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AIIMS2019Chemistry-Chemical Kinetics

AIIMS 2019 Chemistry Rate Laws MCQ Question

Type: MCQ-numerical-Medium-Class 12

For a first order gas phase reaction :

A(g)→2B(g)+C(g)\text{A}_{(\text{g})} \rightarrow 2\text{B}_{(\text{g})} + \text{C}_{(\text{g})}

P0\text{P}_0 be initial pressure of A and Pt\text{P}_t the total pressure at time ‘t’. Integrated rate equation is :

A

2.303tlog⁡(P0P0−Pt)\frac{2.303}{\text{t}}\log\left( \frac{\text{P}_0}{\text{P}_0 - \text{P}_\text{t}} \right)

B

2.303tlog⁡(2P03P0−Pt)\frac{2.303}{\text{t}}\log\left( \frac{2\text{P}_0}{3\text{P}_0 - \text{P}_\text{t}} \right)

C

2.303tlog⁡(P02P0−Pt)\frac{2.303}{\text{t}}\log\left( \frac{\text{P}_0}{2\text{P}_0 - \text{P}_\text{t}} \right)

D

2.303tlog⁡(2P02P0−Pt)\frac{2.303}{\text{t}}\log\left( \frac{2\text{P}_0}{2\text{P}_0 - \text{P}_\text{t}} \right)

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⁡(P0P0−P)k = \frac{2.303}{t} \log \left( \frac{P_0}{P_0 - P} \right). This substitution leads to option B: k=2.303tlog⁡(2P02P0−Pt)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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