In the reaction `Ato2B` the concentration of a falls from 1.0M to 0.98 M in one minute what is the rate of in moles `"litre"^(-1)"sec^(-1)`

Options

A. `1.8xx10^(-3)`
B. `3.3xx10^(-4)`
C. `3.6xx10^(-3)`
D. `6.0xx10^(-5)`

Answer

To solve the problem, we need to find the rate of the reaction \( A \rightarrow 2B \) based on the change in concentration of \( A \) over a given time period.

### Step-by-Step Solution:

1. **Identify Initial and Final Concentration**:
- The initial concentration of \( A \) is \( [A]_0 = 1.0 \, \text{M} \).
- The final concentration of \( A \) after one minute is \( [A]_t = 0.98 \, \text{M} \).

2. **Calculate the Change in Concentration**:
- The change in concentration \( \Delta [A] \) can be calculated as:
\[ \Delta [A] = [A]_t - [A]_0 = 0.98 \, \text{M} - 1.0 \, \text{M} = -0.02 \, \text{M} \]
- Note that the change is negative because the concentration of \( A \) is decreasing.

3. **Determine the Time Interval**:
- The time interval \( \Delta t \) is given as 1 minute. We need to convert this into seconds:
\[ \Delta t = 1 \, \text{minute} = 60 \, \text{seconds} \]

4. **Calculate the Rate of Reaction**:
- The rate of reaction is defined as the change in concentration of \( A \) per unit time. The rate can be expressed as:
\[ \text{Rate} = -\frac{\Delta [A]}{\Delta t} \]
- Substituting the values we calculated:
\[ \text{Rate} = -\frac{-0.02 \, \text{M}}{60 \, \text{s}} = \frac{0.02 \, \text{M}}{60 \, \text{s}} = \frac{0.02}{60} \, \text{mol/L/s} \]

5. **Perform the Calculation**:
- Now, calculate the numerical value:
\[ \text{Rate} = \frac{0.02}{60} \approx 0.0003333 \, \text{mol/L/s} = 3.33 \times 10^{-4} \, \text{mol/L/s} \]

### Final Answer:
The rate of the reaction is approximately \( 3.33 \times 10^{-4} \, \text{mol/L/s} \).

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