Which of the following is the correct expression for Arrhenius
equation ?

Options

A. `In(K_(2))/(K_(1))=(E_(a))/(R)((1)/(T_(1))-(1)/(T_(2)))`
B. `"log"(K_(2))/(K_(1))=(E_(a))/(2.303)((T_(1)T_(2))/(T_(2)-T_(1)))`
C. `In(K_(2))/(K_(1))=(E_(a))/(2.303R)((1)/(T_(1))-(1)/(T_(2)))`
D. `(K_(2))/(K_(1))=(E_(a))/(2.303R)((T_(2)-T_(1))/(T_(1)T_(2)))`

Answer

To solve the question regarding the correct expression for the Arrhenius equation, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Arrhenius Equation**:
The Arrhenius equation describes how the rate constant (k) of a chemical reaction depends on temperature (T) and is given by:
\[ k = A e^{-\frac{E_a}{RT}} \]
where:
- \( k \) = rate constant
- \( A \) = frequency factor (pre-exponential factor)
- \( E_a \) = activation energy
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

2. **Take the Natural Logarithm**:
To linearize the equation, take the natural logarithm of both sides:
\[ \ln k = \ln A - \frac{E_a}{RT} \]

3. **Consider Two Different Temperatures**:
If we consider the rate constants at two different temperatures \( T_1 \) and \( T_2 \), we can write:
\[ \ln k_1 = \ln A - \frac{E_a}{RT_1} \quad \text{and} \quad \ln k_2 = \ln A - \frac{E_a}{RT_2} \]

4. **Subtract the Two Equations**:
By subtracting the second equation from the first, we get:
\[ \ln k_2 - \ln k_1 = -\frac{E_a}{RT_2} + \frac{E_a}{RT_1} \]
This simplifies to:
\[ \ln \frac{k_2}{k_1} = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]

5. **Rearranging the Equation**:
The equation can also be rearranged in a different form:
\[ \ln \frac{k_2}{k_1} = \frac{E_a}{R} \left( \frac{T_2 - T_1}{T_1 T_2} \right) \]

6. **Final Expression**:
Thus, the correct expression for the Arrhenius equation in terms of the rate constants at two different temperatures is:
\[ \ln \frac{k_2}{k_1} = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]
or equivalently,
\[ \ln \frac{k_2}{k_1} = \frac{E_a}{R} \frac{T_2 - T_1}{T_1 T_2} \]

### Conclusion:
The correct expression for the Arrhenius equation is:
\[ \ln \frac{k_2}{k_1} = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]