When the temperature changes from 20,^oC to 5 | vedclass

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(A) Using the Arrhenius equation:$\log_{10} \frac{K_2}{K_1} = \frac{E_a}{2.303 	imes R} \left[ \frac{T_2 - T_1}{T_1 	imes T_2} ight]$Given:$\frac{

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$28.81$

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(A) Using the Arrhenius equation:$\log_{10} \frac{K_2}{K_1} = \frac{E_a}{2.303 	imes R} \left[ \frac{T_2 - T_1}{T_1 	imes T_2} ight]$Given:$\frac{K_2}{K_1} = 3$,$R = 8.314 \, J \, K^{-1} \, mol^{-1}$,$T_1 = 20 + 273 = 293 \, K$,$T_2 = 50 + 273 = 323 \, K$,$\log_{10} 3 \approx 0.4771$.Substituting the values:$0.4771 = \frac{E_a}{2.303 	imes 8.314} \left[ \frac{323 - 293}{323 	imes 293} ight]$$E_a = \frac{0.4771 	imes 2.303 	imes 8.314 	imes 323 	imes 293}{30}$$E_a \approx 28811.8 \, J \, mol^{-1} = 28.81 \, kJ \, mol^{-1}$.

When the temperature changes from $20\,^oC$ to $50\,^oC$,the rate of reaction becomes three times. The activation energy $(E_a)$ for the reaction is $.... \, kJ \, mol^{-1}$ $(R = 8.314 \, J \, K^{-1} \, mol^{-1})$.

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