The activation energy for most of the reaction is approximately `50 kJ mol^(-1)`. The rate for temperature coefficient for such reaction will be

Options

A. `~~2`
B. `~~3`
C. `lt1`
D. `gt 4`

Answer

To find the rate temperature coefficient for a reaction with an activation energy of approximately 50 kJ mol^(-1), we can follow these steps:

### Step 1: Understand the Temperature Coefficient
The temperature coefficient (Q10) is defined as the ratio of the rate constants (k2 and k1) of a reaction at two temperatures that differ by 10 degrees Celsius (or 10 Kelvin).

### Step 2: Set the Temperatures
We can choose two temperatures:
- \( T_1 = 298 \, \text{K} \) (which is 25°C)
- \( T_2 = 308 \, \text{K} \) (which is 35°C)

### Step 3: Use the Arrhenius Equation
The relationship between the rate constants and activation energy can be expressed using the Arrhenius equation:
\[ \log \left( \frac{k_2}{k_1} \right) = \frac{E_a}{2.303 R} \cdot \frac{(T_2 - T_1)}{T_1 T_2} \]

### Step 4: Substitute the Values
Given:
- Activation energy \( E_a = 50 \, \text{kJ mol}^{-1} = 50 \times 10^3 \, \text{J mol}^{-1} \)
- Universal gas constant \( R = 8.314 \times 10^{-3} \, \text{kJ mol}^{-1} \text{K}^{-1} \)

Now, substituting the values into the equation:
\[ \log \left( \frac{k_2}{k_1} \right) = \frac{50 \times 10^3}{2.303 \times 8.314 \times 10^{-3}} \cdot \frac{10}{298 \times 308} \]

### Step 5: Calculate the Denominator
Calculate \( T_1 \times T_2 \):
\[ T_1 \times T_2 = 298 \times 308 = 91784 \, \text{K}^2 \]

### Step 6: Calculate the Right Side
Now calculate the right side:
\[ \log \left( \frac{k_2}{k_1} \right) = \frac{50000}{2.303 \times 8.314 \times 10^{-3}} \cdot \frac{10}{91784} \]

Calculating the denominator:
\[ 2.303 \times 8.314 \times 10^{-3} \approx 0.0192 \]

Now substituting:
\[ \log \left( \frac{k_2}{k_1} \right) = \frac{50000}{0.0192} \cdot \frac{10}{91784} \]
\[ \log \left( \frac{k_2}{k_1} \right) \approx 2604.17 \cdot 0.0001089 \approx 0.283 \]

### Step 7: Calculate the Ratio
Now, we can find \( \frac{k_2}{k_1} \):
\[ \frac{k_2}{k_1} = 10^{0.283} \approx 1.9 \approx 2 \]

### Conclusion
Thus, the temperature coefficient for the reaction is approximately 2.