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. `-494 kJ`
C. `146 kJ`
D. `-98 kJ`

Answer

To find the enthalpy change for the reaction:

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

we will use the given heats of formation for water (\(H_2O\)) and hydrogen peroxide (\(H_2O_2\)).

### Step 1: Identify the enthalpy of formation values
- The heat of formation of \(H_2O\) is \(-188 \, \text{kJ/mol}\).
- The heat of formation of \(H_2O_2\) is \(-286 \, \text{kJ/mol}\).

### Step 2: Write the enthalpy change formula
The enthalpy change (\(\Delta H\)) for the reaction can be calculated using the formula:

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

### Step 3: Calculate the total enthalpy for reactants
For the reactants, we have 2 moles of \(H_2O_2\):

\[ \text{Enthalpy of reactants} = 2 \times \text{(Enthalpy of } H_2O_2) = 2 \times (-286 \, \text{kJ/mol}) = -572 \, \text{kJ} \]

### Step 4: Calculate the total enthalpy for products
For the products, we have 2 moles of \(H_2O\) and 1 mole of \(O_2\). The heat of formation of \(O_2\) is 0 (as it is in its elemental form):

\[ \text{Enthalpy of products} = 2 \times \text{(Enthalpy of } H_2O) + 1 \times \text{(Enthalpy of } O_2) = 2 \times (-188 \, \text{kJ/mol}) + 0 = -376 \, \text{kJ} \]

### Step 5: Substitute values into the enthalpy change formula
Now we can substitute the values into the enthalpy change formula:

\[ \Delta H = \text{Enthalpy of products} - \text{Enthalpy of reactants} \]
\[ \Delta H = (-376 \, \text{kJ}) - (-572 \, \text{kJ}) \]
\[ \Delta H = -376 + 572 = 196 \, \text{kJ} \]

### Step 6: Finalize the answer
Since the reaction is exothermic, we represent the enthalpy change as:

\[ \Delta H = -196 \, \text{kJ} \]

Thus, the enthalpy change for the reaction \(2H_2O_2 \rightarrow 2H_2O + O_2\) is \(-196 \, \text{kJ}\).

### Final Answer:
\(-196 \, \text{kJ}\)

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