The rate of a reaction quadruples when the temperature changes from 300 to 310 K. The activation energy of this reaction is `:`
( Assume Activation energy and pre-exponential factor are independent of temperature, `ln(2) = 0.693, R = 8.314J - mol^(-1) K^(-1))`

Options

A. `53.6 kJ mol^(-1)`
B. `214.4 kJ mol^(-1)`
C. `107 .2 kJ mol^(-1)`
D. `52.6 kJ mol^(-1)`

Answer

To solve the problem of finding the activation energy (EA) when the rate of a reaction quadruples with a temperature change from 300 K to 310 K, we can use the Arrhenius equation in the logarithmic form. Here’s the step-by-step solution:

### Step 1: Understand the relationship between rate constants and temperature
The rate of a reaction is related to the rate constants (K1 and K2) at two different temperatures (T1 and T2). Given that the rate quadruples, we can express this as:
\[ K_2 = 4K_1 \]

### Step 2: Use the logarithmic form of the Arrhenius equation
The logarithmic form of the Arrhenius equation is given by:
\[ \ln \left( \frac{K_2}{K_1} \right) = \frac{E_A}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]
However, since we are given that the rate quadruples, we can express this as:
\[ \ln(4) = \frac{E_A}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]

### Step 3: Substitute known values
We know:
- \( T_1 = 300 \, K \)
- \( T_2 = 310 \, K \)
- \( R = 8.314 \, J \cdot mol^{-1} \cdot K^{-1} \)
- \( \ln(4) = 2 \ln(2) = 2 \times 0.693 = 1.386 \)

Now we can substitute these values into the equation:
\[ 1.386 = \frac{E_A}{8.314} \left( \frac{1}{300} - \frac{1}{310} \right) \]

### Step 4: Calculate the difference in reciprocals of temperatures
Calculate \( \frac{1}{300} - \frac{1}{310} \):
\[ \frac{1}{300} - \frac{1}{310} = \frac{310 - 300}{300 \times 310} = \frac{10}{93000} = \frac{1}{9300} \]

### Step 5: Substitute back into the equation
Now substitute this back into the equation:
\[ 1.386 = \frac{E_A}{8.314} \times \frac{1}{9300} \]

### Step 6: Solve for EA
Rearranging gives:
\[ E_A = 1.386 \times 8.314 \times 9300 \]

### Step 7: Calculate the activation energy
Calculating the above expression:
\[ E_A = 1.386 \times 8.314 \times 9300 \approx 107199.3 \, J/mol \]
Converting to kilojoules:
\[ E_A \approx 107.2 \, kJ/mol \]

### Final Answer
The activation energy of the reaction is approximately:
\[ \boxed{107.2 \, kJ/mol} \]