In a reaction, `2X to Y`, the concentration of X decreases from 0.50M to 0.38 M in 10 min. What is the rate of reaction in M `s^(-1)` during this interval?

Options

A. `2 xx 10^(-4)`
B. `4 xx 10^(-2)`
C. `2 xx 10^(-2)`
D. `1 xx 10^(-2)`

Explanation

To solve the problem, we need to calculate the rate of reaction for the given chemical equation \(2X \rightarrow Y\) based on the change in concentration of \(X\) over a specified time period.

### Step-by-Step Solution:

1. **Identify Initial and Final Concentrations:**
- Initial concentration of \(X\) (\([X]_0\)) = 0.50 M
- Final concentration of \(X\) (\([X]_t\)) = 0.38 M

2. **Calculate the Change in Concentration:**
- Change in concentration (\(\Delta [X]\)) = Final concentration - Initial concentration
\[ \Delta [X] = [X]_t - [X]_0 = 0.38 \, \text{M} - 0.50 \, \text{M} = -0.12 \, \text{M} \]

3. **Determine the Time Interval:**
- Time interval (\(\Delta t\)) = 10 minutes
- Convert time from minutes to seconds:
\[ \Delta t = 10 \, \text{minutes} \times 60 \, \text{seconds/minute} = 600 \, \text{seconds} \]

4. **Calculate the Rate of Reaction:**
- The rate of reaction is defined as the change in concentration of reactants (with a negative sign since the concentration is decreasing) divided by the change in time:
\[ \text{Rate} = -\frac{\Delta [X]}{\Delta t} = -\frac{-0.12 \, \text{M}}{600 \, \text{s}} = \frac{0.12 \, \text{M}}{600 \, \text{s}} = 2 \times 10^{-4} \, \text{M/s} \]

5. **Final Answer:**
- The rate of reaction during this interval is \(2 \times 10^{-4} \, \text{M/s}\).