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(D) Given that $K_1 = K_2$,we have:$10^{15} \exp \left( \frac{-2000}{T} \right) = 10^{14} \exp \left( \frac{-1000}{T} \right)$Dividing both sides by $

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$434.22$

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(D) Given that $K_1 = K_2$,we have:$10^{15} \exp \left( \frac{-2000}{T} \right) = 10^{14} \exp \left( \frac{-1000}{T} \right)$Dividing both sides by $10^{14}$:$10 \exp \left( \frac{-2000}{T} \right) = \exp \left( \frac{-1000}{T} \right)$Rearranging the exponential terms:$10 = \frac{\exp \left( \frac{-1000}{T} \right)}{\exp \left( \frac{-2000}{T} \right)} = \exp \left( \frac{-1000}{T} - \left( \frac{-2000}{T} \right) \right) = \exp \left( \frac{1000}{T} \right)$Taking the natural logarithm $(\ln)$ on both sides:$\ln(10) = \frac{1000}{T}$Using $\ln(10) \approx 2.303$:$2.303 = \frac{1000}{T}$$T = \frac{1000}{2.303} \approx 434.22 \ K$

For the reaction,following data is given,
$A \rightarrow B$; $K_1 = 10^{15} \exp \left( \frac{-2000}{T} \right)$
$C \rightarrow D$; $K_2 = 10^{14} \exp \left( \frac{-1000}{T} \right)$
The temperature at which $K_1 = K_2$ is ........... $K$ $(exp. = e)$

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For the reaction,following data is given,$A \rightarrow B$; $K_1 = 10^{15} \exp \left( \frac{-2000}{T} \right)$$C \rightarrow D$; $K_2 = 10^{14} \exp \left( \frac{-1000}{T} \right)$The temperature at which $K_1 = K_2$ is ........... $K$ $(exp. = e)$