If `x_(1)` , `x_(2)` and `x_(3)` are enthalpies of `H-H` , `O=O` and `O-H` bonds respective, and `x_(4)` is the enthaply of vaporisation of water, estimate the standard enthalpy opf combustion of bydrogen.

Options

A. `x_(1)+(x_(2))/(2)-2x_(3)+x_(4)`
B. `x_(1)+(x_(2))/(2)-2x_(3)-x_(4)`
C. `x_(1)+(x_(2))/(2)-x_(3)-x_(4)`
D. `2x_(3)-x_(1) -(x_(2))/(2)-x_(4)`

Explanation

To estimate the standard enthalpy of combustion of hydrogen, we will follow these steps:

### Step 1: Write the balanced chemical equation for the combustion of hydrogen.
The combustion of hydrogen can be represented by the following reaction:
\[ \text{H}_2(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{H}_2O(l) \]

### Step 2: Identify the bond enthalpies involved.
- Let \( x_1 \) be the bond enthalpy of the H-H bond.
- Let \( x_2 \) be the bond enthalpy of the O=O bond.
- Let \( x_3 \) be the bond enthalpy of the O-H bond.
- Let \( x_4 \) be the enthalpy of vaporization of water.

### Step 3: Apply the bond enthalpy concept.
According to the bond enthalpy concept, the standard enthalpy change (\( \Delta H \)) for the combustion of hydrogen can be calculated using the formula:
\[ \Delta H = \text{(Sum of bond enthalpies of reactants)} - \text{(Sum of bond enthalpies of products)} \]

### Step 4: Calculate the bond enthalpies for the reactants.
For the reactants, we have:
- 1 H-H bond: \( x_1 \)
- 1/2 O=O bond: \( \frac{1}{2} x_2 \)

So, the total bond enthalpy for the reactants is:
\[ \text{Total bond enthalpy of reactants} = x_1 + \frac{1}{2} x_2 \]

### Step 5: Calculate the bond enthalpies for the products.
For the products, we have:
- 2 O-H bonds in water: \( 2 x_3 \)

### Step 6: Include the enthalpy of vaporization of water.
Since the product is in the liquid state, we need to account for the enthalpy of vaporization of water (\( x_4 \)):
\[ \text{Total bond enthalpy of products} = 2 x_3 + x_4 \]

### Step 7: Combine the equations to find \( \Delta H \).
Now, substituting the values into the equation for \( \Delta H \):
\[ \Delta H = \left( x_1 + \frac{1}{2} x_2 \right) - \left( 2 x_3 + x_4 \right) \]
This simplifies to:
\[ \Delta H = x_1 + \frac{1}{2} x_2 - 2 x_3 - x_4 \]

### Conclusion
Thus, the standard enthalpy of combustion of hydrogen is given by:
\[ \Delta H = x_1 + \frac{1}{2} x_2 - 2 x_3 - x_4 \]