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For reaction ArarrB, the rate constant k...

For reaction `ArarrB,` the rate constant `k_(1)=A_(1)^(-Ea_(1//(RT)))` and for the reactio `XrarrY,` the rate constant `k_(2)=A_(2)^(-Ea_(1//(RT)))` . If `A_(1)=10^(8),A_(2)=10^(10)` and `E_(a_(1))=600` cal /mol, `E_(a_(2))=1800` cal/mol then the temperature at which `k_(1)=k_(2)` is (Given :R=2 cal/K-mol)

A

1200 K

B

`1200xx4.606 K`

C

`(1200)/(4.606)K`

D

`(600)/(4.606)K`

Text Solution

AI Generated Solution

The correct Answer is:
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 can use the Arrhenius equation for both reactions. ### Step-by-step Solution: 1. **Write the Arrhenius equation for both reactions:** \[ k_1 = A_1 e^{-\frac{E_{a1}}{RT}} \] \[ k_2 = A_2 e^{-\frac{E_{a2}}{RT}} \] 2. **Set \( k_1 \) equal to \( k_2 \):** \[ A_1 e^{-\frac{E_{a1}}{RT}} = A_2 e^{-\frac{E_{a2}}{RT}} \] 3. **Rearrange the equation:** \[ \frac{A_1}{A_2} = e^{-\frac{E_{a2} - E_{a1}}{RT}} \] 4. **Substitute the values:** Given: - \( A_1 = 10^8 \) - \( A_2 = 10^{10} \) - \( E_{a1} = 600 \) cal/mol - \( E_{a2} = 1800 \) cal/mol - \( R = 2 \) cal/(K·mol) Substitute these values into the equation: \[ \frac{10^8}{10^{10}} = e^{-\frac{1800 - 600}{2T}} \] Simplifying the left side: \[ 10^{-2} = e^{-\frac{1200}{2T}} \] 5. **Take the natural logarithm of both sides:** \[ \ln(10^{-2}) = -\frac{1200}{2T} \] 6. **Calculate \( \ln(10^{-2}) \):** \[ \ln(10^{-2}) = -2 \ln(10) \quad \text{(where } \ln(10) \approx 2.303\text{)} \] \[ \ln(10^{-2}) = -2 \times 2.303 = -4.606 \] 7. **Substitute back into the equation:** \[ -4.606 = -\frac{1200}{2T} \] Simplifying gives: \[ 4.606 = \frac{1200}{2T} \] 8. **Solve for \( T \):** \[ T = \frac{1200}{2 \times 4.606} \] \[ T = \frac{1200}{9.212} \approx 130.5 \text{ K} \] ### Final Answer: The temperature at which \( k_1 = k_2 \) is approximately **130.5 K**.
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