Integrated rate law equation for a first order gas phase reaction is given by (where $\mathrm{P}_{\mathrm{i}}$ is initial pressure and $\mathrm{P}_{\mathrm{t}}$ is total pressure at time t )

(A) $\mathrm{k}=\frac{2.303}{\mathrm{t}} \times \log \frac{\mathrm{P}_{\mathrm{i}}}{\left(2 \mathrm{P}_{\mathrm{i}}-\mathrm{P}_{\mathrm{t}}\right)}$

(B) $\mathrm{k}=\frac{2.303}{\mathrm{t}} \times \log \frac{2 \mathrm{P}_{\mathrm{i}}}{\left(2 \mathrm{P}_{\mathrm{i}}-\mathrm{P}_{\mathrm{t}}\right)}$

(C) $\mathrm{k}=\frac{2.303}{\mathrm{t}} \times \log \frac{\left(2 \mathrm{P}_{\mathrm{i}}-\mathrm{P}_{\mathrm{t}}\right)}{\mathrm{P}_{\mathrm{i}}}$

(D) $\mathrm{k}=\frac{2.303}{\mathrm{t}} \times \frac{\mathrm{P}_{\mathrm{i}}}{\left(2 \mathrm{P}_{\mathrm{i}}-\mathrm{P}_{\mathrm{t}}\right)}$

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Chemical Kinetics

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Integrated rate law equation for a first order gas phase reaction is given by (where

P_{i}

is initial pressure and

P_{t}

is total pressure at time t )

(A) $k = \frac{2.303}{t} \times lo g \frac{P _{i}}{( 2 P _{i} - P _{t} )}$ ✓ Correct
(B) $k = \frac{2.303}{t} \times lo g \frac{2 P _{i}}{( 2 P _{i} - P _{t} )}$
(C) $k = \frac{2.303}{t} \times lo g \frac{( 2 P _{i} - P _{t} )}{P _{i}}$
(D) $k = \frac{2.303}{t} \times \frac{P _{i}}{( 2 P _{i} - P _{t} )}$

Answer: (A)

Solution

For a first-order gas-phase reaction

A (g) \to B (g) + C (g)

, let the initial pressure be

P_{i}

and, at time

t

, the pressure of unreacted A be

(P_{i} - x)

P_{t} = (P_{i} - x) + x + x = P_{i} + x ⟹ x = P_{t} - P_{i} ⟹ P_{i} - x = 2 P_{i} - P_{t}

For a first-order reaction,

k = \frac{2.303}{t} lo g \frac{[ A ] _{0}}{[ A ] _{t}}

, so

k = \frac{2.303}{t} lo g \frac{P _{i}}{2 P _{i} - P _{t}}

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Single correct option

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JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Chemical Kinetics

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