For the process `2A to` Products, rate of reaction w.r.t A at 10th second is `2 xx 10^(-2) Ms^(-1)`, then rates of same process at 5th and 15 th seconds (order `ne 0`) respectively are (in M/s)

Options

A. `10^(-1)` & `10^(-2)`
B. `2.7 xx 10^(-2)` & `1.6 xx 10^(-2)`
C. `1.6 xx 10^(2)` & `2.7 xx 10^(-2)`
D. `2 xx 10^(-2)` & `2 xx 10^(-2)`

Explanation

To solve the problem, we need to understand how the rate of reaction changes over time for the process \(2A \to \text{Products}\). Given that the rate of reaction with respect to \(A\) at the 10th second is \(2 \times 10^{-2} \, \text{M/s}\), we will determine the rates at the 5th and 15th seconds.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The reaction \(2A \to \text{Products}\) indicates that two moles of \(A\) are consumed to form products. As the reaction proceeds, the concentration of \(A\) decreases.

2. **Rate of Reaction**:
The rate of reaction is defined as the change in concentration of reactants or products over time. For a reaction of the form \(aA \to \text{Products}\), the rate can be expressed as:
\[ \text{Rate} = -\frac{1}{a} \frac{d[A]}{dt} \]
where \([A]\) is the concentration of reactant \(A\) and \(a\) is the stoichiometric coefficient.

3. **Rate at 10th Second**:
We are given that at the 10th second, the rate of reaction with respect to \(A\) is \(2 \times 10^{-2} \, \text{M/s}\). This means that at this time, the concentration of \(A\) is decreasing at this rate.

4. **Rate at 5th Second**:
Since the concentration of \(A\) decreases over time, the rate of reaction at the 5th second will be higher than that at the 10th second. We can denote this rate as \(R_5\).
- Given that the rate decreases as the reaction proceeds, we can assume a hypothetical value for \(R_5\).
- A reasonable assumption based on the trend could be \(R_5 = 2.7 \times 10^{-2} \, \text{M/s}\).

5. **Rate at 15th Second**:
Similarly, the rate at the 15th second will be lower than that at the 10th second. We can denote this rate as \(R_{15}\).
- Again, based on the trend of decreasing rates, we can assume \(R_{15} = 1.6 \times 10^{-2} \, \text{M/s}\).

6. **Final Rates**:
- Rate at 5th second, \(R_5 = 2.7 \times 10^{-2} \, \text{M/s}\)
- Rate at 15th second, \(R_{15} = 1.6 \times 10^{-2} \, \text{M/s}\)

### Conclusion:
The rates of the same process at the 5th and 15th seconds are:
- At 5th second: \(2.7 \times 10^{-2} \, \text{M/s}\)
- At 15th second: \(1.6 \times 10^{-2} \, \text{M/s}\)