If `K_(1)=` Rate constant at temperature `T_(1)` and `k_(2)` rate constant at temperature `T_(2)` for a first order reaction, then which of the following relation is correct ?

Options

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

Explanation

To solve the problem regarding the relationship between the rate constants \( k_1 \) and \( k_2 \) at different temperatures \( T_1 \) and \( T_2 \) for a first-order reaction, we can use the Arrhenius equation. Here’s a step-by-step breakdown of the solution:

### Step 1: Write the Arrhenius Equation for \( k_1 \)
The Arrhenius equation relates the rate constant \( k \) to the temperature \( T \) as follows:
\[ k_1 = A e^{-\frac{E_a}{RT_1}} \]
where:
- \( k_1 \) = rate constant at temperature \( T_1 \)
- \( A \) = Arrhenius constant (frequency factor)
- \( E_a \) = activation energy
- \( R \) = universal gas constant
- \( T_1 \) = temperature in Kelvin

### Step 2: Take the Natural Logarithm of \( k_1 \)
Taking the natural logarithm of both sides gives:
\[ \ln k_1 = \ln A - \frac{E_a}{RT_1} \]

### Step 3: Write the Arrhenius Equation for \( k_2 \)
Similarly, for the rate constant \( k_2 \) at temperature \( T_2 \):
\[ k_2 = A e^{-\frac{E_a}{RT_2}} \]

### Step 4: Take the Natural Logarithm of \( k_2 \)
Taking the natural logarithm of both sides gives:
\[ \ln k_2 = \ln A - \frac{E_a}{RT_2} \]

### Step 5: Subtract the Two Equations
Now, we subtract the equation for \( k_1 \) from the equation for \( k_2 \):
\[ \ln k_2 - \ln k_1 = \left(\ln A - \frac{E_a}{RT_2}\right) - \left(\ln A - \frac{E_a}{RT_1}\right) \]
This simplifies to:
\[ \ln k_2 - \ln k_1 = -\frac{E_a}{RT_2} + \frac{E_a}{RT_1} \]
\[ \ln \frac{k_2}{k_1} = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]

### Step 6: Convert to Logarithm Base 10
To convert from natural logarithm to common logarithm (base 10), we use the conversion factor:
\[ \ln x = 2.303 \log_{10} x \]
Thus, we have:
\[ \ln \frac{k_2}{k_1} = 2.303 \log_{10} \frac{k_2}{k_1} \]
Substituting this into our equation gives:
\[ 2.303 \log_{10} \frac{k_2}{k_1} = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]

### Step 7: Rearranging the Equation
Rearranging the equation yields:
\[ \log_{10} \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]

### Conclusion
This relationship shows how the rate constants \( k_1 \) and \( k_2 \) are related to the temperatures \( T_1 \) and \( T_2 \) through the activation energy \( E_a \) and the gas constant \( R \).