For a certain gaseous reaction rise of temperature from `25^(@)C` to `35^(@)C` doubles the rate of reaction. What is the value of activation energy :-

Options

A. `10/(2.30R times 298 times 308)`
B. `(0.693R times 298 times 308)/10`
C. `(2.303 times 10)/(298 times 308R)`
D. `(0.693R times 10)/(290 times 308)`

Answer

To find the activation energy (Ea) for the given reaction, we can use the Arrhenius equation and the information provided in the question. Here’s a step-by-step solution:

### Step 1: Understand the relationship between rate constants and temperature
Given that the rate of reaction doubles when the temperature increases from 25°C to 35°C, we can express this relationship in terms of the rate constants (k1 and k2) at these temperatures.

- Let \( T_1 = 25°C = 298 K \)
- Let \( T_2 = 35°C = 308 K \)
- Since the rate doubles, we have \( k_2 = 2k_1 \).

### Step 2: Use the Arrhenius equation
The Arrhenius equation relates the rate constants to temperature and activation energy:
\[ \ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]
Since \( k_2 = 2k_1 \), we can rewrite this as:
\[ \ln(2) = \frac{E_a}{R} \left(\frac{1}{298} - \frac{1}{308}\right) \]

### Step 3: Calculate the temperature difference
Now, we need to compute \( \frac{1}{T_1} - \frac{1}{T_2} \):
\[ \frac{1}{298} - \frac{1}{308} = \frac{308 - 298}{298 \times 308} = \frac{10}{298 \times 308} \]

### Step 4: Substitute values into the equation
Now substitute this back into the equation:
\[ \ln(2) = \frac{E_a}{R} \left(\frac{10}{298 \times 308}\right) \]
Where \( R \) (the gas constant) is approximately \( 8.314 \, \text{J/(mol K)} \).

### Step 5: Solve for activation energy (Ea)
Rearranging gives:
\[ E_a = \ln(2) \cdot R \cdot \frac{298 \times 308}{10} \]
Calculating \( \ln(2) \):
\[ \ln(2) \approx 0.693 \]
Now substituting \( R \):
\[ E_a = 0.693 \cdot 8.314 \cdot \frac{298 \times 308}{10} \]

### Step 6: Calculate the final value
Calculating the numerical values:
\[ E_a = 0.693 \cdot 8.314 \cdot \frac{298 \times 308}{10} \approx 0.693 \cdot 8.314 \cdot 9144 \approx 0.693 \cdot 75802.5 \approx 52500 \, \text{J/mol} \approx 52.5 \, \text{kJ/mol} \]

### Final Answer
The activation energy \( E_a \) is approximately \( 52.5 \, \text{kJ/mol} \).

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