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

Answer

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 enthalpy values provided. The bond enthalpy values are as follows:

- Bond enthalpy of H-H = 438 kJ/mol
- Bond enthalpy of O=O = 498 kJ/mol
- Bond enthalpy of O-H = 464 kJ/mol
- Bond enthalpy of O-O = 138 kJ/mol

### Step-by-Step Solution:

1. **Identify the bonds broken and formed:**
- In the reactants, we have:
- 1 H-H bond in \( H_2 \)
- 1 O=O bond in \( O_2 \)
- In the product \( H_2O_2 \), we have:
- 2 O-H bonds (since \( H_2O_2 \) has two hydrogen atoms bonded to oxygen).

2. **Write the formula for ΔH:**
\[ \Delta H = \text{(Bond enthalpies of bonds broken)} - \text{(Bond enthalpies of bonds formed)} \]

3. **Calculate the total bond enthalpy of the reactants:**
- Total bond enthalpy of reactants = Bond enthalpy of H-H + Bond enthalpy of O=O
\[ = 438 \, \text{kJ/mol} + 498 \, \text{kJ/mol} = 936 \, \text{kJ/mol} \]

4. **Calculate the total bond enthalpy of the products:**
- Total bond enthalpy of products = 2 × Bond enthalpy of O-H
\[ = 2 \times 464 \, \text{kJ/mol} = 928 \, \text{kJ/mol} \]

5. **Substituting the values into the ΔH equation:**
\[ \Delta H = \text{(936 kJ/mol)} - \text{(928 kJ/mol)} = 936 \, \text{kJ/mol} - 928 \, \text{kJ/mol} \]
\[ = 8 \, \text{kJ/mol} \]

6. **Adjust for the O-O bond:**
- Since we also have an O-O bond in the product, we need to subtract the bond enthalpy of the O-O bond:
\[ \Delta H = 8 \, \text{kJ/mol} - 138 \, \text{kJ/mol} = -130 \, \text{kJ/mol} \]

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