Question

For a certain first order reaction 32% of the reactant is left after 570s. The rate constant of this reaction is _________ \(\times 10^{-3}\ \mathrm{s^{-1}}\). (Round off to the Nearest Integer). [Given : \(\log_{10}2 = 0.301\), \(\ln 10 = 2.303\)]

Hint:

Use \(k = \dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]}\) with \([A]_0 = 100\), \([A] = 32\).

Answer 1

Correct Answer: 2

\(k = \dfrac{2.303}{t}\log_{10}\dfrac{[A]_0}{[A]} = \dfrac{2.303}{570}\log_{10}\dfrac{100}{32}\).
\(\log_{10}\dfrac{100}{32} = 2 - 5\log_{10}2 = 2 - 5(0.301) = 0.495\).
\(k = \dfrac{2.303 \times 0.495}{570} = \dfrac{1.140}{570} = 2.0 \times 10^{-3}\ s^{-1}\).
Answer: 2