Read the passage carefully and answer the questions.

The chemical reactions proceed at different rates which depend on various factors like concentration, temperature, etc. Rate law is an expression which relates the rate of reaction with concentration of various reacting species in the reaction. For a first order reaction, the concentration of the reactant after time $t$ is related with its initial concentration by the relation, $C = C_0e^{-kt}$. The rate constants for different orders have different expressions. For a first order reaction, $k =\frac{2.303}{t}\log\frac{a}{a-x}$, where $a$ is the initial concentration and $x$ is the extent of reaction.

A reaction which is first order with respect to the reactant A has a rate constant of $6\, min^{-1}$. If the reaction is started with $[A] = 5.0\, mol\, L^{-1}$ when would [A] reach the value of $0.05\, mol\, L^{-1}$

0.7676 min

The correct answer is Option (1) → 0.7676 min

We are dealing with a first-order reaction, so we use the formula:

$[A] = [A]_0 e^{-kt}$

Where:

• $[A]_0 = 5.0 \, \text{mol/L}$
• $[A] = 0.05 \, \text{mol/L}$
• $k = 6 \, \text{min}^{-1}$

Step 1: Solve for time t

$t = \frac{1}{k} \ln \frac{[A]_0}{[A]}$

$t = \frac{1}{6} \ln \frac{5.0}{0.05} = \frac{1}{6} \ln 100$

$\ln 100 = \ln(10^2) = 2 \ln 10 \approx 2 \times 2.3026 = 4.6052$

$t = \frac{4.6052}{6} \approx 0.7675 \, \text{min}$