AIIMS2019Chemistry-Chemical Kinetics

AIIMS 2019 Chemistry Rate Laws MCQ Question

Type: MCQ-numerical-Medium-Class 12

A(g) → 2B(g) + C(g) Initial: P₀ 0 0 P₀ - P 2P P Pₜ = P₀ - P + 2P + P = P₀ + 2P Therefore, P = (Pₜ + P₀) / 2 Apply the expression shown below to calculate rate constant. k = (2.303 / t) log [P₀ / (P₀ - P)] Substitute the value of P in above equation.

A

k = 2.303t\frac{2.303}{t} log P0PPt+P02\frac{P_0}{P_⁰⁻ \frac{P_t + P_0}{2}}

B

k = 2.303t\frac{2.303}{t} log 2P02PPt+P0\frac{2P_0}{2P_⁰⁻ P_t + P_0}

C

k = 2.303t\frac{2.303}{t} log 2P03PPt\frac{2P_0}{3P_⁰⁻ P_t}

Correct Answer

Option B

Detailed Explanation

To derive the rate constant kk for the reaction A(g)2B(g)+C(g)A(g) \rightarrow 2B(g) + C(g), we substitute P=Pt+P02P = \frac{P_t + P_0}{2} into the rate equation k=2.303tlog(P0P0P)k = \frac{2.303}{t} \log \left( \frac{P_0}{P_0 - P} \right). This substitution leads to option B: k=2.303tlog(2P02P0Pt)k = \frac{2.303}{t} \log \left( \frac{2P_0}{2P_0 - P_t} \right), which correctly accounts for the stoichiometry of the reaction and the changes in pressure.

Options A and C are incorrect because they misrepresent the relationship between P0P_0, PtP_t, and PP, leading to invalid logarithmic expressions that do not accurately reflect the system's behavior. Understanding these relationships is crucial for correctly applying the integrated rate laws in chemical kinetics.

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