i. In a first order reaction, the concentration of reactant decreases from 20 mmol dm-3 to 8 mmol dm-3 in 38 minutes.

chapter 6. CHEMICAL KINETICS class 12 chemistry textbook solution

4. Solve

i. In a first order reaction, the concentration of reactant decreases from 20 mmol dm-3 to 8 mmol dm-3 in 38 minutes. What is the half life of reaction? (28.7 min)

Answer:-

Given:

[A]₀ = 20 mmol dm^-3, [A]_t = 8 mmol dm^-3, t = 38 min

To find:

Half-life of the reaction t_1/2

Formulas:

i. \(k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}\)

ii. \(t_{1/2} = \frac{0.693}{k}\)

Calculation:

Substituting given values into (i):

\(k = \frac{2.303}{38 \text{ min}} \log_{10} \frac{20}{8} = \frac{2.303}{38 \text{ min}} \cdot 0.3979 = 0.0241\) min^-1

Using formula (ii):

\(t_{1/2} = \frac{0.693}{0.0241} = 28.7\) min

The half-life of the reaction is 28.7 minutes.