If the enthalpies of formation of `C_3H_8(g), CO_2(g) and H_2O(l)` at standard conditions are `-104 kJ mol^-1, -394 kJ mol^-1 and -286 kJ mol^-1` respectively then the standard anthalpy of combustion of 66 g of `C_3H_8(g)` will be

Options

A. `- 2222 kJ`
B. ` - 2500 kJ`
C. `- 784 kJ`
D. `- 3333 kJ`

Answer

To find the standard enthalpy of combustion of 66 g of \( C_3H_8(g) \), we will follow these steps:

### Step 1: Calculate the number of moles of \( C_3H_8 \)
The molar mass of \( C_3H_8 \) (propane) can be calculated as follows:
- Carbon (C): 12.01 g/mol × 3 = 36.03 g/mol
- Hydrogen (H): 1.008 g/mol × 8 = 8.064 g/mol
- Total molar mass of \( C_3H_8 \) = 36.03 g/mol + 8.064 g/mol = 44.094 g/mol

Now, we can calculate the number of moles of \( C_3H_8 \) in 66 g:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{66 \text{ g}}{44.094 \text{ g/mol}} \approx 1.496 \text{ mol} \]

### Step 2: Write the balanced combustion reaction for \( C_3H_8 \)
The balanced equation for the combustion of propane is:
\[ C_3H_8(g) + 5 O_2(g) \rightarrow 3 CO_2(g) + 4 H_2O(l) \]

### Step 3: Calculate the enthalpy change for the combustion reaction
Using the enthalpies of formation provided:
- \( \Delta H_f \) of \( C_3H_8(g) = -104 \text{ kJ/mol} \)
- \( \Delta H_f \) of \( CO_2(g) = -394 \text{ kJ/mol} \)
- \( \Delta H_f \) of \( H_2O(l) = -286 \text{ kJ/mol} \)

The standard enthalpy change of combustion (\( \Delta H_{comb} \)) can be calculated using the formula:
\[ \Delta H_{comb} = \sum \Delta H_f \text{(products)} - \sum \Delta H_f \text{(reactants)} \]
For the products:
- For \( 3 \, CO_2 \): \( 3 \times (-394) = -1182 \text{ kJ} \)
- For \( 4 \, H_2O \): \( 4 \times (-286) = -1144 \text{ kJ} \)

Total for products:
\[ -1182 \text{ kJ} + (-1144 \text{ kJ}) = -2326 \text{ kJ} \]

For the reactants:
- For \( C_3H_8 \): \( -104 \text{ kJ} \)
- For \( 5 \, O_2 \): \( 0 \text{ kJ} \) (since elemental oxygen has an enthalpy of formation of 0)

Total for reactants:
\[ -104 \text{ kJ} + 0 = -104 \text{ kJ} \]

Now, substituting into the enthalpy change formula:
\[ \Delta H_{comb} = -2326 \text{ kJ} - (-104 \text{ kJ}) = -2326 + 104 = -2222 \text{ kJ} \]

### Step 4: Calculate the total enthalpy of combustion for 66 g of \( C_3H_8 \)
Now, we need to find the enthalpy of combustion for the number of moles calculated:
\[ \Delta H_{comb} \text{ per mole of } C_3H_8 = -2222 \text{ kJ/mol} \]
Total for 1.496 moles:
\[ \Delta H_{total} = 1.496 \text{ mol} \times (-2222 \text{ kJ/mol}) \approx -3325.5 \text{ kJ} \]

### Final Answer
The standard enthalpy of combustion of 66 g of \( C_3H_8(g) \) is approximately **-3325.5 kJ**.

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