For the reaction : `H_2O_2 + 2I^– + 2H^+ to I_2 + 2H_2O`. If the concentration of `H_2O_2` reduces from 0.10 to 0.08 (moles/ litre) in 20 minutes, what is average rate of reaction?

Options

A. `10^-3 mol.litre^-1 minute^-1`
B. `10^-2 mol.litre^-1 minute^-1`
C. `10^-1 mol.litre^-1 minute^-1`
D. `10^-4 mol.litre^-1 minute^-1`

Answer

To solve the problem, we need to find the average rate of the reaction given the change in concentration of hydrogen peroxide (H₂O₂) over a specific time period.

### Step-by-Step Solution:

1. **Identify the Initial and Final Concentrations**:
- Initial concentration of H₂O₂ = 0.10 moles/litre
- Final concentration of H₂O₂ = 0.08 moles/litre

2. **Calculate the Change in Concentration**:
- Change in concentration (Δ[H₂O₂]) = Final concentration - Initial concentration
- Δ[H₂O₂] = 0.08 - 0.10 = -0.02 moles/litre

3. **Determine the Time Interval**:
- Time interval (Δt) = 20 minutes

4. **Calculate the Average Rate of Reaction**:
- The average rate of reaction is defined as:
\[ \text{Average Rate} = -\frac{\Delta[H₂O₂]}{\Delta t} \]
- Substituting the values:
\[ \text{Average Rate} = -\frac{-0.02 \text{ moles/litre}}{20 \text{ minutes}} = \frac{0.02}{20} = 0.001 \text{ moles/litre/minute} \]

5. **Express the Rate in Scientific Notation**:
- 0.001 moles/litre/minute can be expressed as \(1 \times 10^{-3}\) moles/litre/minute.

### Final Answer:
The average rate of the reaction is \(1 \times 10^{-3}\) moles/litre/minute.

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