Concentration of a reactant ‘A’ is changed from 0.044M to 0.032M in 25 minutes, the average rate of the reaction during this interval is

Options

A. `0.0048 "mol"."lit"^(-1)."min"^(-1)`
B. `0.00048 mol.lit^(-1) sec^(-1)`
C. `4.8 xx 10^(-4) mol.lit^(-1).min^(-1)`
D. `0.0048 mol.lit^(-1)sec^(-1)`

Explanation

To find the average rate of the reaction when the concentration of reactant 'A' changes from 0.044 M to 0.032 M over a period of 25 minutes, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial and Final Concentrations**:
- Initial concentration of A, [A]₀ = 0.044 M
- Final concentration of A, [A]₁ = 0.032 M

2. **Calculate the Change in Concentration**:
- Change in concentration, Δ[A] = [A]₀ - [A]₁
- Δ[A] = 0.044 M - 0.032 M
- Δ[A] = 0.012 M

3. **Determine the Time Interval**:
- Time interval, Δt = 25 minutes

4. **Convert Time to Appropriate Units**:
- Since we want the rate in terms of minutes, we can keep it as is. However, if we need to convert it to seconds, we would multiply by 60 (1 minute = 60 seconds).
- For this problem, we will keep it in minutes.

5. **Calculate the Average Rate of Reaction**:
- Average rate = Change in concentration / Time interval
- Average rate = Δ[A] / Δt
- Average rate = 0.012 M / 25 min
- Average rate = 0.00048 M/min

6. **Express in Scientific Notation**:
- 0.00048 M/min can be written as 4.8 × 10⁻⁴ M/min.

### Final Answer:
The average rate of the reaction is **4.8 × 10⁻⁴ M/min**.

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