Question

For a reaction, $A \longrightarrow B ;$ if $\log _{10}\, K\left( sec ^{-1}\right)=14-\frac{1.25 \times 10^{4}}{T} K$, the frequency factor and energy of activation for the reaction are

• A. $10^{14}\, sec ^{-1}, 239.34 \,kJ$
• B. $14,57.6 kcal$
• C. $10^{14} \,sec ^{-1}, 23.93 \,kJ$
• D. $10^{14} \,sec , 5.76 \,kcal$

Answer 1

Answer: (A)

$\log _{10}\, K\left( sec ^{-1}\right)=14-\frac{1.25 \times 10^{4}}{T}$

$\log _{10}\, K=\log _{10} A-\frac{E_{a}}{2.303 \,R T}$

$\log _{10} A=14$

$A=10^{14} \sec ^{-1}$ and $\frac{E_{a}}{2.303\, R}=1.25 \times 10^{4}$

$E_{a}=1.25 \times 2.303 \times 10^{4} \times 8.314 \,J \approx 239 \times 10^{3} \,J =239\, kJ$