Calculate the standard enthalpy change (in kJ `"mol"^(-1)`) for the reaction `H_(2)(g)+O_(2)(g)toH_(2)O_(2)(g)`, given that bond enthalpy of H-H, O=O,O-H and O-O (in kJ `"mol"^(-1)`) are respectively 438, 498, 464 and 138.

Options

A. `130`
B. `-65`
C. `+130`
D. `-334`

Explanation

To calculate the standard enthalpy change (ΔH) for the reaction \( H_2(g) + O_2(g) \rightarrow H_2O_2(g) \), we will use the bond enthalpies provided. The bond enthalpies are as follows:

- H-H bond: 438 kJ/mol
- O=O bond: 498 kJ/mol
- O-H bond: 464 kJ/mol
- O-O bond: 138 kJ/mol

### Step-by-Step Solution:

1. **Identify Bonds in Reactants**:
- In the reactants \( H_2 \) and \( O_2 \):
- \( H_2 \) has 1 H-H bond.
- \( O_2 \) has 1 O=O bond.

The total bond enthalpy for the reactants is:
\[ \text{Total bond enthalpy of reactants} = \text{Bond enthalpy of H-H} + \text{Bond enthalpy of O=O} \]
\[ = 438 \, \text{kJ/mol} + 498 \, \text{kJ/mol} = 936 \, \text{kJ/mol} \]

2. **Identify Bonds in Products**:
- In the product \( H_2O_2 \):
- There are 2 O-H bonds.
- There is 1 O-O bond.

The total bond enthalpy for the products is:
\[ \text{Total bond enthalpy of products} = 2 \times \text{Bond enthalpy of O-H} + \text{Bond enthalpy of O-O} \]
\[ = 2 \times 464 \, \text{kJ/mol} + 138 \, \text{kJ/mol} = 928 \, \text{kJ/mol} + 138 \, \text{kJ/mol} = 1066 \, \text{kJ/mol} \]

3. **Calculate the Enthalpy Change (ΔH)**:
The enthalpy change for the reaction is given by the formula:
\[ \Delta H = \text{Total bond enthalpy of reactants} - \text{Total bond enthalpy of products} \]
\[ = 936 \, \text{kJ/mol} - 1066 \, \text{kJ/mol} = -130 \, \text{kJ/mol} \]

### Final Answer:
The standard enthalpy change (ΔH) for the reaction \( H_2(g) + O_2(g) \rightarrow H_2O_2(g) \) is \(-130 \, \text{kJ/mol}\).

---