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 find the rate of the reaction where the concentration of A decreases from 0.8 mole/L to 0.3 mole/L in 20 minutes, we can follow these steps:

### Step 1: Identify the change in concentration
The initial concentration of A is 0.8 mole/L, and the final concentration after 20 minutes is 0.3 mole/L.

\[ \Delta [A] = [A]_{final} - [A]_{initial} = 0.3 \, \text{mol/L} - 0.8 \, \text{mol/L} = -0.5 \, \text{mol/L} \]

### Step 2: Calculate the change in time
The time interval given is 20 minutes.

### Step 3: Write the rate expression
The rate of reaction can be expressed as:

\[ \text{Rate} = -\frac{\Delta [A]}{\Delta t} \]

Where:
- \(\Delta [A]\) is the change in concentration of A
- \(\Delta t\) is the change in time

### Step 4: Substitute the values into the rate expression
Substituting the values we have:

\[ \text{Rate} = -\frac{-0.5 \, \text{mol/L}}{20 \, \text{min}} = \frac{0.5 \, \text{mol/L}}{20 \, \text{min}} \]

### Step 5: Perform the calculation
Now, we perform the division:

\[ \text{Rate} = \frac{0.5}{20} = 0.025 \, \text{mol/L/min} \]

### Step 6: Adjust for stoichiometry
Since the reaction is \(2A \to \text{Product}\), we need to account for the stoichiometry:

\[ \text{Rate} = \frac{1}{2} \times 0.025 \, \text{mol/L/min} = 0.0125 \, \text{mol/L/min} \]

### Final Answer
Thus, the rate of the reaction during this interval of time is:

\[ \text{Rate} = 0.0125 \, \text{mol/L/min} \]