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 first order reaction is one-fifth completed in 40 minutes. Calculate the time required for its 100% completion

infinite

The correct answer is Option (2) → infinite

For a first-order reaction, the integrated rate law is:

$t = \frac{2.303}{k} \log \left( \frac{a}{a - x} \right)$

Given: The reaction is one-fifth completed in 40 minutes.

This means:

• Fraction completed = 1/5 = 0.2
• Fraction remaining = 1 – 0.2 = 0.8
• So, $\frac{a}{a - x} = \frac{1}{0.8} = 1.25$

Now calculate k:

$40 = \frac{2.303}{k} \log(1.25)$

$\log(1.25) \approx 0.0969$

$40 = \frac{2.303}{k} \times 0.0969$

$k = \frac{2.303 \times 0.0969}{40} \approx 0.00558 \, \text{min}^{-1}$

Now, time for 100% completion:

100% completion means a – x = 0 → $\frac{a}{a - x} \rightarrow \infty$

$t = \frac{2.303}{k} \log(\infty) = \infty$

A first-order reaction never goes to 100% completion in finite time.