(B) Step $1$: Convert time to seconds. $t = 35 \text{ min} = 35 \times 60 \text{ s} = 2100 \text{ s}$.Step $2$: Calculate the rate constant $k$ using

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(B) Step $1$: Convert time to seconds. $t = 35 \text{ min} = 35 \times 60 \text{ s} = 2100 \text{ s}$.Step $2$: Calculate the rate constant $k$ using $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{900 \text{ s}} = 7.7 \times 10^{-4} \text{ s}^{-1}$.Step $3$: Use the first-order integrated rate equation: $[A] = [A]_0 e^{-kt}$.$[A] = 0.08 \times e^{-(7.7 \times 10^{-4} \times 2100)} = 0.08 \times e^{-1.617} = 0.08 \times 0.1985 = 0.01588 \text{ mol dm}^{-3} \approx 0.0159 \text{ mol dm}^{-3}$.

Class 12 Chemistry Chemical Kinetics First Order reaction

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The half-life of a first-order reaction is $900 \text{ s}$. If the initial concentration of the reactant is $0.08 \text{ mol dm}^{-3}$, find the concentration that remains after $35 \text{ minutes}$?

MHT CET 2026 All MHT CET

A
$0.159 \text{ mol dm}^{-3}$
B
$0.0159 \text{ mol dm}^{-3}$
C
$0.05 \text{ mol dm}^{-3}$
D
$0.759 \text{ mol dm}^{-3}$

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The half-life of a first-order reaction is $900 \text{ s}$. If the initial concentration of the reactant is $0.08 \text{ mol dm}^{-3}$, find the concentration that remains after $35 \text{ minutes}$?

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If the rate constant of a reaction is $0.03 \ s^{-1}$,how much time does it take for $7.2 \ mol \ L^{-1}$ concentration of the reactant to get reduced to $0.9 \ mol \ L^{-1}$ (in $s$)? (Given: $\log 2 = 0.301$)

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