The rate constant of the reaction `A to B` is `0.6xx10^(3)` mole per litre 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"`

Explanation

To solve the problem, we need to determine the concentration of product B after a certain time, given the rate constant and initial concentration of reactant A. The reaction is a zero-order reaction, which means the rate of reaction is constant and does not depend on the concentration of reactants.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Rate constant (K) = \(0.6 \times 10^{-3}\) mol/L/s
- Initial concentration of A ([A]₀) = 5 M
- Time (t) = 20 minutes = \(20 \times 60 = 1200\) seconds

2. **Use the Zero-Order Reaction Formula:**
For a zero-order reaction, the concentration of the reactant decreases linearly over time. The formula is:
\[ [A] = [A]_0 - Kt \]
Where:
- [A] is the concentration of A at time t
- [A]_0 is the initial concentration of A
- K is the rate constant
- t is the time in seconds

3. **Substitute the Values into the Formula:**
\[ [A] = 5 \, \text{M} - (0.6 \times 10^{-3} \, \text{mol/L/s}) \times (1200 \, \text{s}) \]

4. **Calculate the Change in Concentration:**
\[ Kt = 0.6 \times 10^{-3} \times 1200 = 0.72 \, \text{mol/L} \]

5. **Calculate the Final Concentration of A:**
\[ [A] = 5 \, \text{M} - 0.72 \, \text{mol/L} = 4.28 \, \text{M} \]

6. **Determine the Concentration of B:**
Since the reaction starts with A and produces B, the amount of B formed is equal to the amount of A that has reacted:
\[ [B] = [A]_0 - [A] = 5 \, \text{M} - 4.28 \, \text{M} = 0.72 \, \text{M} \]

### Final Answer:
The concentration of B after 20 minutes is **0.72 M**.