The rate constant of a first order reaction at `27^(@)C "is" 10^(-3)"min"^(-1)`.The 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 of finding the rate constant of a first-order reaction at 17°C given the rate constant at 27°C, we can use the Arrhenius equation or the temperature dependence of the rate constant. Here’s a step-by-step solution:

### Step 1: Identify the given data
- Rate constant at \(27^\circ C\) (T2) = \(10^{-3} \, \text{min}^{-1}\)
- Temperature 1 (T1) = \(17^\circ C\)
- Temperature 2 (T2) = \(27^\circ C\)

### Step 2: Calculate the temperature difference
- \(\Delta T = T2 - T1 = 27^\circ C - 17^\circ C = 10^\circ C\)

### Step 3: Use the temperature dependence of the rate constant for first-order reactions
For a first-order reaction, the relationship between the rate constants at two different temperatures can be expressed as:
\[ \frac{k_2}{k_1} = 2^{\frac{\Delta T}{10}} \]
Where:
- \(k_2\) = rate constant at \(T2\) (27°C)
- \(k_1\) = rate constant at \(T1\) (17°C)

### Step 4: Rearranging the equation to find \(k_1\)
We can rearrange the equation to solve for \(k_1\):
\[ k_1 = k_2 \cdot 2^{-\frac{\Delta T}{10}} \]

### Step 5: Substitute the known values
Substituting \(k_2 = 10^{-3} \, \text{min}^{-1}\) and \(\Delta T = 10\):
\[ k_1 = 10^{-3} \cdot 2^{-\frac{10}{10}} = 10^{-3} \cdot 2^{-1} = 10^{-3} \cdot \frac{1}{2} = \frac{10^{-3}}{2} = 0.5 \times 10^{-3} \, \text{min}^{-1} \]

### Step 6: Final result
Thus, the rate constant at \(17^\circ C\) is:
\[ k_1 = 5 \times 10^{-4} \, \text{min}^{-1} \]

### Conclusion
The rate constant at \(17^\circ C\) is \(5 \times 10^{-4} \, \text{min}^{-1}\).

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