Heat of formation of `H_(2)O` is -188kJ/mol and `H_(2)O_(2)` is -286 kJ/mol. The enthalpy change for the reaction,
`2H_(2)O_(2) to 2H_(2)O+O_(2)`

Options

A. `196 kJ`
B. `-196kJ`
C. `984kJ`
D. `-984kJ`

Explanation

To find the enthalpy change for the reaction:

\[ 2H_2O_2 \rightarrow 2H_2O + O_2 \]

we can use the standard enthalpy of formation values given for \( H_2O \) and \( H_2O_2 \).

### Step-by-Step Solution:

1. **Identify the Enthalpy of Formation Values**:
- The enthalpy of formation of \( H_2O \) is given as \( \Delta H_f^\circ (H_2O) = -188 \, \text{kJ/mol} \).
- The enthalpy of formation of \( H_2O_2 \) is given as \( \Delta H_f^\circ (H_2O_2) = -286 \, \text{kJ/mol} \).

2. **Write the Reaction**:
The balanced reaction is:
\[ 2H_2O_2 \rightarrow 2H_2O + O_2 \]

3. **Apply the Enthalpy Change Formula**:
The enthalpy change for the reaction can be calculated using:
\[ \Delta H_{reaction} = \sum \Delta H_f^\circ (\text{products}) - \sum \Delta H_f^\circ (\text{reactants}) \]

4. **Calculate the Enthalpy of Products**:
- For \( 2H_2O \):
\[ \Delta H_f^\circ (2H_2O) = 2 \times (-188) = -376 \, \text{kJ} \]
- For \( O_2 \):
\[ \Delta H_f^\circ (O_2) = 0 \, \text{kJ} \quad (\text{since it's in its elemental form}) \]
- Total for products:
\[ \Delta H_f^\circ (\text{products}) = -376 + 0 = -376 \, \text{kJ} \]

5. **Calculate the Enthalpy of Reactants**:
- For \( 2H_2O_2 \):
\[ \Delta H_f^\circ (2H_2O_2) = 2 \times (-286) = -572 \, \text{kJ} \]

6. **Substitute into the Enthalpy Change Formula**:
\[ \Delta H_{reaction} = (-376) - (-572) \]
\[ \Delta H_{reaction} = -376 + 572 = 196 \, \text{kJ} \]

7. **Final Result**:
The enthalpy change for the reaction \( 2H_2O_2 \rightarrow 2H_2O + O_2 \) is:
\[ \Delta H_{reaction} = +196 \, \text{kJ} \]