What is the energy of activation of a reaction is its rate doubles when the temperature is raised from 290 K to 300 K ?

Options

A. 12 kcal
B. 15 kcal
C. 10 kcal
D. 20 kcal

Explanation

To solve the problem of finding the energy of activation (Ea) for a reaction where the rate doubles when the temperature increases from 290 K to 300 K, we can use the Arrhenius equation in its logarithmic form. Here’s a step-by-step solution:

### Step 1: Understand the Given Information
We know that:
- The rate doubles when the temperature increases from 290 K to 300 K.
- We need to find the activation energy (Ea).

### Step 2: Use the Arrhenius Equation
The Arrhenius equation in logarithmic form is given by:
\[ \log \left( \frac{K_2}{K_1} \right) = \frac{E_a}{2.303 R} \left( \frac{T_2 - T_1}{T_1 T_2} \right) \]
Where:
- \( K_1 \) is the rate constant at temperature \( T_1 \).
- \( K_2 \) is the rate constant at temperature \( T_2 \).
- \( R \) is the universal gas constant (approximately 8.314 J/mol·K).
- \( T_1 = 290 \, K \) and \( T_2 = 300 \, K \).

### Step 3: Substitute Known Values
Since the rate doubles, we have:
\[ \frac{K_2}{K_1} = 2 \]
Thus, we can rewrite the equation as:
\[ \log(2) = \frac{E_a}{2.303 R} \left( \frac{300 - 290}{290 \times 300} \right) \]

### Step 4: Calculate the Temperature Difference and Product
Calculate \( T_2 - T_1 \):
\[ T_2 - T_1 = 300 - 290 = 10 \, K \]

Calculate \( T_1 \times T_2 \):
\[ T_1 \times T_2 = 290 \times 300 = 87000 \, K^2 \]

### Step 5: Substitute and Solve for Ea
Substituting the values into the equation:
\[ \log(2) = \frac{E_a}{2.303 \times 8.314} \left( \frac{10}{87000} \right) \]

Now, calculate \( \log(2) \):
\[ \log(2) \approx 0.301 \]

Substituting this into the equation:
\[ 0.301 = \frac{E_a}{2.303 \times 8.314} \left( \frac{10}{87000} \right) \]

Now, rearranging to solve for \( E_a \):
\[ E_a = 0.301 \times 2.303 \times 8.314 \times \frac{87000}{10} \]

### Step 6: Calculate the Activation Energy
Calculating the right-hand side:
\[ E_a \approx 0.301 \times 2.303 \times 8.314 \times 8700 \]
Calculating this gives:
\[ E_a \approx 12.062 \, kcal \]

### Final Answer
Thus, the activation energy \( E_a \) is approximately:
\[ E_a \approx 12 \, kcal \]