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viii. The rate constant for the first order reaction is given by log10 k = 14.34 – 1.25 × 104 T

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

### 4. Solve

viii. The rate constant for the first order reaction is given by log10 k = 14.34 – 1.25 × 104 T. Calculate activation energy of the reaction. (239.3 kJ/mol)

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

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- viii. The rate constant for the first order reaction is given by log10 k = 14.34 – 1.25 × 104 T

The given rate constant equation is

log_{10}k = 14.34 – 1.25 × 10^{4} T ...(1)

Arrhenius equation is

k = "Ae"^((-E_{a})/(RT))

Therefore, In k = In A - (E_{a})/(2.303 RT) ...(2)

Comparing (1) and (2),

(E_{a})/(2.303 RT) = 1.25 × 10^{4} T

Therefore, (E_{a})/(2.303 R) = 1.25 × 10^{4}

Therefore, (E_{a})/(2.303 × 8.314) = 1.25 × 10^{4}

Therefore, E_{a} = 1.25 × 10^{4} × 2.303 × 8.314 = 239339 K mol^{-1} = 239.3 kJ mol^{-1}

The energy of activation of the reaction is 239.3 kJ mol^{-1}