For the reaction,aA + bB rightarrow cC + dD,t | vedclass

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(B) The Arrhenius equation is given by $\log k = \log A - \frac{E_a}{2.303 RT}$.Comparing this with the equation of a straight line $y = mx + c$,the s

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$526$

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(B) The Arrhenius equation is given by $\log k = \log A - \frac{E_a}{2.303 RT}$.Comparing this with the equation of a straight line $y = mx + c$,the slope is $m = -\frac{E_a}{2.303 R}$.From the given graph,the slope is $-10,000 \ K$.Therefore,$\frac{E_a}{2.303 R} = 10,000 \ K$.Using the Arrhenius equation in the form $\log \left( \frac{k_2}{k_1} \right) = \frac{E_a}{2.303 R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$:Given $k_1 = 10^{-5} \ s^{-1}$ at $T_1 = 500 \ K$ and $k_2 = 10^{-4} \ s^{-1}$ at $T_2 = ?$.Substituting the values: $\log \left( \frac{10^{-4}}{10^{-5}} \right) = 10,000 \left( \frac{1}{500} - \frac{1}{T_2} \right)$.$\log(10) = 1 = 10,000 \left( \frac{1}{500} - \frac{1}{T_2} \right)$.$\frac{1}{10,000} = \frac{1}{500} - \frac{1}{T_2}$.$\frac{1}{T_2} = \frac{1}{500} - \frac{1}{10,000} = \frac{20 - 1}{10,000} = \frac{19}{10,000}$.$T_2 = \frac{10,000}{19} \approx 526.31 \ K$.Rounding to the nearest integer,$T_2 = 526 \ K$.

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