At 290K the rate constant of a reaction is `3.2 times 10^-3`. Its value of 310K will be approximately:

Options

A. `1.28 times 10^-2`
B. `3.2 times 10^-4`
C. `9.6 times 10^-4`
D. `6.4 times 10^-3`

Answer

To find the rate constant of a reaction at 310 K given its value at 290 K, we can use the rule of thumb that states the rate constant approximately doubles for every 10-degree Celsius increase in temperature.

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- Given:
- Rate constant at 290 K, \( k_{290} = 3.2 \times 10^{-3} \, \text{s}^{-1} \)
- We need to find the rate constant at 310 K.

2. **Calculate the Temperature Increase:**
- The temperature increase from 290 K to 310 K is:
\[ 310 \, \text{K} - 290 \, \text{K} = 20 \, \text{K} \]
- This corresponds to a change of \( 20 \, \text{K} / 10 \, \text{K} = 2 \) increments of 10 K.

3. **Apply the Doubling Rule:**
- For each 10 K increase, the rate constant doubles. Therefore, for a 20 K increase, the rate constant will double twice:
- First increment (from 290 K to 300 K):
\[ k_{300} = 2 \times k_{290} = 2 \times (3.2 \times 10^{-3}) = 6.4 \times 10^{-3} \, \text{s}^{-1} \]
- Second increment (from 300 K to 310 K):
\[ k_{310} = 2 \times k_{300} = 2 \times (6.4 \times 10^{-3}) = 12.8 \times 10^{-3} \, \text{s}^{-1} \]

4. **Convert to Standard Scientific Notation:**
- Convert \( 12.8 \times 10^{-3} \) to proper scientific notation:
\[ 12.8 \times 10^{-3} = 1.28 \times 10^{-2} \, \text{s}^{-1} \]

5. **Final Result:**
- The approximate value of the rate constant at 310 K is:
\[ k_{310} \approx 1.28 \times 10^{-2} \, \text{s}^{-1} \]