The velocity constant of a reaction at 290 K was found to be `3.2 xx 10^(-3) s^(-1)`. When the temperature is raised to 310 K, it will be about

Options

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

Explanation

To determine the velocity constant (k) of a reaction at 310 K given that the velocity constant at 290 K is \(3.2 \times 10^{-3} \, s^{-1}\), we can use the rule of thumb that for every 10°C (or 10 K) increase in temperature, the rate constant approximately doubles.

### Step-by-Step Solution:

1. **Identify the initial conditions**:
- Initial temperature (T1) = 290 K
- Initial rate constant (k1) = \(3.2 \times 10^{-3} \, s^{-1}\)
- Final temperature (T2) = 310 K

2. **Calculate the temperature increase**:
- The increase in temperature (ΔT) = T2 - T1 = 310 K - 290 K = 20 K.

3. **Determine the number of 10 K intervals**:
- Since the temperature increase is 20 K, we have 2 intervals of 10 K (i.e., 290 K to 300 K and 300 K to 310 K).

4. **Calculate the new rate constant after the first 10 K increase**:
- For the first 10 K increase, the new rate constant (k2) will be approximately double the initial rate constant:
\[ k2 = 2 \times k1 = 2 \times (3.2 \times 10^{-3}) = 6.4 \times 10^{-3} \, s^{-1} \]

5. **Calculate the new rate constant after the second 10 K increase**:
- For the second 10 K increase, the new rate constant (k3) will again be approximately double the rate constant after the first increase:
\[ k3 = 2 \times k2 = 2 \times (6.4 \times 10^{-3}) = 1.28 \times 10^{-2} \, s^{-1} \]

6. **Final answer**:
- The velocity constant at 310 K is approximately \(1.28 \times 10^{-2} \, s^{-1}\).