Question

For the first order reaction,
$2H_2{O}_2\to 2H_2+O_2\log k=14.34-1.25\times {10}^{4}K/T$. The energy of activation for the above reaction if its half life period is 256 min.

• A. 400 KJ / mole
• B. 230 KJ / mole
• C. 480 KJ / mole
• D. 239 KJ / mole

Answer 1

Answer: (D)

239 KJ / mole

Given order of the reaction= First order

$t_{1/2}=256min$

$K=\frac{0.693}{t_{1/2}}=\frac{0.693}{256\times 60}=8.67\times {10}^{-8}$

Given, $\log K=14.34-\frac{1.25\times {10}^4}{T}K⟶\left(1\right)$

By Arrhenius equation,

$\log K=\log A-\frac{E_a}{2.303RT}⟶\left(2\right)$

(1) & (2) $⟶$

$\Rightarrow \frac{E_a}{2.303RT}=\frac{1.25\times {10}^4}{T}$

$\Rightarrow E_a=1.25\times {10}^4\times 2.303\times 8.314$

$=2.39\times {10}^5$

$=239kJ/mole$