The initial rate of zero order reaction of the gaseous equation `A(g)to2B (g)` is `10^(-2) M min^(-1)` If the initial concentration of A is 0.1 M what would be concentration of B after 60 seconds ?

Options

A. 0.09 M
B. 0.01 M
C. 0.02 M
D. 0.03 M

Explanation

To solve the problem, we need to determine the concentration of B after 60 seconds in a zero-order reaction given by the equation \( A(g) \rightarrow 2B(g) \).

### Step-by-Step Solution:

1. **Understand the Reaction Order**:
- The reaction is a zero-order reaction, which means the rate of reaction is constant and does not depend on the concentration of the reactants.

2. **Identify the Given Data**:
- Initial rate of reaction \( = 10^{-2} \, \text{M min}^{-1} \)
- Initial concentration of \( A \) \( = 0.1 \, \text{M} \)
- Time \( t = 60 \, \text{seconds} = 1 \, \text{minute} \)

3. **Relate Rate to Concentration Change**:
- For a zero-order reaction, the rate \( r \) can be expressed as:
\[ r = -\frac{d[A]}{dt} = k \]
- Since \( 2B \) is produced for every mole of \( A \) consumed, we can express the change in concentration of \( B \):
\[ \frac{d[B]}{dt} = 2k \]

4. **Integrate the Rate Equation**:
- Rearranging gives:
\[ d[B] = 2k \, dt \]
- Integrating from \( t = 0 \) to \( t = 1 \) minute:
\[ [B] = 2k \cdot t \]
- Here, \( k = 10^{-2} \, \text{M min}^{-1} \) and \( t = 1 \, \text{minute} \).

5. **Substitute Values**:
- Substitute \( k \) and \( t \) into the equation:
\[ [B] = 2 \cdot (10^{-2}) \cdot (1) = 0.02 \, \text{M} \]

6. **Final Concentration of B**:
- The concentration of \( B \) after 60 seconds is \( 0.02 \, \text{M} \).

### Conclusion:
The concentration of \( B \) after 60 seconds is \( 0.02 \, \text{M} \).