Activation energy `(E_a)` and rate constants `(k_1 and k_2)` of a chemical reaction at two different temperature `(T_1 and T_2)` are related by

Options

A. In `k_2/k_1=-E_a/R(1/T_1-1/T_2)`
B. In `k_2/k_1=-E_a/R(1/T_2-1/T_1)`
C. In `k_2/k_1=-E_a/R(1/T_2+1/T_1)`
D. In `k_2/k_1=E_a/R(1/T_1-1/T_2)`

Answer

To solve the problem of how activation energy \(E_a\) and rate constants \(k_1\) and \(k_2\) of a chemical reaction at two different temperatures \(T_1\) and \(T_2\) are related, we can use the Arrhenius equation. Here’s a step-by-step breakdown of the solution:

### Step 1: Write the Arrhenius Equation
The Arrhenius equation relates the rate constant \(k\) to the activation energy \(E_a\), the pre-exponential factor \(A\), the gas constant \(R\), and the temperature \(T\):
\[ k = A e^{-\frac{E_a}{RT}} \]

### Step 2: Write the Equation for Two Different Temperatures
For the two different temperatures \(T_1\) and \(T_2\), we can express the rate constants \(k_1\) and \(k_2\) as:
\[ k_1 = A e^{-\frac{E_a}{RT_1}} \]
\[ k_2 = A e^{-\frac{E_a}{RT_2}} \]

### Step 3: Take the Natural Logarithm of Both Equations
Taking the natural logarithm of both sides for each equation gives us:
\[ \ln k_1 = \ln A - \frac{E_a}{RT_1} \quad \text{(1)} \]
\[ \ln k_2 = \ln A - \frac{E_a}{RT_2} \quad \text{(2)} \]

### Step 4: Subtract the Two Equations
Now, we can subtract equation (1) from equation (2):
\[ \ln k_2 - \ln k_1 = \left(-\frac{E_a}{RT_2}\right) - \left(-\frac{E_a}{RT_1}\right) \]
This simplifies to:
\[ \ln k_2 - \ln k_1 = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]

### Step 5: Rewrite the Equation
We can rewrite the left side using the properties of logarithms:
\[ \ln \left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]

### Step 6: Final Form
Thus, the relationship between the activation energy, rate constants, and temperatures can be expressed as:
\[ \ln \left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]

### Conclusion
This equation shows how the rate constants at two different temperatures are related to the activation energy of the reaction.