From the given data for the reaction H_2 + I_ | vedclass

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(A) Using the Arrhenius equation: $\log \frac{K_2}{K_1} = \frac{E_a}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$ Given $R \approx 8.314 \ J

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$4 	imes 10^4 \ J \ mol^{-1}$

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(A) Using the Arrhenius equation: $\log \frac{K_2}{K_1} = \frac{E_a}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} ight)$ Given $R \approx 8.314 \ J \ K^{-1} \ mol^{-1}$,but using $R = 2 \ cal \ K^{-1} \ mol^{-1}$ as per the provided calculation steps: $\log K_2 - \log K_1 = \frac{E_a}{2.303 	imes 2} [ (1.5 	imes 10^{-3}) - (1.3 	imes 10^{-3}) ]$ $1.1 - 2.9 = \frac{E_a}{4.606} [ 0.2 	imes 10^{-3} ]$ $-1.8 = \frac{E_a 	imes 0.2 	imes 10^{-3}}{4.606}$ $E_a = \frac{1.8 	imes 4.606}{0.2 	imes 10^{-3}} = 9 	imes 4.606 	imes 10^3 \approx 4.14 	imes 10^4 \ J \ mol^{-1}$. Rounding to the nearest provided option,the correct value is $4 	imes 10^4 \ J \ mol^{-1}$.

From the given data for the reaction $H_2 + I_2 \rightarrow 2HI$,calculate the activation energy $(E_a)$:
$T_1 = 769 \ K, \ 1/T_1 = 1.3 \times 10^{-3} \ K^{-1}, \ \log_{10} K_1 = 2.9$
$T_2 = 667 \ K, \ 1/T_2 = 1.5 \times 10^{-3} \ K^{-1}, \ \log_{10} K_2 = 1.1$

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From the given data for the reaction $H_2 + I_2 \rightarrow 2HI$,calculate the activation energy $(E_a)$: $T_1 = 769 \ K, \ 1/T_1 = 1.3 \times 10^{-3} \ K^{-1}, \ \log_{10} K_1 = 2.9$ $T_2 = 667 \ K, \ 1/T_2 = 1.5 \times 10^{-3} \ K^{-1}, \ \log_{10} K_2 = 1.1$