Question

For a first order reaction A $\rightarrow$ products, the rate of reaction at [A] = 0.2 M is 1 x $10^{-2}$ mol $lit^{-1}$ $min^{-1}$. The half life period for the reaction is

• A. 832 min
• B. 440 sec.
• C. 416 min.
• D. 14 min

Answer 1

Answer: (D)

14 min

Answer 2

For a first order reaction, the rate law is given by: Rate = $k[A]$ where $k$ is the rate constant and $[A]$ is the concentration of the reactant.

Given: Rate = $1 \times 10^{-2}$ mol $lit^{-1}$ $min^{-1}$ at $[A] = 0.2$ M.

We can calculate the rate constant $k$ using the rate law: $1 \times 10^{-2} = k \times 0.2$ $k = \frac{1 \times 10^{-2}}{0.2} = \frac{0.01}{0.2} = \frac{0.1}{2} = 0.05 \text{ min}^{-1}$.

For a first order reaction, the half-life period ($t_{1/2}$) is related to the rate constant ($k$) by the formula: $t_{1/2} = \frac{\ln(2)}{k}$

Using the common approximation $\ln(2) \approx 0.693$: $t_{1/2} = \frac{0.693}{0.05} = \frac{0.693 \times 100}{0.05 \times 100} = \frac{69.3}{5} = 13.86 \text{ min}$.

Using the approximation $\ln(2) \approx 0.7$ (which is sometimes used in problems to give round numbers in options): $t_{1/2} = \frac{0.7}{0.05} = \frac{70}{5} = 14 \text{ min}$.

Comparing the calculated value with the given options, 14 min is the closest value and matches exactly when using the approximation $\ln(2) \approx 0.7$.