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Q.

What is the activation energy (kJ/mol) for a reaction if its rate constant doubles when the temperature is raised from 300 K to 400 K ? $(R = 8.314 J m o l^{- 1} K^{- 1})$

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a

6.91

b

34.4

c

3.44

d

69.1

answer is B.

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Detailed Solution

Given, $T_{1} = 300 K, T_{2} = 400 K$
$R = 8.314 J m o l^{- 1} K^{- 1}$
$\Rightarrow \frac{k_{2}}{k_{1}} = \frac{2}{1} \Rightarrow \log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 \times R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$
Thus, $\log \frac{2}{1} = \frac{E_{a}}{2.303 \times 8.314} [\frac{1}{300} - \frac{1}{400}]$
$E_{a} = 0.3010 \times 2.303 \times 8.314 \times 3 \times 400$
$E_{a} = 6.91 k J m o l^{- 1}$

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