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_(2)` will increase faster than `k_(1)`
B. `k_(1)` will increase faster than `k_(2)` but will always remain less than `k_(2)`
C. `k_(1)` will increase faster than `k_(2)` and become equal to `k_(2)`.
D. `k_(1)` will increase faster than `k_(2)` and become 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 as temperature increases, given the activation energies.

### Step-by-Step Solution:

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

2. **Given Information**:
- At a certain temperature, \( k_1 < k_2 \) (the first-order rate constant is smaller than the second-order rate constant).
- 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 due to the exponential factor in the Arrhenius equation.
- However, because \( E_{a1} > E_{a2} \), the first-order reaction is less sensitive to temperature changes compared to the second-order reaction.

4. **Comparing the Rate Constants**:
- Since \( k_2 \) is initially greater than \( k_1 \) and the second-order reaction is more temperature-dependent (due to lower activation energy), \( k_2 \) will increase at a faster rate than \( k_1 \) as temperature rises.
- Therefore, even though both constants increase, \( k_2 \) will remain greater than \( k_1 \) as the temperature continues to increase.

5. **Conclusion**:
- As temperature is raised, \( k_1 \) will increase, but \( k_2 \) will increase at a faster rate, maintaining the inequality \( k_2 > k_1 \).

### Final Answer:
As temperature is raised, \( k_2 \) will increase faster than \( k_1 \), and \( k_2 \) will remain greater than \( k_1 \).

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