Read the passage carefully and answer the Questions.
The concentration dependence of the rate of a reaction is called the differential rate equation. It is not always convenient to determine the instantaneous rate, as it is measured by calculating the slope of the tangent at point 't' in the concentration vs time plot. This makes it difficult to determine the rate and hence the order of the reaction. This difficulty is overcome by integrating the differential rate equation. This integrated rate equation gives a direct relation between concentrations at different times and the rate constant. The integrated rate equations are different for the reactions having different reaction orders.
A certain amount of \( N_2O_5 \) decomposes to half of its initial amount in 50 minutes. If the decomposition is a first order reaction, the rate constant of the reaction is.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(2.31 \times 10^{-4}\, s^{-1}\)
For a first order reaction the half-life does not depend on the initial amount. It follows from the integrated rate equation by putting \( [\mathrm{R}] = [\mathrm{R}]_0/2 \) at \( t = t_{1/2} \):
\( \displaystyle k = \frac{2.303}{t}\log\frac{[\mathrm{R}]_0}{[\mathrm{R}]} \)
\( \displaystyle k = \frac{2.303}{t_{1/2}}\log 2 = \frac{2.303 \times 0.3010}{t_{1/2}} = \frac{0.693}{t_{1/2}} \)
Here \( t_{1/2} \) = 50 min. Since three options are in s−1, convert it to seconds first:
\( \displaystyle t_{1/2} = 50 \times 60\,\mathrm{s} = 3000\,\mathrm{s} \)
\( \displaystyle k = \frac{0.693}{3000\,\mathrm{s}} = 2.31 \times 10^{-4}\,\mathrm{s^{-1}} \)
So the rate constant is 2.31 × 10−4 s−1.