A first order reaction is 50 % completed in 20 min at 27° C and in 5 min. at 47° C the energy of activation of the reaction is :-

Question

A first order reaction is 50 % completed in 20 min at 27° C and in 5 min. at 47° C the energy of activation of the reaction is :- (A) 43.85 KJ / mol (B) 55.14 KJ / mol (C) 11.97 KJ / mol (D) 6.65 KJ / mol

Tardigrade Admin · Accepted Answer

t 1 / 2 propto (1/ k ) log ( k 2/ k 1)=( E a /2.303 R )(( T 2- T 1/ T 1 T 2)) log (20/5)=( E a /2.303 ×((8.314)1000))((320-300/300 × 320))

Q. A first order reaction is $50%$ completed in $20 min$ at $2 7^{\circ} C$ and in $5 min$ . at $4 7^{\circ} C$ the energy of activation of the reaction is :-

43.85 KJ / mol

55.14 KJ / mol

11.97 KJ / mol

6.65 KJ / mol

$t_{1/2} \propto \frac{1}{k}$
$lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} (\frac{T _{2} - T _{1}}{T _{1} T _{2}})$
$lo g \frac{20}{5} = \frac{E _{a}}{2.303 \times ( \frac{8.314}{1000} )} (\frac{320 - 300}{300 \times 320})$

Q. A first order reaction is 50% completed in 20min at 27∘C and in 5min. at 47∘C the energy of activation of the reaction is :-