A reaction that is of the first order with respect to reactant A has a rate constant `0.6"min"^(-1)`. If we state 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 the time taken for a first-order reaction to reach a certain concentration is given by:

\[ t = \frac{2.303}{k} \log \left( \frac{[A]_0}{[A]} \right) \]

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

**Step 1: Identify the given values.**
- Initial concentration, \( [A]_0 = 0.5 \, \text{mol/L} \)
- Final concentration, \( [A] = 0.05 \, \text{mol/L} \)
- Rate constant, \( k = 0.6 \, \text{min}^{-1} \)

**Step 2: Substitute the values into the formula.**
\[ t = \frac{2.303}{0.6} \log \left( \frac{0.5}{0.05} \right) \]

**Step 3: Calculate the ratio of concentrations.**
\[ \frac{0.5}{0.05} = 10 \]

**Step 4: Calculate the logarithm.**
\[ \log(10) = 1 \]

**Step 5: Substitute the logarithm back into the equation.**
\[ t = \frac{2.303}{0.6} \cdot 1 \]

**Step 6: Calculate the time.**
\[ t = \frac{2.303}{0.6} \approx 3.838 \, \text{minutes} \]

**Step 7: Round the answer.**
The time taken for the concentration of A to decrease from 0.5 mol/L to 0.05 mol/L is approximately \( 3.84 \, \text{minutes} \).

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