The rate of a reaction doubles when its temperature changes from `300 K` to `310 K`. Activation energy of such a reaction will be:
`(R = 8.314 JK^(-1) mol^(-1) and log 2 = 0.301)`

Options

A. `48.6 kJ mol^(-1)`
B. `58.5 kJ mol^(-1)`
C. `60.5 kJ mol^(-1)`
D. `53.6 kJ mol^(-1)`

Answer

To find the activation energy (Ea) of the reaction given that the rate doubles when the temperature changes from 300 K to 310 K, we can use the Arrhenius equation:

\[ k = A e^{-\frac{E_a}{RT}} \]

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/(K·mol)),
- \( T \) is the temperature in Kelvin.

Given that the rate doubles, we can express this as:

\[ \frac{k_2}{k_1} = 2 \]

This implies:

\[ k_2 = 2k_1 \]

Using the Arrhenius equation, we can write:

\[ \log \left( \frac{k_2}{k_1} \right) = \log(2) = \frac{E_a}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]

Where:
- \( T_1 = 300 \, K \)
- \( T_2 = 310 \, K \)

### Step 1: Calculate \( \frac{1}{T_1} - \frac{1}{T_2} \)

\[ \frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{300} - \frac{1}{310} \]

Calculating this gives:

\[ \frac{1}{300} - \frac{1}{310} = \frac{310 - 300}{300 \times 310} = \frac{10}{93000} = \frac{1}{9300} \]

### Step 2: Substitute values into the equation

Now substituting into the logarithmic form:

\[ \log(2) = 0.301 \]

So we have:

\[ 0.301 = \frac{E_a}{2.303 \times 8.314} \times \frac{1}{9300} \]

### Step 3: Rearranging to solve for \( E_a \)

Rearranging gives:

\[ E_a = 0.301 \times 2.303 \times 8.314 \times 9300 \]

### Step 4: Calculate \( E_a \)

Now calculating the right side:

1. Calculate \( 2.303 \times 8.314 \):

\[ 2.303 \times 8.314 \approx 19.187 \]

2. Now multiply by \( 9300 \):

\[ 19.187 \times 9300 \approx 178,000 \]

3. Finally multiply by \( 0.301 \):

\[ E_a \approx 0.301 \times 178,000 \approx 53,598.59 \, J/mol \]

### Step 5: Convert to kJ/mol

To convert from J/mol to kJ/mol:

\[ E_a \approx 53.59859 \, kJ/mol \approx 53.6 \, kJ/mol \]

### Final Answer

The activation energy \( E_a \) is approximately **53.6 kJ/mol**.

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