A chemical reaction was carried out at 300 K and 280 K. The rate constants were found to be `k_(1) and k_(2)` respectively. Then

Options

A. `k_(2)=4k_(1)`
B. `k_(2)=2k_(1)`
C. `k_(2)=0.25 k_(1)`
D. `k_(2)=0.5 k_(1)`

Explanation

To solve the problem, we need to establish the relationship between the rate constants \( k_1 \) and \( k_2 \) at two different temperatures, \( T_1 = 300 \, K \) and \( T_2 = 280 \, K \).

### Step-by-Step Solution:

1. **Identify the temperatures and rate constants:**
- Let \( T_1 = 300 \, K \) with rate constant \( k_1 \).
- Let \( T_2 = 280 \, K \) with rate constant \( k_2 \).

2. **Determine the temperature difference:**
- Calculate \( \Delta T = T_2 - T_1 = 280 \, K - 300 \, K = -20 \, K \).

3. **Use the Arrhenius equation concept:**
- According to the Arrhenius equation, the rate constant \( k \) increases with temperature. A common rule of thumb is that for every 10 degrees Celsius increase in temperature, the rate constant approximately doubles.
- Therefore, we can express the relationship between \( k_2 \) and \( k_1 \) as:
\[ \frac{k_2}{k_1} = 2^{\frac{\Delta T}{10}} \]

4. **Substitute the temperature difference into the equation:**
- Substitute \( \Delta T = -20 \):
\[ \frac{k_2}{k_1} = 2^{\frac{-20}{10}} = 2^{-2} \]

5. **Calculate \( 2^{-2} \):**
- \( 2^{-2} = \frac{1}{4} \), so:
\[ \frac{k_2}{k_1} = \frac{1}{4} \]

6. **Rearrange to find the relationship between \( k_2 \) and \( k_1 \):**
- This implies:
\[ k_2 = \frac{1}{4} k_1 \]
- Alternatively, we can express this as:
\[ k_1 = 4 k_2 \]

7. **Conclusion:**
- The relationship between the rate constants is:
\[ k_2 = 0.25 k_1 \]
- Therefore, the correct option is that \( k_2 \) is equal to \( 0.25 k_1 \).

### Final Answer:
The relationship between the rate constants is:
\[ k_2 = 0.25 k_1 \]