Q.16 For the reaction mechanism,

2X ⇌ Y For this step, assume K_eq = [Y] / [X]²

Y → P : rate constant for this step

The rate law is:

(A) d[P]/dt = K_eq[Y]

(B) d[P]/dt = k[X]²

(C) d[P]/dt = kK_eq[Y]

(D) d[P]/dt = kK_eq[X]²

Rate Law Determination Using Pre-Equilibrium Approximation

Given Reaction Mechanism

The reaction proceeds in two steps:

Step 1 (Fast, reversible – pre-equilibrium):

2X ⇆ Y

For this step, it is given:

K_eq = [Y] / [X]²

Step 2 (Slow, rate-determining step):

Y → P

The rate constant for this step is k.

Step 1: Identify the Rate-Determining Step

Since the second step is slow, it controls the overall reaction rate.

Rate = d[P]/dt = k[Y]

So, the rate depends on the concentration of the intermediate Y.

Step 2: Eliminate the Intermediate Using Pre-Equilibrium

From the equilibrium expression of the first step:

K_eq = [Y] / [X]²

Rearranging:

[Y] = K_eq[X]²

Step 3: Substitute into the Rate Equation

d[P]/dt = k[Y]

d[P]/dt = kK_eq[X]²

Final Rate Law

d[P]/dt = kK_eq[X]²

Correct Answer

✅ Option (D)

d[P]/dt = kK_eq[X]²

Explanation of All Options

Option (A):

d[P]/dt = K_eq[Y]

• ❌ Missing the rate constant k
• ❌ Rate law must depend on the slow step kinetics

Option (B):

d[P]/dt = k[X]²

• ❌ Ignores the equilibrium constant
• ❌ Does not account for intermediate formation

Option (C):

d[P]/dt = kK_eq[Y]

• ❌ Uses both equilibrium constant and intermediate concentration
• ❌ Intermediate Y must be eliminated from final rate law

Option (D):

d[P]/dt = kK_eq[X]²

• ✅ Correctly applies pre-equilibrium approximation
• ✅ Expresses rate in terms of reactant concentration only

Key Takeaways

• In pre-equilibrium approximation, the fast reversible step is treated as equilibrium
• The slow step determines the rate
• The final rate law must be written in terms of stable reactants, not intermediates

Conclusion

Using the pre-equilibrium approximation, the correct rate law for the given mechanism is:

d[P]/dt = kK_eq[X]²

This makes Option (D) the correct answer.