The rate constant of the reaction `A rarr B` is `0.6 xx 10^(-3)` mole per second. If the concentration of `A` is `5 M`, then concentration of `B` after `20` minutes is:

Options

A. 0.36 M
B. 0.72 M
C. 1.08 M
D. 3.60 M

Answer

To solve the problem, we need to find the concentration of B after 20 minutes given the rate constant and the initial concentration of A. 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.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Rate constant (k) = \(0.6 \times 10^{-3}\) moles per second
- Initial concentration of A ([A]₀) = 5 M
- Time (t) = 20 minutes

2. **Convert Time to Seconds:**
- Since the rate constant is given in moles per second, we need to convert the time from minutes to seconds.
\[ t = 20 \text{ minutes} \times 60 \text{ seconds/minute} = 1200 \text{ seconds} \]

3. **Use the Zero-Order Reaction Formula:**
- For a zero-order reaction, the change in concentration (x) can be calculated using the formula:
\[ x = k \times t \]

4. **Substitute the Values:**
- Now, substitute the values of k and t into the formula:
\[ x = (0.6 \times 10^{-3} \text{ moles/second}) \times (1200 \text{ seconds}) \]

5. **Calculate x:**
\[ x = 0.6 \times 10^{-3} \times 1200 = 0.72 \text{ moles} \]

6. **Find the Concentration of B:**
- The concentration of B after the reaction can be found by considering that the amount of A that has reacted is equal to the amount of B that has been formed. Therefore, the concentration of B is:
\[ [B] = x = 0.72 \text{ M} \]

7. **Final Answer:**
- The concentration of B after 20 minutes is \(0.72 \text{ M}\).