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JEE Main · 2024 · 30 Jan 2024 · Shift 1 · Chemical Kinetics

The rate of first order reaction is

0.04 mol L^{- 1} s^{- 1}

at 10 minutes and

0.03 mol L^{- 1} s^{- 1}

at 20 minutes after initiation. Half life of the reaction is ____ minutes. (Given

lo g 2 = 0.3010, lo g 3 = 0.4771

)

Answer: 24

Solution

For a first-order reaction, rate

= k [A]

, so

\frac{rate _{1}}{rate _{2}} = e^{k (t_{2} - t_{1})}

.

\frac{0.04}{0.03} = e^{600 k} \Rightarrow ln \frac{4}{3} = 600 k

t_{1/2} = \frac{ln 2}{k} = 600 \cdot \frac{lo g 2}{lo g 4 - lo g 3} = 600 \cdot \frac{0.3010}{0.6020 - 0.4771} = 600 \cdot \frac{0.3010}{0.1249} s \approx 1446 s \approx 24 min

JEE Main · 2024 · 30 Jan 2024 · Shift 1 · Integer answer

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