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Q.

It has been found that for a chemical reaction with rise in temperature by 9K the rate constant gets doubled. Assuming a reaction to be occurring at 300K, the value of activation energy is found to be _____ ${kJmol}^{- 1}$ . [nearest integer]
(Given $\ln 10 = 2.3, R = 8.3 {JK}^{- 1} {mol}^{- 1}, \log 2 = 0.30$ )

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answer is 59.

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Detailed Solution

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$\log_{10} \frac{K_{2}}{K_{1}} = \frac{E_{a}}{2.303 R} (\frac{1}{300} - \frac{1}{309})$
$0.3 = \frac{E_{a}}{2.303 \times 8.3} (\frac{9}{300 \times 309})$

$E_{a} = \frac{0.3 \times 2.303 \times 8.3 \times 300 \times 309}{9} = 59065.04 J E_{a} = 59.06 kJ$

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It has been found that for a chemical reaction with rise in temperature by 9K the rate constant gets doubled. Assuming a reaction to be occurring at 300K, the value of activation energy is found to be _____ kJmol-1. [nearest integer](Given ln10=2.3,R=8.3JK-1 mol-1,log2=0.30 )