The rate constant of a reaction is `3.2 times 10^–4` at 280 K, what will be its value at 300 K approximately?

Options

A. `9.6 times 10^-4`
B. `3.2 times 10^-5`
C. `6.4 times 10^-4`
D. `1.28 times 10^-3`

Answer

To find the rate constant of a reaction at a different temperature using the given rate constant at a known temperature, we can apply the principle derived from the Arrhenius equation. According to this principle, for every 10 degrees Celsius increase in temperature, the rate constant approximately doubles.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Rate constant at 280 K, \( k_{280} = 3.2 \times 10^{-4} \)
- Initial temperature, \( T_1 = 280 \, K \)
- Final temperature, \( T_2 = 300 \, K \)

2. **Calculate the Temperature Change:**
- The change in temperature is \( 300 \, K - 280 \, K = 20 \, K \).
- Since the temperature change is 20 K, we can break it down into two increments of 10 K.

3. **Apply the Doubling Principle:**
- For the first 10 K increase (from 280 K to 290 K):
\[ k_{290} = 2 \times k_{280} = 2 \times (3.2 \times 10^{-4}) = 6.4 \times 10^{-4} \]

- For the second 10 K increase (from 290 K to 300 K):
\[ k_{300} = 2 \times k_{290} = 2 \times (6.4 \times 10^{-4}) = 12.8 \times 10^{-4} \]

4. **Convert to Scientific Notation:**
- The value \( 12.8 \times 10^{-4} \) can be expressed in scientific notation as:
\[ k_{300} = 1.28 \times 10^{-3} \]

5. **Final Answer:**
- The approximate value of the rate constant at 300 K is:
\[ k_{300} \approx 1.28 \times 10^{-3} \]