If temperature changes from `27^@C to 42^@C` then no. of molecules having energy greater than thershold energy become five times, then find activation energy of reaction in KJ. Given ln 5= 1.6094

Answer

To solve the problem, we will use the Arrhenius equation and the information provided about the change in temperature and the increase in the number of molecules with energy greater than the threshold energy.

### Step-by-Step Solution:

1. **Convert Temperatures to Kelvin:**
- Initial temperature \( T_1 = 27^\circ C = 27 + 273 = 300 \, K \)
- Final temperature \( T_2 = 42^\circ C = 42 + 273 = 315 \, K \)

2. **Understand the Relationship Between Rate Constants:**
- The problem states that the number of molecules with energy greater than the threshold energy becomes five times. Therefore, the rate constant \( K_2 \) at the final temperature is:
\[ K_2 = 5 K_1 \]

3. **Write the Arrhenius Equation for Both Temperatures:**
- For \( T_1 \):
\[ K_1 = A e^{-\frac{E_a}{RT_1}} \]
- For \( T_2 \):
\[ K_2 = A e^{-\frac{E_a}{RT_2}} \]

4. **Substituting \( K_2 \) into the Equation:**
- Substitute \( K_2 = 5 K_1 \) into the second equation:
\[ 5 K_1 = A e^{-\frac{E_a}{RT_2}} \]
- Now, we can write:
\[ 5 A e^{-\frac{E_a}{RT_1}} = A e^{-\frac{E_a}{RT_2}} \]

5. **Dividing Both Sides by \( A \):**
- Since \( A \) is a common factor, we can cancel it out:
\[ 5 e^{-\frac{E_a}{RT_1}} = e^{-\frac{E_a}{RT_2}} \]

6. **Taking Natural Logarithm on Both Sides:**
- Taking the natural logarithm gives:
\[ \ln(5) = -\frac{E_a}{RT_2} + \frac{E_a}{RT_1} \]
- Rearranging gives:
\[ \ln(5) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]

7. **Substituting Known Values:**
- We know \( \ln(5) = 1.6094 \), \( R = 8.314 \, J/(mol \cdot K) \), \( T_1 = 300 \, K \), and \( T_2 = 315 \, K \):
\[ 1.6094 = \frac{E_a}{8.314} \left( \frac{1}{300} - \frac{1}{315} \right) \]

8. **Calculating \( \frac{1}{300} - \frac{1}{315} \):**
- Finding a common denominator (which is 94500):
\[ \frac{1}{300} - \frac{1}{315} = \frac{315 - 300}{94500} = \frac{15}{94500} = \frac{1}{6300} \]

9. **Substituting Back to Find \( E_a \):**
- Now substituting back:
\[ 1.6094 = \frac{E_a}{8.314} \cdot \frac{1}{6300} \]
- Rearranging gives:
\[ E_a = 1.6094 \cdot 8.314 \cdot 6300 \]

10. **Calculating \( E_a \):**
- Performing the multiplication:
\[ E_a = 1.6094 \cdot 8.314 \cdot 6300 \approx 84.297 \, kJ \]

### Final Answer:
The activation energy \( E_a \) is approximately **84.297 kJ**.