For a first-order reaction, 32% of the reactant remains after 570 s. Calculate the rate constant k (in times10^ -3 s^ -1 , rounded to nearest integer). [Given: log_ 10 2 = 0.301, ln 10 = 2.303]

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JEE MAIN Chemistry QUESTION #5542

Question 1

For a first-order reaction, 32% of the reactant remains after 570 s. Calculate the rate constant $k$ (in $\times10^{-3}\ s^{-1}$, rounded to nearest integer). [Given: $\log_{10}2 = 0.301$, $\ln 10 = 2.303$]

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Correct Answer Explanation

For first-order: $k = \frac{2.303}{t}\log\frac{[A]_0}{[A]} = \frac{2.303}{570}\log\frac{100}{32} = \frac{2.303}{570}\log\left(\frac{25}{8}\right)$. $\log\frac{25}{8} = \log 25 - \log 8 = (2\log 5)-(3\log 2) = 2(1-0.301)-3(0.301) = 2(0.699)-0.903 = 1.398-0.903 = 0.495$. $k = \frac{2.303\times0.495}{570} = \frac{1.14}{570} \approx 2\times10^{-3}\ s^{-1}$. Nearest integer = $2\times10^{-3}\ s^{-1}$.

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