For the reaction of H_2 with I_2,the rate con | vedclass

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(B) 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:$K_1 = 2.5 	imes

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$166$

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(B) 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:$K_1 = 2.5 	imes 10^{-4} \ dm^3 \ mol^{-1} \ s^{-1}$,$T_1 = 327 + 273 = 600 \ K$$K_2 = 1.0 \ dm^3 \ mol^{-1} \ s^{-1}$,$T_2 = 527 + 273 = 800 \ K$$R = 8.314 \ J \ K^{-1} \ mol^{-1}$Substituting the values:$\log \left( \frac{1}{2.5 	imes 10^{-4}} ight) = \frac{E_a}{2.303 	imes 8.314} \left( \frac{800 - 600}{600 	imes 800} ight)$$\log (4000) = \frac{E_a}{19.147} \left( \frac{200}{480000} ight)$$3.602 = \frac{E_a}{19.147} 	imes 4.167 	imes 10^{-4}$$E_a = \frac{3.602 	imes 19.147}{4.167 	imes 10^{-4}} \approx 165500 \ J \ mol^{-1} = 165.5 \ kJ \ mol^{-1} \approx 166 \ kJ \ mol^{-1}$

For the reaction of $H_2$ with $I_2$,the rate constant is $2.5 \times 10^{-4} \ dm^3 \ mol^{-1} \ s^{-1}$ at $327 \ ^oC$ and $1.0 \ dm^3 \ mol^{-1} \ s^{-1}$ at $527 \ ^oC$. The activation energy for the reaction,in $kJ \ mol^{-1}$ is: $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$

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