The Question is from neet-chemistry surface-chemistry , and it was created by the author Mr. S M HUSSAIN

Question

In the Lindemann-Hinshelwood mechanism for unimolecular gas-phase reactions, the rate-determining step at high pressures is:

Mr. S M HUSSAIN · Accepted Answer

## CORE INFORMATION
✓ **Answer**: C
## CONCEPT FRAMEWORK
 **Primary Concept**: Lindemann-Hinshelwood Mechanism for Unimolecular Reactions
 **Related Topics**: Reaction kinetics, rate-determining steps, collision theory, gas-phase reactions
 **Prerequisites**: Understanding of reaction mechanisms, rate laws, and activation energy.
## SOLUTION PATHWAY
**Given Information**:
  The question asks about the rate-determining step in the Lindemann-Hinshelwood mechanism at high pressures.
  The Lindemann-Hinshelwood mechanism proposes a two-step process for unimolecular reactions.
**Method & Steps**:
 1. **Understanding the Lindemann-Hinshelwood Mechanism**: The mechanism involves an activation step where a reactant molecule ($A$) collides with another molecule ($M$) to gain enough energy to become activated ($A^*$), followed by decomposition of the activated molecule into products ($P$).  The steps are:
		$A + M ightleftharpoons A^* + M$ (fast equilibrium)
		$A^* ightarrow P$ (slow)
 2. **High Pressure Regime**: At high pressures, the concentration of $M$ (the collision partner) is high. This means the activation step ($A + M ightleftharpoons A^* + M$) is very efficient, and the concentration of activated molecules ($A^*$) is relatively high and is essentially determined by the equilibrium constant of the first step.
 3. **Rate-Determining Step**: Since the second step ($A^* ightarrow P$) is slow, it becomes the rate-determining step at high pressures. The rate of the overall reaction is determined by the rate of decomposition of the energized reactant molecule.
 4. **Rate Law**: The rate law under high-pressure conditions is approximately first-order in the reactant concentration [A].
  Result: The decomposition of the energized reactant molecule ($A^*$) is the rate-determining step at high pressures.
  Verification: This aligns with the Lindemann-Hinshelwood mechanism's prediction under high-pressure conditions.
## OPTION ANALYSIS
A :x: At high pressures, the activation step is fast and not rate-limiting.
B :x: Deactivation is also fast at high pressures; it doesn't control the overall reaction rate.
C ✓: The decomposition of the energized molecule is the slow step and thus rate-determining at high pressures.
D :x: The formation of the product from the activated complex is essentially the same as the decomposition of the energized reactant molecule. However, the phrasing in option C is more precise and directly relates to the Lindemann-Hinshelwood mechanism.
## LEARNING AIDS
 **Common Mistakes**:
  1. Confusing the rate-determining step at high and low pressures: At low pressures, the activation step becomes rate-limiting.
  2. Incorrectly applying the rate law: The overall rate law changes depending on pressure conditions.
 **Quick Tips**:
  1. Remember that the rate-determining step is the slowest step in a reaction mechanism.
  2. At high pressures in the Lindemann-Hinshelwood mechanism, the concentration of the activated complex is high, making its decomposition the bottleneck.

Mr. S M HUSSAIN · Answer

The activation of the reactant molecule by collision.

Mr. S M HUSSAIN · Answer

The deactivation of the energized reactant molecule by collision.

Mr. S M HUSSAIN · Answer

The correct Option is "The decomposition of the energized reactant molecule."

Mr. S M HUSSAIN · Answer

The formation of the product from the activated complex.

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