JEE PYQ: Chemical Kinetics Question 3
Question 3 - 2021 (17 Mar Shift 1)
For a certain first order reaction 32% of the reactant is left after 570 s. The rate constant of this reaction is ____ $\times 10^{-3},\text{s}^{-1}$. (Round off to the Nearest Integer).
[Given: $\log_{10} 2 = 0.301$, $\ln 10 = 2.303$]
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Answer: 2
Solution
For 1st order reaction, $K = \frac{2.303}{t} \cdot \log\frac{[A_0]}{[A_t]} = \frac{2.303}{570} \cdot \log\left(\frac{100}{32}\right) = 1.999 \times 10^{-3} \approx 2 \times 10^{-3},\text{s}^{-1}$.
BETA
