Question

A first order reaction is 50% completed in 20 minutes at ${27}^{\circ }C$ and in 5 min at ${47}^{\circ }C.$ The energy of activation of the reaction is:

• A. 43.85 kJ/mol
• B. 55.34 kJ/mol
• C. 11.97 kJ/mol
• D. 6.65 kJ/mol

Answer 1

Answer: (B)

55.34 kJ/mol

$E_a=?$ [ $R$ = $8.314$]

_{$t_{1/2}$ = $\frac{0.693}{K}$}

$K_1$ = $\frac{0.693}{20};$ $T_1=300k$

$K_2$ = $\frac{0.693}{5};$ $T_2=320k$

$log\frac{K_2}{K_1}$ = $\frac{E_a}{2.303R}\times \frac{T_2-T_1}{T_1\times T_2}$

$log\frac{K_2}{K_1}$ = $\frac{E_a}{2.303\times 8.314}\times \frac{320-300}{300\times 320}$

$log\frac{0.693\times 20}{5\times 0.693}$ =$\frac{E_a}{19.15}\times \frac{20}{96000}$

$log$ $4$ = $\frac{E_a}{19.15}\times \frac{1}{4800}$

$0.6021$ = $\frac{E_a}{19.15}\times \frac{1}{4800}$

$E_a$ = $0.6021\times 19.15\times 4800$

$E_a$ = $55,345.032$

$E_a$ = $5.534\times {10}^{4}J$

$E_a$ = $55.34$ $kJ/mol$