Change in enthalpy for reaction `2H_(2)O_(2)(l)rarr2H_(2)O(l)+O_(2)(g)`
if heat of formation of `H_(2)O_(2)(l)` and `H_(2)O(l)` are `-188` and `-286KJ//mol` respectively is

Options

A. `-196` KJ/mol
B. `+196` KJ/mol
C. `+948` KJ/mol
D. `-948` KJ/mol

Explanation

To find the change in enthalpy (ΔH) for the reaction

\[ 2H_2O_2(l) \rightarrow 2H_2O(l) + O_2(g) \]

we will use the formula for the change in enthalpy of a reaction:

\[ \Delta H = \sum (\Delta H_f \text{ of products}) - \sum (\Delta H_f \text{ of reactants}) \]

### Step-by-Step Solution:

1. **Identify the heat of formation values:**
- Given:
- Heat of formation of \( H_2O_2(l) = -188 \, \text{kJ/mol} \)
- Heat of formation of \( H_2O(l) = -286 \, \text{kJ/mol} \)
- Heat of formation of \( O_2(g) = 0 \, \text{kJ/mol} \) (since it is in its standard state)

2. **Write the reaction and identify products and reactants:**
- Reaction:
\[ 2H_2O_2(l) \rightarrow 2H_2O(l) + O_2(g) \]
- Products: \( 2H_2O(l) \) and \( O_2(g) \)
- Reactants: \( 2H_2O_2(l) \)

3. **Calculate the total heat of formation for products:**
- For \( 2H_2O(l) \):
\[ \Delta H_f \text{ of products} = 2 \times (-286 \, \text{kJ/mol}) = -572 \, \text{kJ} \]
- For \( O_2(g) \):
\[ \Delta H_f \text{ of products} = 0 \, \text{kJ} \]
- Total heat of formation for products:
\[ \Delta H_f \text{ of products} = -572 + 0 = -572 \, \text{kJ} \]

4. **Calculate the total heat of formation for reactants:**
- For \( 2H_2O_2(l) \):
\[ \Delta H_f \text{ of reactants} = 2 \times (-188 \, \text{kJ/mol}) = -376 \, \text{kJ} \]

5. **Substitute values into the enthalpy change formula:**
\[ \Delta H = \Delta H_f \text{ of products} - \Delta H_f \text{ of reactants} \]
\[ \Delta H = (-572 \, \text{kJ}) - (-376 \, \text{kJ}) \]
\[ \Delta H = -572 + 376 = -196 \, \text{kJ} \]

### Final Answer:
The change in enthalpy for the reaction is:
\[ \Delta H = -196 \, \text{kJ} \]

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