For any reaction `A(g) rarr B(g)`, rate constant `k = 8.21 xx 10^(-2)` atm/min at 300 K. If initial concentration of A is 2M then what is the half life (in hr.)?

Options

A. 5
B. 6
C. 7
D. 8

Explanation

To solve the problem of finding the half-life of the reaction \( A(g) \rightarrow B(g) \) with the given rate constant and initial concentration, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction Order**:
The problem states that the reaction is zero-order because the rate constant \( k \) is given in units of pressure/time (atm/min). In zero-order reactions, the rate of reaction is constant and does not depend on the concentration of the reactant.

2. **Convert Initial Concentration to Pressure**:
The initial concentration of \( A \) is given as 2 M. To convert this to pressure in atm, we can use the ideal gas law:
\[ P = C \cdot R \cdot T \]
where:
- \( C = 2 \, \text{mol/L} \)
- \( R = 0.0821 \, \text{L atm/(K mol)} \)
- \( T = 300 \, \text{K} \)

Substituting the values:
\[ P = 2 \, \text{mol/L} \cdot 0.0821 \, \text{L atm/(K mol)} \cdot 300 \, \text{K} = 49.26 \, \text{atm} \]

3. **Use the Half-Life Formula for Zero-Order Reactions**:
The half-life \( t_{1/2} \) for a zero-order reaction is given by the formula:
\[ t_{1/2} = \frac{C_0}{2k} \]
where \( C_0 \) is the initial concentration in atm and \( k \) is the rate constant.

Substituting the values:
\[ t_{1/2} = \frac{49.26 \, \text{atm}}{2 \cdot 8.21 \times 10^{-2} \, \text{atm/min}} = \frac{49.26}{0.1642} \approx 300 \, \text{min} \]

4. **Convert Minutes to Hours**:
To convert the half-life from minutes to hours:
\[ t_{1/2} = \frac{300 \, \text{min}}{60} = 5 \, \text{hours} \]

### Final Answer:
The half-life of the reaction is **5 hours**.