Question

The half-life period for first order reaction having activation energy $39.3kcal.{mol}^{-1}$ at ${300}^{o}C$ and frequency constant $1.11\times {10}^{11}{s}^{-1}$ will be:

• A. $1h$
• B. $1.68h$
• C. $1.28h$
• D. $1.11h$

Answer 1

Answer: (B)

$1.68h$

Given, $A=1.11\times {10}^{11}{s}^{-1};T=573K$

$E_a=39.3\times {10}^{3}cal{mol}^{-1};R=1.987cal$

$\because k=A{e}^{-E_a/RT}$

$\therefore \ln k=\ln A-\frac{E_a}{RT}$

${\log }_{10}k={\log }_{10}A-\frac{E_a}{2.303RT}$

${\log }_{10}k={\log }_{10}(1.11\times {10}^{11})-\left\{\frac{39.3\times {10}^3}{2.303\times 1.987\times 573}\right\}$

$k=1.14\times {10}^{-4}{s}^{-1}$

$t_{1/2}=\frac{0.693}{k}=\frac{0.693}{1.14\times {10}^{-4}}=6074s=1.68h$