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)/(2.303R xx 298 xx 308)`
B. `(2.303xx10)/(298xx308 R)`
C. `(0.693 R xx 10)/(290xx308)`
D. `(0.693Rxx298xx308)/(10)`

Explanation

To find the activation energy (Ea) for the given reaction, we can use the Arrhenius equation and the information provided about the temperature change and the doubling of the reaction rate. Here’s a step-by-step solution:

### Step 1: Understand the Arrhenius Equation
The Arrhenius equation is given by:
\[ k = A e^{-\frac{E_a}{RT}} \]
where:
- \( k \) = rate constant
- \( A \) = Arrhenius factor (frequency factor)
- \( E_a \) = activation energy
- \( R \) = universal gas constant (8.314 J/mol·K)
- \( T \) = temperature in Kelvin

### Step 2: Set Up the Problem
We know that the rate of reaction doubles when the temperature increases from \( 25^\circ C \) to \( 35^\circ C \). We can denote the rate constants at these temperatures as \( k_1 \) and \( k_2 \):
- \( k_1 \) at \( T_1 = 25^\circ C = 298 \, K \)
- \( k_2 \) at \( T_2 = 35^\circ C = 308 \, K \)

Given that \( k_2 = 2k_1 \), we can express this relationship using the natural logarithm:
\[ \ln\left(\frac{k_2}{k_1}\right) = \ln(2) \]

### Step 3: Use the Arrhenius Equation
Using the Arrhenius equation for both temperatures, we have:
\[ \ln(k_2) - \ln(k_1) = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]

Substituting \( k_2 = 2k_1 \):
\[ \ln(2) = -\frac{E_a}{R} \left(\frac{1}{308} - \frac{1}{298}\right) \]

### Step 4: Calculate the Temperature Difference
Calculate \( \frac{1}{T_2} - \frac{1}{T_1} \):
\[ \frac{1}{T_2} - \frac{1}{T_1} = \frac{1}{308} - \frac{1}{298} \]
Finding a common denominator (which is \( 308 \times 298 \)):
\[ \frac{298 - 308}{308 \times 298} = \frac{-10}{308 \times 298} \]

### Step 5: Substitute into the Equation
Now substituting back into the equation:
\[ \ln(2) = -\frac{E_a}{R} \left(\frac{-10}{308 \times 298}\right) \]
This simplifies to:
\[ \ln(2) = \frac{E_a \cdot 10}{R \cdot (308 \times 298)} \]

### Step 6: Solve for Activation Energy (Ea)
Rearranging gives:
\[ E_a = \frac{\ln(2) \cdot R \cdot (308 \times 298)}{10} \]
Substituting \( R = 8.314 \, J/(mol \cdot K) \) and \( \ln(2) \approx 0.693 \):
\[ E_a = \frac{0.693 \cdot 8.314 \cdot (308 \times 298)}{10} \]

### Step 7: Calculate the Final Value
Calculating \( 308 \times 298 \):
\[ 308 \times 298 = 91684 \]
Now substituting:
\[ E_a = \frac{0.693 \cdot 8.314 \cdot 91684}{10} \]
Calculating the multiplication:
\[ E_a \approx \frac{0.693 \cdot 8.314 \cdot 91684}{10} \approx 5070.5 \, J/mol \approx 50.7 \, kJ/mol \]

### Final Answer
The activation energy \( E_a \) is approximately **50.7 kJ/mol**.