Question

For a reaction, A ¾® B ; if

log₁₀ K (sec^–1) = 14 – $\frac{1.25 \times 10^{4}}{T}$ K, the frequency factor and energy of activation for the reaction are -

• A. 10¹⁴ sec^–1, 239.34 kJ
• B. 14, 57.6 kcal
• C. 10¹⁴ sec^–1, 23.93 kJ
• D. 10¹⁴ sec, 5.76 kcal

Answer 1

Answer: (A)

10¹⁴ sec^–1, 239.34 kJ

Answer 2

Sol. log₁₀ K(sec^–1) = 14 – $\frac{1.25 \times 10^{4}}{T}$

log₁₀ K = log₁₀ A – $\frac{E_{a}}{2.303RT}$

log₁₀ A = 14

A = 10¹⁴ sec^–1 and $\frac{E_{a}}{2.303R}$= 1.25 × 10⁴ ;

E_a = 1.25 × 2.303 × 10⁴ × 8.314 J

» 239 × 10³ J