The rate of a reaction doubles when its tempe | vedclass

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(A) Using the Arrhenius equation: $\log \frac{K_{2}}{K_{1}} = \frac{E_a}{2.303 \ R} \left( \frac{T_{2} - T_{1}}{T_{1} T_{2}} \right)$Given: $K_{2} = 2

Vedclass · Accepted Answer

$53.6$

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(A) Using the Arrhenius equation: $\log \frac{K_{2}}{K_{1}} = \frac{E_a}{2.303 \ R} \left( \frac{T_{2} - T_{1}}{T_{1} T_{2}} ight)$Given: $K_{2} = 2K_{1}$,$T_{1} = 300 \ K$,$T_{2} = 310 \ K$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.$\log 2 = \frac{E_a}{2.303 	imes 8.314} \left( \frac{310 - 300}{300 	imes 310} ight)$$0.3010 = \frac{E_a}{19.147} \left( \frac{10}{93000} ight)$$E_a = \frac{0.3010 	imes 19.147 	imes 93000}{10} \ J/mol$$E_a \approx 53598 \ J/mol = 53.6 \ kJ/mol$.

The rate of a reaction doubles when its temperature changes from $300 \ K$ to $310 \ K$. The activation energy of such a reaction will be ........ $kJ/mol$.

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