At a certain temperature, the first order rate constant `k_(1)` is found to be smaller than the second order rate constant `k_(2)`. If `E_(a)(1)` of the first order reaction is greater than `E_(a)(2)` of the second order reaction, then as temperature is raised:

Options

A. `k_1` will increase faster than `k_2`, but always will remain less than `k_2`
B. `k_2` will increase fasther than `k_1`
C. `k_1` will increase than fasther than `k_2` and becomes equal to `k_2`
D. `k_1` will increase faster `k_2` and becomes greater than `k_2`

Explanation

To solve the problem, we need to analyze the relationship between the rate constants of first-order and second-order reactions in relation to their activation energies and how they change with temperature.

### Step-by-Step Solution:

1. **Understanding Rate Constants**:
- The rate constant \( k \) of a reaction is influenced by temperature and activation energy (\( E_a \)). According to the Arrhenius equation, the rate constant can be expressed as:
\[ k = A e^{-\frac{E_a}{RT}} \]
where \( A \) is the pre-exponential factor, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.

2. **Given Information**:
- At a certain temperature, the first-order rate constant \( k_1 \) is smaller than the second-order rate constant \( k_2 \) (i.e., \( k_1 < k_2 \)).
- The activation energy for the first-order reaction \( E_{a1} \) is greater than that for the second-order reaction \( E_{a2} \) (i.e., \( E_{a1} > E_{a2} \)).

3. **Effect of Temperature on Rate Constants**:
- As temperature increases, both \( k_1 \) and \( k_2 \) will increase, but the rate of increase will depend on their respective activation energies.
- Since \( E_{a1} > E_{a2} \), the first-order reaction (with higher activation energy) will be less sensitive to temperature changes compared to the second-order reaction.

4. **Comparing the Rate Constants**:
- Because \( k_1 < k_2 \) at the initial temperature and \( E_{a1} > E_{a2} \), as the temperature increases, \( k_2 \) will increase at a faster rate than \( k_1 \).
- Therefore, \( k_1 \) will always remain less than \( k_2 \) even as both constants increase with temperature.

5. **Conclusion**:
- The correct conclusion is that \( k_1 \) will increase faster than \( k_2 \) but will always remain less than \( k_2 \).

### Final Answer:
**Option A**: \( k_1 \) will increase faster than \( k_2 \) but always will remain less than \( k_2 \).