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 find the average rate of the reaction given the change in concentration of \( H_2O_2 \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial and Final Concentrations**:
- Initial concentration of \( H_2O_2 \) = 0.10 moles/litre
- Final concentration of \( H_2O_2 \) = 0.08 moles/litre

2. **Calculate the Change in Concentration**:
- Change in concentration (\( \Delta [H_2O_2] \)) = Initial concentration - Final concentration
\[ \Delta [H_2O_2] = 0.10 - 0.08 = 0.02 \text{ moles/litre} \]

3. **Determine the Time Interval**:
- Time interval (\( \Delta t \)) = 20 minutes

4. **Calculate the Average Rate of Reaction**:
- The average rate of reaction can be calculated using the formula:
\[ \text{Average Rate} = -\frac{\Delta [H_2O_2]}{\Delta t} \]
- Substituting the values:
\[ \text{Average Rate} = -\frac{0.02 \text{ moles/litre}}{20 \text{ minutes}} = -0.001 \text{ moles/litre/minute} \]
- Since we are interested in the rate, we take the absolute value:
\[ \text{Average Rate} = 0.001 \text{ moles/litre/minute} = 1.0 \times 10^{-3} \text{ moles/litre/minute} \]

5. **Final Result**:
- The average rate of the reaction is \( 1.0 \times 10^{-3} \) moles/litre/minute.