For the gaseous reactions:A to B k_1 = 10 | vedclass

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(A) Given the rate constants $k_1 = 10^{15} e^{-25000 / 8.34 T}$ and $k_2 = 10^{14} e^{-15000 / 8.34 T}$.Setting $k_1 = k_2$:$10^{15} e^{-25000 / 8.34

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$520$

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(A) Given the rate constants $k_1 = 10^{15} e^{-25000 / 8.34 T}$ and $k_2 = 10^{14} e^{-15000 / 8.34 T}$.Setting $k_1 = k_2$:$10^{15} e^{-25000 / 8.34 T} = 10^{14} e^{-15000 / 8.34 T}$$\frac{10^{15}}{10^{14}} = \frac{e^{-15000 / 8.34 T}}{e^{-25000 / 8.34 T}}$$10 = e^{(25000 - 15000) / 8.34 T} = e^{10000 / 8.34 T}$Taking the natural logarithm on both sides:$\ln(10) = \frac{10000}{8.34 T}$$2.303 = \frac{10000}{8.34 T}$$T = \frac{10000}{8.34 	imes 2.303} \approx \frac{10000}{19.207} \approx 520.6 \ K$Rounding to the nearest integer,$T = 520 \ K$.Hence,the correct answer is $(A)$.

For the gaseous reactions:
$A \to B$ $k_1 = 10^{15} e^{-25000 / 8.34 T}$
$C \to D$ $k_2 = 10^{14} e^{-15000 / 8.34 T}$
Calculate the approximate temperature at which $k_1 = k_2$ in $K$.

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For the gaseous reactions:$A \to B$ $k_1 = 10^{15} e^{-25000 / 8.34 T}$$C \to D$ $k_2 = 10^{14} e^{-15000 / 8.34 T}$Calculate the approximate temperature at which $k_1 = k_2$ in $K$.