The enthalpy of combustion of propance `(C_(3)H_(8))` gas in temes of given of geven data is , Bond energy (kJ/mol)
`underset(+x_(1))(.^(epsi)C-H)underset(+x_(2))(.^(epsi)O-O)underset(+x_(3))(.^(epsi)C-O)underset(+x_(4))(.^(epsi)O-H)underset(+x_(5))(.^(epsi)C-C)`
[Resonance energy of `CO_(2)` is -z KJ/mol and `DeltaH_("vaporization")[H_(2)O(l)" is y " KJ//mol]`

Options

A. `8x_(1)+2x_(5)+5x_(2)-6x_(3)-8x_(4)-4y-3z`
B. `6x_(1)+x_(5)+5x_(2)-3x_(3)-4x_(4)-4y-3z`
C. `8x_(1)+2x_(5)+5x_(2)-6x_(3)-8x_(4)-y-z`
D. `8x_(1)+x_(5)+5x_(2)-6x_(3)-8x_(4)-4y+3z`

Explanation

To find the enthalpy of combustion of propane (C₃H₈), we will use the bond energies provided and apply Hess's law. The combustion reaction of propane can be represented as follows:

### Step 1: Write the balanced combustion reaction
The balanced equation for the combustion of propane is:
\[ C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O \]

### Step 2: Identify bonds broken in the reactants
In the reactants, we have:
- 3 Carbon-Carbon (C-C) bonds
- 8 Carbon-Hydrogen (C-H) bonds
- 5 Oxygen-Oxygen (O=O) bonds

The total energy required to break these bonds can be calculated as follows:
- Energy to break C-C bonds: \( 2 \times X_5 \) (since there are 2 C-C bonds in propane)
- Energy to break C-H bonds: \( 8 \times X_1 \)
- Energy to break O=O bonds: \( 5 \times X_2 \)

### Step 3: Identify bonds formed in the products
In the products, we have:
- 6 Carbon-Oxygen (C=O) bonds from 3 CO₂ (2 bonds per CO₂)
- 8 Oxygen-Hydrogen (O-H) bonds from 4 H₂O

The total energy released when these bonds are formed is:
- Energy released from C=O bonds: \( 6 \times X_3 \)
- Energy released from O-H bonds: \( 8 \times X_4 \)

### Step 4: Consider resonance energy and enthalpy of vaporization
- The resonance energy of CO₂ is given as \(-Z\) kJ/mol, which means when we form CO₂, we need to account for this energy.
- The enthalpy of vaporization of water is given as \(Y\) kJ/mol. Since we have 4 moles of water, we need to account for \(4Y\).

### Step 5: Apply Hess's Law
According to Hess's law, the enthalpy change for the combustion of propane can be expressed as:
\[ \Delta H_{combustion} = \text{Energy of bonds broken} - \text{Energy of bonds formed} \]

Substituting the expressions from steps 2 and 3, we get:
\[ \Delta H_{combustion} = (8X_1 + 2X_5 + 5X_2) - (6X_3 + 8X_4) + Z + 4Y \]

### Step 6: Final expression for the enthalpy of combustion
Thus, the final expression for the enthalpy of combustion of propane is:
\[ \Delta H_{combustion} = 8X_1 + 2X_5 + 5X_2 - 6X_3 - 8X_4 + Z + 4Y \]