The time required for a first-order reaction to be a certain percentage complete can be determined using the first-order integrated rate equation:
$\ln\left(\frac{[A]_t}{[A]_0}\right) = -kt$
where $[A]_t$ is the concentration of the reactant at time t, $[A]_0$ is the initial concentration, k is the rate constant, and t is the time.
Let's consider the time required for the reaction to be 99% complete $(t_{99})$ and the time required for the reaction to be 90% complete $(t_{90})$.
For 99% completion, we can write:
$\ln\left(\frac{[A]_{99}}{[A]_0}\right) = -k \cdot t_{99}$
Similarly, for 90% completion, we can write:
$\ln\left(\frac{[A]_{90}}{[A]_0}\right) = -k \cdot t_{90}$
To compare the times, let's divide the two equations:
$\frac{\ln\left(\frac{[A]_{99}}{[A]_0}\right)}{\ln\left(\frac{[A]_{90}}{[A]_0}\right)} = \frac{-k \cdot t_{99}}{-k \cdot t_{90}}$
Simplifying further:
$\frac{\ln\left(\frac{[A]_{99}}{[A]_0}\right)}{\ln\left(\frac{[A]_{90}}{[A]_0}\right)} = \frac{t_{99}}{t_{90}}$
We know that for a first-order reaction, the concentration of the reactant decreases exponentially with time. Therefore, the ratio of the natural logarithms of the concentration ratios is equal to the ratio of the times. However, the actual values of the concentration ratios do not affect this ratio.
The ratio$ t_{99} / t_{90}$ is independent of the concentration values and is determined solely by the nature of the first-order reaction. Since it is independent of the concentrations, the answer is (b) Time taken is double. |
