For the reaction A to B,K_1 = 10^8 , e^{-6000 | vedclass

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(D) Given the rate constants for the two reactions are:$K_1 = 10^8 \, e^{-6000/8.34T}$$K_2 = 10^{10} \, e^{-8000/8.34T}$Setting $K_1 = K_2$:$10^8 \, e

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$52$

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(D) Given the rate constants for the two reactions are:$K_1 = 10^8 \, e^{-6000/8.34T}$$K_2 = 10^{10} \, e^{-8000/8.34T}$Setting $K_1 = K_2$:$10^8 \, e^{-6000/8.34T} = 10^{10} \, e^{-8000/8.34T}$Divide both sides by $10^8$:$e^{-6000/8.34T} = 10^2 \, e^{-8000/8.34T}$Rearranging the exponential terms:$e^{(-6000/8.34T) + (8000/8.34T)} = 10^2$$e^{2000/8.34T} = 10^2$Taking the natural logarithm $(\ln)$ on both sides:$2000 / (8.34 \, T) = \ln(10^2) = 2 \ln(10)$Using $\ln(10) \approx 2.303$:$2000 / (8.34 \, T) = 2 	imes 2.303 = 4.606$$T = 2000 / (8.34 	imes 4.606) \approx 2000 / 38.414 \approx 52.06 \ K$Thus,the temperature is approximately $52 \ K$.

For the reaction $A \to B$,$K_1 = 10^8 \, e^{-6000/8.34T}$ and for the reaction $P \to Q$,$K_2 = 10^{10} \, e^{-8000/8.34T}$. At what temperature $T$ will $K_1 = K_2$ (in $K$)?

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