If `DeltaH_(f)^(@)` for `H_(2)O_(2)` and `H_(2)O` are `-188` kJ/mole and `-286` kJ/mole, what will be the enthalpy change of the reaction

Options

A. `-196` kJ/mole
B. `-494` kJ/mole
C. `-146` kJ/mole
D. `-98` kJ/mole

Explanation

To find the enthalpy change of the reaction where hydrogen peroxide (H₂O₂) decomposes to form water (H₂O) and oxygen (O₂), we can use the standard enthalpy of formation values provided for H₂O₂ and H₂O.

### Step-by-Step Solution:

1. **Write the Formation Reactions**:
- The formation reaction for hydrogen peroxide (H₂O₂) is:
\[ H_2 + \frac{1}{2} O_2 \rightarrow H_2O_2 \quad \Delta H_f^\circ = -188 \text{ kJ/mol} \]
- The formation reaction for water (H₂O) is:
\[ H_2 + \frac{1}{2} O_2 \rightarrow H_2O \quad \Delta H_f^\circ = -286 \text{ kJ/mol} \]

2. **Reverse the Reaction for H₂O₂**:
- To find the enthalpy change for the decomposition of H₂O₂, we need to reverse the first reaction:
\[ H_2O_2 \rightarrow H_2 + \frac{1}{2} O_2 \quad \Delta H = +188 \text{ kJ/mol} \]

3. **Combine the Reactions**:
- Now, we can add the reversed reaction for H₂O₂ to the formation reaction for H₂O:
\[ H_2O_2 \rightarrow H_2 + \frac{1}{2} O_2 \quad \Delta H = +188 \text{ kJ/mol} \]
\[ H_2 + \frac{1}{2} O_2 \rightarrow H_2O \quad \Delta H = -286 \text{ kJ/mol} \]
- Adding these two reactions together:
\[ H_2O_2 + H_2 + \frac{1}{2} O_2 \rightarrow H_2 + \frac{1}{2} O_2 + H_2O \]

4. **Cancel Out Common Terms**:
- The H₂ on both sides cancels out, and we are left with:
\[ H_2O_2 \rightarrow H_2O + \frac{1}{2} O_2 \]

5. **Calculate the Overall Enthalpy Change**:
- The overall enthalpy change (ΔH) for the reaction is the sum of the enthalpy changes from the two reactions:
\[ \Delta H = +188 \text{ kJ/mol} - 286 \text{ kJ/mol} = -98 \text{ kJ/mol} \]

### Final Answer:
The enthalpy change of the reaction is:
\[ \Delta H = -98 \text{ kJ/mol} \]