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)

Options

A. 300K
B. `300 xx 2.303 K`
C. `(300)/(2.303 )K`
D. None of these

Answer

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 Arrhenius equation for both reactions:

1. **Write the Arrhenius equations:**
\[ K_1 = A_1 e^{-\frac{E_{a1}}{RT}} \]
\[ K_2 = A_2 e^{-\frac{E_{a2}}{RT}} \]

2. **Set \( K_1 \) equal to \( K_2 \):**
\[ A_1 e^{-\frac{E_{a1}}{RT}} = A_2 e^{-\frac{E_{a2}}{RT}} \]

3. **Substitute the given values:**
- \( A_1 = 10^9 \)
- \( A_2 = 10^{10} \)
- \( E_{a1} = 1200 \) cal/mol
- \( E_{a2} = 1800 \) cal/mol
- \( R = 2 \) cal/K-mol

The equation becomes:
\[ 10^9 e^{-\frac{1200}{2T}} = 10^{10} e^{-\frac{1800}{2T}} \]

4. **Divide both sides by \( 10^9 \):**
\[ e^{-\frac{1200}{2T}} = 10 e^{-\frac{1800}{2T}} \]

5. **Rearranging gives:**
\[ \frac{e^{-\frac{1200}{2T}}}{e^{-\frac{1800}{2T}}} = 10 \]
\[ e^{\left(-\frac{1200}{2T} + \frac{1800}{2T}\right)} = 10 \]
\[ e^{\frac{600}{2T}} = 10 \]

6. **Taking the natural logarithm of both sides:**
\[ \frac{600}{2T} = \ln(10) \]

7. **Solving for \( T \):**
\[ T = \frac{600}{2 \ln(10)} = \frac{300}{\ln(10)} \]

8. **Using the approximation \( \ln(10) \approx 2.303 \):**
\[ T \approx \frac{300}{2.303} \approx 130.0 \, \text{K} \]

Thus, the temperature at which \( K_1 = K_2 \) is approximately \( 130.0 \, \text{K} \).