The rate of a reaction doubles when its tempe | vedclass

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(A) The Arrhenius equation is given by $\log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} \left( \frac{T_{2} - T_{1}}{T_{1} T_{2}} \right).$Given: $\fr

Vedclass · Accepted Answer

$53.6$

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(A) The Arrhenius equation is given by $\log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} \left( \frac{T_{2} - T_{1}}{T_{1} T_{2}} ight).$Given: $\frac{k_{2}}{k_{1}} = 2, T_{1} = 300 \, K, T_{2} = 310 \, K, R = 8.314 \, J \, K^{-1} \, mol^{-1}, \log 2 = 0.301.$Substituting the values:$0.301 = \frac{E_{a}}{2.303 	imes 8.314} \left( \frac{310 - 300}{310 	imes 300} ight)$$0.301 = \frac{E_{a}}{19.147} 	imes \frac{10}{93000}$$E_{a} = \frac{0.301 	imes 19.147 	imes 93000}{10}$$E_{a} = 53598.6 \, J \, mol^{-1} = 53.6 \, kJ \, mol^{-1}.$

The rate of a reaction doubles when its temperature changes from $300 \, K$ to $310 \, K.$ Activation energy of such a reaction will be .......... $kJ \, mol^{-1}$. $(R= 8.314 \, J \, K^{-1} \, mol^{-1}$ and $\log 2=0.301)$

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