A first order reaction with respect to reactant A , has a rate constant `6" min"^(-1)` . If we start with `[A]=0.5 "mol L"^(-1)` , when would [A] reach the value `0.05" mol L"^(-1)`

Options

A. 0.384 min
B. 0.15 min
C. 3 min
D. 3.84 min

Answer

To solve the problem, we will use the first-order reaction kinetics formula. The formula for a first-order reaction is given by:

\[ \ln \left( \frac{[A]_0}{[A]} \right) = k \cdot t \]

Where:
- \([A]_0\) is the initial concentration of the reactant.
- \([A]\) is the concentration of the reactant at time \(t\).
- \(k\) is the rate constant.
- \(t\) is the time.

### Step 1: Identify the given values
- Initial concentration, \([A]_0 = 0.5 \, \text{mol L}^{-1}\)
- Final concentration, \([A] = 0.05 \, \text{mol L}^{-1}\)
- Rate constant, \(k = 6 \, \text{min}^{-1}\)

### Step 2: Rearrange the formula
We need to find \(t\), so we rearrange the formula:

\[ t = \frac{1}{k} \ln \left( \frac{[A]_0}{[A]} \right) \]

### Step 3: Substitute the values into the equation
Substituting the known values into the rearranged formula:

\[ t = \frac{1}{6} \ln \left( \frac{0.5}{0.05} \right) \]

### Step 4: Calculate the ratio
Calculate the ratio:

\[ \frac{0.5}{0.05} = 10 \]

### Step 5: Calculate the natural logarithm
Now, calculate the natural logarithm:

\[ \ln(10) \approx 2.303 \]

### Step 6: Substitute back to find \(t\)
Now substitute back into the equation for \(t\):

\[ t = \frac{1}{6} \cdot 2.303 \]

### Step 7: Calculate \(t\)
Now calculate \(t\):

\[ t \approx 0.384 \, \text{min} \]

### Final Answer
Thus, the time when the concentration of \(A\) reaches \(0.05 \, \text{mol L}^{-1}\) is approximately **0.384 minutes**.

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