In a reaction 2A `to`Product the concentration of A decreases from 0.8 mole/L to 0.3 mole/L in 20 min. The rate of the reaction during this interval of time is

Options

A. `-0.01 mol l^-1 min^-1`
B. `0.01 mol l^-1 min^-1`
C. `-0.005 mol l^-1 min^-1`
D. `0.012 mol l^-1 min^-1`

Answer

To solve the problem, we need to calculate the rate of the reaction based on the change in concentration of reactant A over a given time interval.

### Step-by-Step Solution:

1. **Identify the initial and final concentrations of A:**
- Initial concentration (\[A_0\]) = 0.8 mole/L
- Final concentration (\[A_f\]) = 0.3 mole/L

2. **Calculate the change in concentration (\( \Delta [A] \)):**
\[ \Delta [A] = [A_f] - [A_0] = 0.3 \, \text{mole/L} - 0.8 \, \text{mole/L} = -0.5 \, \text{mole/L} \]
(The negative sign indicates a decrease in concentration.)

3. **Determine the time interval (\( \Delta t \)):**
- Time interval = 20 minutes

4. **Calculate the rate of the reaction:**
The rate of the reaction is defined as:
\[ \text{Rate} = -\frac{1}{2} \frac{\Delta [A]}{\Delta t} \]
Substituting the values we calculated:
\[ \text{Rate} = -\frac{1}{2} \frac{-0.5 \, \text{mole/L}}{20 \, \text{min}} = \frac{0.5}{2 \times 20} = \frac{0.5}{40} = 0.0125 \, \text{mole/L/min} \]

5. **Final Answer:**
The rate of the reaction during this interval is approximately:
\[ \text{Rate} \approx 0.0125 \, \text{mole/L/min} \]