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 `H_(2)O_(2) rarr H_(2)O+1/2 O_(2)`

Options

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

Answer

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

\[ H_2O_2 \rightarrow H_2O + \frac{1}{2} O_2 \]

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

### Step 1: Write the formula for ΔH of the reaction
The enthalpy change for a reaction can be calculated using the formula:

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

### Step 2: Identify the products and reactants
In our reaction:

- Reactant: \( H_2O_2 \)
- Products: \( H_2O \) and \( \frac{1}{2} O_2 \)

### Step 3: Write down the ΔH_f values
From the question, we have:

- \( \Delta H_f (H_2O_2) = -188 \, \text{kJ/mol} \)
- \( \Delta H_f (H_2O) = -286 \, \text{kJ/mol} \)
- \( \Delta H_f (O_2) = 0 \, \text{kJ/mol} \) (since it is in its elemental form)

### Step 4: Calculate the ΔH for the products
For the products, we have:

\[ \Delta H_f \text{(products)} = \Delta H_f (H_2O) + \Delta H_f \left(\frac{1}{2} O_2\right) \]

Calculating this gives:

\[ \Delta H_f \text{(products)} = -286 \, \text{kJ/mol} + 0 = -286 \, \text{kJ/mol} \]

### Step 5: Calculate the ΔH for the reactants
For the reactants, we have:

\[ \Delta H_f \text{(reactants)} = \Delta H_f (H_2O_2) = -188 \, \text{kJ/mol} \]

### Step 6: Substitute the values into the ΔH formula
Now we can substitute the values into the ΔH formula:

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

\[ \Delta H = (-286 \, \text{kJ/mol}) - (-188 \, \text{kJ/mol}) \]

### Step 7: Simplify the expression
This simplifies to:

\[ \Delta H = -286 + 188 = -98 \, \text{kJ/mol} \]

### Final Answer
Thus, the enthalpy change for the reaction \( H_2O_2 \rightarrow H_2O + \frac{1}{2} O_2 \) is:

\[ \Delta H = -98 \, \text{kJ/mol} \]