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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)

**chapter 6. CHEMICAL KINETICS class 12 chemistry textbook solution page 137**

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

**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.