The rate of a reaction quadruples when temperature changes from 27°C to 57°C. Calculate the energy of activation. Given R = 8.314 J , log 4 = 0.6021 [2024]

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Solution

Q.
The rate of a reaction quadruples when temperature changes from 27°C to 57°C. Calculate the energy of activation.

Given R = 8.314 J $K^{- 1}$ ${mol}^{- 1}$ , log 4 = 0.6021 [2024]

1 38.04 kJ/mol

2 380.4 kJ/mol

3 3.80 kJ/mol

4 3804 kJ/mol

Ans.
(1)

$T_{1} = 27 ° C = 300$ $K, T_{2} = 57 ° C = 330$ $K$

It is given that, $k_{2} = 4 k_{1}$

We know, $\log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$

Therefore, $\log \frac{4 k_{1}}{k_{1}} = \frac{E_{a}}{2.303 \times 8.314} [\frac{330 - 300}{300 \times 330}]$

$\Rightarrow 0.6021 = \frac{E_{a} \times 30}{2.303 \times 8.314 \times 300 \times 330}$

$\Rightarrow E_{a} = 38044.03$ $J$ ${mol}^{- 1} = 38.04$ $kJ$ ${mol}^{- 1}$

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