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 find the activation energy of the reaction given the temperature change and the increase in the number of molecules with energy greater than the threshold energy, we can follow these steps:

### Step 1: Convert temperatures from Celsius to Kelvin
- The initial temperature \( T_1 = 27^\circ C = 27 + 273 = 300 \, K \)
- The final temperature \( T_2 = 42^\circ C = 42 + 273 = 315 \, K \)

### Step 2: Use the Arrhenius equation
The Arrhenius equation is given by:
\[ k = A e^{-\frac{E_a}{RT}} \]
where:
- \( k \) is the rate constant,
- \( A \) is the pre-exponential factor,
- \( E_a \) is the activation energy,
- \( R \) is the universal gas constant (8.314 J/mol·K),
- \( T \) is the temperature in Kelvin.

### Step 3: Relate the rate constants at two different temperatures
Given that the number of molecules with energy greater than the threshold energy becomes five times, we can express this as:
\[ \frac{k_2}{k_1} = 5 \]
Taking the natural logarithm of both sides:
\[ \ln\left(\frac{k_2}{k_1}\right) = \ln(5) \]
Thus,
\[ \ln(k_2) - \ln(k_1) = \ln(5) \]
Given \( \ln(5) = 1.6094 \).

### Step 4: Substitute into the Arrhenius equation
From the Arrhenius equation, we can write:
\[ \ln(k_2) = \ln(A) - \frac{E_a}{R T_2} \]
\[ \ln(k_1) = \ln(A) - \frac{E_a}{R T_1} \]
Subtracting these two equations gives:
\[ \ln(k_2) - \ln(k_1) = -\frac{E_a}{R T_2} + \frac{E_a}{R T_1} \]
Substituting \( \ln(k_2) - \ln(k_1) = \ln(5) \):
\[ 1.6094 = -\frac{E_a}{R T_2} + \frac{E_a}{R T_1} \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ 1.6094 = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]
Substituting \( R = 8.314 \, J/mol·K \):
\[ 1.6094 = \frac{E_a}{8.314} \left(\frac{1}{300} - \frac{1}{315}\right) \]

### Step 6: Calculate \( \frac{1}{T_1} - \frac{1}{T_2} \)
Calculating \( \frac{1}{300} - \frac{1}{315} \):
\[ \frac{1}{300} = 0.003333 \, K^{-1} \]
\[ \frac{1}{315} = 0.003175 \, K^{-1} \]
Thus,
\[ \frac{1}{300} - \frac{1}{315} = 0.003333 - 0.003175 = 0.000158 \, K^{-1} \]

### Step 7: Substitute back to find \( E_a \)
Substituting back into the equation:
\[ 1.6094 = \frac{E_a}{8.314} \cdot 0.000158 \]
Rearranging gives:
\[ E_a = \frac{1.6094 \cdot 8.314}{0.000158} \]
Calculating this gives:
\[ E_a = \frac{13.379}{0.000158} \approx 84666.46 \, J/mol \]
Converting to kJ:
\[ E_a \approx 84.67 \, kJ/mol \]

### Final Answer
The activation energy \( E_a \) is approximately **84.67 kJ/mol**.