`A overset(k) rarrB, t_(1//2)=10" min "3A rarrC`
Both reaction have same rate constant and each occurring following first order kinetics. Choose the correct option for second reaction.

Options

A. `t_(1//2)=10//3" min"`
B. `t_(1//2)=30" min"`
C. `t_(1//2)=10" min"`
D. Data insufficient

Answer

To solve the problem, we need to determine the half-life of the second reaction, which is given as \(3A \rightarrow C\), knowing that both reactions follow first-order kinetics and have the same rate constant \(k\).

### Step-by-Step Solution:

1. **Understanding First-Order Kinetics**:
- For a first-order reaction, the half-life (\(t_{1/2}\)) is given by the formula:
\[ t_{1/2} = \frac{\ln 2}{k} \]
- The half-life is independent of the initial concentration.

2. **Given Information**:
- The half-life for the reaction \(A \rightarrow B\) is given as \(10 \text{ minutes}\). Thus:
\[ t_{1/2} = 10 \text{ minutes} = \frac{\ln 2}{k} \]
- From this, we can find the rate constant \(k\):
\[ k = \frac{\ln 2}{10 \text{ minutes}} \]

3. **Analyzing the Second Reaction**:
- The second reaction is \(3A \rightarrow C\). This is also a first-order reaction, but the stoichiometry indicates that three moles of \(A\) are consumed to produce one mole of \(C\).
- For a reaction of the form \(nA \rightarrow \text{products}\), the rate of reaction can be expressed as:
\[ -\frac{1}{n} \frac{d[A]}{dt} = k[A] \]
- For our case, \(n = 3\), so:
\[ -\frac{1}{3} \frac{d[A]}{dt} = k[A] \]

4. **Integrating the Rate Equation**:
- Rearranging gives:
\[ d[A] = -3k[A]dt \]
- Integrating from \(t = 0\) to \(t\) and from \([A]_0\) to \([A]_t\):
\[ \int_{[A]_0}^{[A]_t} \frac{d[A]}{[A]} = -3k \int_0^t dt \]
- This results in:
\[ \ln \left(\frac{[A]_0}{[A]_t}\right) = -3kt \]

5. **Finding Half-Life for the Second Reaction**:
- At half-life, \([A]_t = \frac{[A]_0}{2}\):
\[ \ln \left(\frac{[A]_0}{\frac{[A]_0}{2}}\right) = -3k t_{1/2} \]
- This simplifies to:
\[ \ln(2) = -3k t_{1/2} \]
- Solving for \(t_{1/2}\):
\[ t_{1/2} = \frac{\ln(2)}{3k} \]

6. **Substituting the Value of \(k\)**:
- We already found \(k = \frac{\ln 2}{10}\):
\[ t_{1/2} = \frac{\ln(2)}{3 \cdot \frac{\ln(2)}{10}} = \frac{10}{3} \text{ minutes} \]

### Final Answer:
The half-life for the second reaction \(3A \rightarrow C\) is:
\[ t_{1/2} = \frac{10}{3} \text{ minutes} \approx 3.33 \text{ minutes} \]