To solve the problem, we need to determine the rate of disappearance of reactant A in the reaction \(3A(g) \overset{k}{\rightarrow} B(g) + C(g)\) given the rate constant \(k = 10^{-4} \, \text{L/mol.min}\) and the concentration of A, \([A] = 0.5 \, \text{M}\).
### Step-by-Step Solution:
1. **Identify the Order of the Reaction**:
- The reaction is given as \(3A \rightarrow B + C\).
- The stoichiometry suggests that for every 3 moles of A that react, 1 mole of B and 1 mole of C are produced.
- Since the rate constant \(k\) has units of \(\text{L/mol.min}\), this indicates that the reaction is second order with respect to A.
2. **Write the Rate Law**:
- For a second-order reaction, the rate law can be expressed as:
\[
\text{Rate} = k[A]^2
\]
- Here, \(k = 10^{-4} \, \text{L/mol.min}\) and \([A] = 0.5 \, \text{M}\).
3. **Calculate the Rate**:
- Substitute the values into the rate equation:
\[
\text{Rate} = 10^{-4} \, \text{L/mol.min} \times (0.5 \, \text{M})^2
\]
- Calculate \((0.5)^2 = 0.25\):
\[
\text{Rate} = 10^{-4} \times 0.25 = 0.25 \times 10^{-4} \, \text{mol/min}
\]
4. **Relate Rate to \(-\frac{d[A]}{dt}\)**:
- The rate of reaction can also be expressed in terms of the change in concentration of A:
\[
\text{Rate} = -\frac{1}{3} \frac{d[A]}{dt}
\]
- Therefore, we can express \(-\frac{d[A]}{dt}\) as:
\[
-\frac{d[A]}{dt} = 3 \times \text{Rate}
\]
- Substitute the calculated rate:
\[
-\frac{d[A]}{dt} = 3 \times (0.25 \times 10^{-4}) = 0.75 \times 10^{-4} \, \text{mol/min}
\]
5. **Convert to \(\text{mol/s}\)**:
- To convert from \(\text{mol/min}\) to \(\text{mol/s}\), divide by 60:
\[
-\frac{d[A]}{dt} = \frac{0.75 \times 10^{-4}}{60} \, \text{mol/s}
\]
- Calculate:
\[
-\frac{d[A]}{dt} = 0.0125 \times 10^{-4} \, \text{mol/s}
\]
6. **Final Result**:
- Thus, the value of \(-\frac{d[A]}{dt}\) in \(\text{mol/s}\) is:
\[
-\frac{d[A]}{dt} = 1.25 \times 10^{-6} \, \text{mol/s}
\]
