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For reaction A rarr B , the rate constan...

For reaction A `rarr` B , the rate constant `K_(1)=A_(1)(e^(-E_(a_(1))//RT))` and the reaction `X rarr Y` , the rate constant `K_(2)=A_(2)(e^(-E_(a_(2))//RT))` . If `A_(1)=10^(9)`, `A_(2)=10^(10)` and `E_(a_(1))`=1200 cal/mol and `E_(a_(2))` =1800 cal/mol , then the temperature at which `K_(1)=K_(2)` is : (Given , R=2 cal/K-mol)

A

300K

B

`300 xx 2.303 K`

C

`(300)/(2.303 )K`

D

None of these

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To find the temperature at which the rate constants \( K_1 \) and \( K_2 \) are equal for the reactions \( A \rightarrow B \) and \( X \rightarrow Y \), we start with the given equations for the rate constants: 1. \( K_1 = A_1 e^{-\frac{E_{a1}}{RT}} \) 2. \( K_2 = A_2 e^{-\frac{E_{a2}}{RT}} \) We are given the following values: - \( A_1 = 10^9 \) - \( A_2 = 10^{10} \) - \( E_{a1} = 1200 \) cal/mol - \( E_{a2} = 1800 \) cal/mol - \( R = 2 \) cal/K·mol ### Step 1: Set \( K_1 \) equal to \( K_2 \) We start by setting the two rate constants equal to each other: \[ A_1 e^{-\frac{E_{a1}}{RT}} = A_2 e^{-\frac{E_{a2}}{RT}} \] ### Step 2: Substitute the values of \( A_1 \) and \( A_2 \) Substituting the values of \( A_1 \) and \( A_2 \): \[ 10^9 e^{-\frac{1200}{2T}} = 10^{10} e^{-\frac{1800}{2T}} \] ### Step 3: Divide both sides by \( 10^9 \) Dividing both sides by \( 10^9 \): \[ e^{-\frac{1200}{2T}} = 10 e^{-\frac{1800}{2T}} \] ### Step 4: Rearranging the equation Rearranging gives: \[ \frac{1}{10} = e^{-\frac{1800}{2T} + \frac{1200}{2T}} \] This simplifies to: \[ \frac{1}{10} = e^{\frac{-600}{2T}} \] ### Step 5: Take the natural logarithm of both sides Taking the natural logarithm of both sides: \[ \ln\left(\frac{1}{10}\right) = \frac{-600}{2T} \] ### Step 6: Solve for \( T \) We know that \( \ln\left(\frac{1}{10}\right) = -\ln(10) \): \[ -\ln(10) = \frac{-600}{2T} \] This simplifies to: \[ \ln(10) = \frac{600}{2T} \] Thus: \[ T = \frac{600}{2 \ln(10)} = \frac{300}{\ln(10)} \] ### Step 7: Convert \( \ln(10) \) to base 10 logarithm Using the conversion \( \ln(10) = 2.303 \): \[ T = \frac{300}{2.303} \] ### Final Answer Calculating this gives: \[ T \approx 130.3 \text{ K} \]
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