The overall rate is not solely determined by the slow step; the equilibrium constant of the fast step influences it.
B :x: The overall rate is dependent on the concentration of the intermediate, which in turn is dependent on the reactant concentrations.
 C ✓: The overall rate law derived above shows that the rate depends on the concentrations of both reactants, A and B.
D :x: The equilibrium constant of the first step directly affects the overall rate constant ($k_{overall} = k_2 K_{eq}$).
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Ignoring the equilibrium of the fast step:  Failing to express the intermediate concentration in terms of reactants leads to an incomplete rate law.
  2.Assuming the overall rate is solely determined by the slow step:  The fast equilibrium step influences the concentration of the intermediate, thus affecting the overall rate.
 **Quick Tips**:
  1.Always identify the rate-determining step.
  2.Use the equilibrium expression of fast steps to express the concentration of intermediates in terms of reactants.

A.RAMA NAGESWARARAO · Answer

The overall rate will be determined solely by the rate of the slow step.

A.RAMA NAGESWARARAO · Answer

The overall rate will be independent of the concentration of the intermediate.

A.RAMA NAGESWARARAO · Answer

The correct Option is "The overall rate will be dependent on the concentrations of the reactants in both steps."

A.RAMA NAGESWARARAO · Answer

The equilibrium constant of the first step will have no effect on the overall rate.

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