Question

The rate of the reaction triples when temperature changes from 293k to 323k . calculate the energy of activation for such reaction.

Answer 1

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Que :-

The rate of the reaction triples when temperature changes from 293k to 323k . calculate the energy of activation for such reaction.

Ans :-

28.8118 kJ {mol}^{ - 1}

Solution :-

Given :

$\frac{k2}{k1} = 3$

$T1 = 293 K$

$T2 = 323 K$

$R = 8.314 \: J{ K }^{ - 1} {mol}^{ - 1}$

By Arrhenius Equation

$log_{10}(\frac{k2}{k1} ) = \frac{ Ea}{R \: \times 2.303} \frac{ (T2 - T1)}{ (T1 T2 )}$

$log_{10}(3) = \frac{Ea}{ 8.314 \times 2.303} \frac{(323 - 293)}{(323 \times 293)}$

$Ea = \frac{8.314 \times 2.303 \times 323 \times 293 \times 0.477}{30}$

$Ea = 28.8118 kJ {mol}^{ - 1}$

Therefore, the energy of activation is

Ea = 28.8118 kJ {mol}^{ - 1}

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