The rate constant is doubled when temperature | vedclass

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(B) The Arrhenius equation is given by: $\log \frac{K_2}{K_1} = \frac{E_a}{2.303R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$Given: $\frac{K_2}{K_1} =

Vedclass · Accepted Answer

$54$

Vedclass · Answer

(B) The Arrhenius equation is given by: $\log \frac{K_2}{K_1} = \frac{E_a}{2.303R} \left[ \frac{T_2 - T_1}{T_1 T_2} ight]$Given: $\frac{K_2}{K_1} = 2$,$T_1 = 300 \, K$,$T_2 = 310 \, K$,$R = 8.314 \, J \, K^{-1} \, mol^{-1}$.Substituting the values: $\log 2 = \frac{E_a}{2.303 	imes 8.314} \left[ \frac{310 - 300}{300 	imes 310} ight]$$0.3010 = \frac{E_a}{19.147} 	imes \frac{10}{93000}$$E_a = \frac{0.3010 	imes 19.147 	imes 93000}{10} \approx 53598 \, J \, mol^{-1} \approx 53.6 \, kJ \, mol^{-1}$.Rounding to the nearest integer,the activation energy is $54 \, kJ \, mol^{-1}$.

The rate constant is doubled when temperature increases from $27\,^oC$ to $37\,^oC.$ Activation energy in $kJ$ is

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