In the Arrhenius equation: `k = A exp(-E_(a)//RT)`, the rate constant

Options

A. Decreases with increaisng activation energy and increases with temperature.
B. Increases with activation energy and temperature.
C. Decreases with activation energy and temperature.
D. Increases with activation energy and decreaisng temperature.

Explanation

To solve the problem regarding the Arrhenius equation and the behavior of the rate constant \( k \) with respect to changes in activation energy \( E_a \) and temperature \( T \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Arrhenius Equation**:
The Arrhenius equation is given by:
\[ k = A \exp\left(-\frac{E_a}{RT}\right) \]
where:
- \( k \) = rate constant
- \( A \) = pre-exponential factor (frequency factor)
- \( E_a \) = activation energy
- \( R \) = universal gas constant
- \( T \) = absolute temperature (in Kelvin)

2. **Effect of Increasing Activation Energy \( E_a \)**:
- If we increase the activation energy \( E_a \), the term \(-\frac{E_a}{RT}\) becomes more negative.
- This means that the exponential term \( \exp\left(-\frac{E_a}{RT}\right) \) decreases.
- Therefore, as \( E_a \) increases, \( k \) decreases.

3. **Effect of Increasing Temperature \( T \)**:
- If we increase the temperature \( T \), the term \(-\frac{E_a}{RT}\) becomes less negative (since \( R \) is constant and \( E_a \) remains the same).
- This results in an increase in the value of the exponential term \( \exp\left(-\frac{E_a}{RT}\right) \).
- Therefore, as \( T \) increases, \( k \) increases.

4. **Conclusion**:
- From the analysis, we conclude that:
- Increasing activation energy \( E_a \) decreases the rate constant \( k \).
- Increasing temperature \( T \) increases the rate constant \( k \).

5. **Final Answer**:
- The correct option is that the rate constant \( k \) decreases with increasing activation energy and increases with increasing temperature.