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. 5 × 10–4
C. 2 × 10–3
D. 10–2

Explanation

To solve the problem, we need to determine the rate constant of a first-order reaction at a lower temperature (17°C) given the rate constant at a higher temperature (27°C) and the temperature coefficient of the reaction.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Rate constant at 27°C (T1): \( k_1 = 10^{-3} \, \text{min}^{-1} \)
- Temperature coefficient (θ): \( θ = 2 \)
- Temperature at which we want to find the rate constant (T2): \( T2 = 17°C \)

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

3. **Use the Temperature Coefficient Formula:**
The relationship between the rate constants at two different temperatures can be expressed as:
\[ \frac{k_2}{k_1} = θ^{\frac{ΔT}{10}} \]
where \( k_2 \) is the rate constant at T2 (17°C).

4. **Substitute the Known Values:**
\[ \frac{k_2}{10^{-3}} = 2^{\frac{10}{10}} = 2^1 = 2 \]

5. **Solve for \( k_2 \):**
\[ k_2 = 2 \times 10^{-3} \]
\[ k_2 = 0.5 \times 10^{-3} = 5 \times 10^{-4} \, \text{min}^{-1} \]

### Final Answer:
The rate constant at 17°C is:
\[ k_2 = 5 \times 10^{-4} \, \text{min}^{-1} \]