The rate constant of a first order reaction at `27^@C` is 10–3 min–1. The temperature coefficient of this reaction is 2. What is the rate constant (in min–1) at `17^@C` for this reaction :-

Options

A. `10^(-3)`
B. `5xx10^(-4)`
C. `2xx10^(-3)`
D. `10^(-2)`

Explanation

To solve the problem, we need to determine the rate constant (k2) at 17°C given the rate constant (k1) at 27°C and the temperature coefficient (θ) of the reaction.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Rate constant at 27°C (k1) = \(10^{-3}\) min\(^{-1}\)
- Temperature coefficient (θ) = 2
- Initial temperature (T1) = 27°C
- Final temperature (T2) = 17°C

2. **Calculate the Change in Temperature (ΔT):**
\[ \Delta T = T2 - T1 = 17°C - 27°C = -10°C \]

3. **Use the Temperature Coefficient Formula:**
The relationship between the rate constants at two different temperatures can be expressed as:
\[ \frac{k2}{k1} = \theta^{\frac{\Delta T}{10}} \]
Substituting the known values:
\[ \frac{k2}{10^{-3}} = 2^{\frac{-10}{10}} = 2^{-1} = \frac{1}{2} \]

4. **Rearranging to Find k2:**
\[ k2 = k1 \times \frac{1}{2} \]
Substituting k1:
\[ k2 = 10^{-3} \times \frac{1}{2} = 5 \times 10^{-4} \text{ min}^{-1} \]

5. **Final Result:**
The rate constant at 17°C is:
\[ k2 = 5 \times 10^{-4} \text{ min}^{-1} \]

### Final Answer:
The rate constant (k2) at 17°C is \(5 \times 10^{-4}\) min\(^{-1}\).

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