For reaction A to B,the rate constant k_1 = A | vedclass

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(D) Given that $k_1 = k_2$,we have $A_1 e^{-E_{a_1} / RT} = A_2 e^{-E_{a_2} / RT}$.Substituting the values: $10^8 e^{-600 / RT} = 10^{10} e^{-1800 / R

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$\frac{600}{4.606} \ K$

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(D) Given that $k_1 = k_2$,we have $A_1 e^{-E_{a_1} / RT} = A_2 e^{-E_{a_2} / RT}$.Substituting the values: $10^8 e^{-600 / RT} = 10^{10} e^{-1800 / RT}$.Rearranging the terms: $\frac{e^{-600 / RT}}{e^{-1800 / RT}} = \frac{10^{10}}{10^8}$.$e^{(1800 - 600) / RT} = 10^2$.$e^{1200 / RT} = 100$.Taking the natural logarithm on both sides: $\frac{1200}{RT} = \ln(100) = 2 \ln(10)$.Using $\ln(10) \approx 2.303$ and $R = 2 \ cal/K \cdot mol$:$\frac{1200}{2 	imes T} = 2 	imes 2.303$.$\frac{600}{T} = 4.606$.$T = \frac{600}{4.606} \ K$.

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