Question

\(\mathrm{N_2O_{5(g)} \rightarrow 2NO_{2(g)} + \frac{1}{2}O_{2(g)}}\)

In the above first order reaction the initial concentration of \(\mathrm{N_2O_5}\) is \(2.40\times 10^{-2}\) mol \(\mathrm{L^{-1}}\) at 318 K. The concentration of \(\mathrm{N_2O_5}\) after 1 hour was \(1.60\times 10^{-2}\) mol \(\mathrm{L^{-1}}\). The rate constant of the reaction at 318 K is ______________ \(\times 10^{-3}\) \(\mathrm{min^{-1}}\). (Nearest integer)

[Given : log 3 = 0.477, log 5 = 0.699]

Hint:

Use \(k = \dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]}\) with t = 60 min, and get log 2 from log 5 using log 2 = 1 - log 5.

Answer 1

Correct Answer: 7

For a first-order reaction \(k = \dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]}\), with t = 1 h = 60 min.
\(\dfrac{[A]_0}{[A]} = \dfrac{2.40}{1.60} = \dfrac{3}{2}\)
\(\log 2 = \log\dfrac{10}{5} = 1 - 0.699 = 0.301\)
\(\log\dfrac{3}{2} = 0.477 - 0.301 = 0.176\)
\(k = \dfrac{2.303\times 0.176}{60} = \dfrac{0.4053}{60} = 6.76\times 10^{-3}\ \mathrm{min^{-1}}\)
Nearest integer = 7
Answer: 7