Question:
The initial concentration of \( N_2O_5 \) in a first order reaction, \( N_2O_5(g) \rightarrow 2NO_2(g) + \frac{1}{2}O_2(g) \), was \( 1.68\times10^{-2}\,\text{mol L}^{-1} \) at \( 310\,K \). The concentration of \( N_2O_5 \) after 10 minutes was \( 0.84\times10^{-2}\,\text{mol L}^{-1} \), what is the rate constant at \( 310\,K \)? ( \( \log 2 = 0.3010 \) )
The initial concentration of \( N_2O_5 \) in a first order reaction, \( N_2O_5(g) \rightarrow 2NO_2(g) + \frac{1}{2}O_2(g) \), was \( 1.68\times10^{-2}\,\text{mol L}^{-1} \) at \( 310\,K \). The concentration of \( N_2O_5 \) after 10 minutes was \( 0.84\times10^{-2}\,\text{mol L}^{-1} \), what is the rate constant at \( 310\,K \)? ( \( \log 2 = 0.3010 \) )
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If concentration halves $\Rightarrow$ use \(\log 2\).
Updated On: Apr 21, 2026
- \(0.0693\,\text{min}^{-1} \)
- \(0.693\,\text{min}^{-1} \)
- \(6.93\,\text{min}^{-1} \)
- \(0.0639\,\text{min}^{-1} \)
- \(0.0963\,\text{min}^{-1} \)
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The Correct Option is A
Solution and Explanation
Concept:
\[
k = \frac{2.303}{t}\log \frac{[A]_0}{[A]}
\]
Step 1: Substitute.
\[ k = \frac{2.303}{10}\log \frac{1.68}{0.84} = \frac{2.303}{10}\log 2 \] \[ = \frac{2.303 \times 0.301}{10} \approx 0.0693 \]
Step 1: Substitute.
\[ k = \frac{2.303}{10}\log \frac{1.68}{0.84} = \frac{2.303}{10}\log 2 \] \[ = \frac{2.303 \times 0.301}{10} \approx 0.0693 \]
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