The standard enthalpies of formation of `H_(2)O_(2(l)) and H_(2)O_((l))` are `-187.8kJ "mole"^(-1) and -285.8 kJ "mole"^(-1)` respectively. The `DeltaH^(0)` for the decomposition of one mole of `H_(2)O_(2(l)) " to" H_(2)O_((l)) and O_(2(g))` is

Options

A. `-473.6kJ "mol^(-1)`
B. `-98.9kJ "mol^(-1)`
C. `+473.6 kJ "mol^(-1)`
D. `+187.8 kJ "mol^(-1)`

Answer

To solve the problem, we need to calculate the standard enthalpy change (ΔH°) for the decomposition of one mole of hydrogen peroxide (H₂O₂) into water (H₂O) and oxygen (O₂). The reaction can be represented as:

\[ 2 \text{H}_2\text{O}_2 (l) \rightarrow 2 \text{H}_2\text{O} (l) + \text{O}_2 (g) \]

### Step-by-Step Solution:

1. **Identify the Reaction**:
The decomposition reaction is:
\[ 2 \text{H}_2\text{O}_2 (l) \rightarrow 2 \text{H}_2\text{O} (l) + \text{O}_2 (g) \]

2. **Write the Standard Enthalpy of Formation Values**:
- For H₂O₂ (l): ΔH°f = -187.8 kJ/mol
- For H₂O (l): ΔH°f = -285.8 kJ/mol
- For O₂ (g): ΔH°f = 0 kJ/mol (as it is in its standard state)

3. **Calculate the Total Enthalpy of Reactants and Products**:
- For 2 moles of H₂O₂:
\[ \text{Total ΔH°f for reactants} = 2 \times (-187.8 \text{ kJ/mol}) = -375.6 \text{ kJ} \]
- For 2 moles of H₂O:
\[ \text{Total ΔH°f for products} = 2 \times (-285.8 \text{ kJ/mol}) = -571.6 \text{ kJ} \]
- For 1 mole of O₂:
\[ \text{Total ΔH°f for products} = 0 \text{ kJ} \]

4. **Apply Hess's Law**:
According to Hess's law, the standard enthalpy change for the reaction can be calculated as:
\[ \Delta H° = \text{(Total ΔH°f of products)} - \text{(Total ΔH°f of reactants)} \]
Substituting the values:
\[ \Delta H° = \left(-571.6 \text{ kJ}\right) - \left(-375.6 \text{ kJ}\right) \]
\[ \Delta H° = -571.6 + 375.6 = -196.0 \text{ kJ} \]

5. **Final Adjustment for One Mole**:
Since the reaction we calculated is for 2 moles of H₂O₂, we need to divide the ΔH° by 2 to find the enthalpy change for the decomposition of one mole of H₂O₂:
\[ \Delta H° = \frac{-196.0 \text{ kJ}}{2} = -98.0 \text{ kJ} \]

### Conclusion:
The standard enthalpy change (ΔH°) for the decomposition of one mole of H₂O₂ into H₂O and O₂ is:
\[ \Delta H° = -98.0 \text{ kJ/mol} \]