The standard enthalpy of combustion of butane is -10.93 kcal per gram of butane when the water is
produced in vapor form.
a) Convert this value into units of kJ per mole of butane.
b) Use this reaction enthalpy to determine the standard enthalpy of formation (Hr) of butane.
The Correct Answer and Explanation is :
To solve this problem, we’ll break it into two parts as requested.
a) Convert the enthalpy of combustion of butane to kJ per mole.
Given:
- The standard enthalpy of combustion of butane (C₄H₁₀) is -10.93 kcal per gram of butane.
- The molar mass of butane is 58.12 g/mol (12.01 g/mol for carbon and 1.008 g/mol for hydrogen).
Step 1: Convert kcal to kJ.
1 kcal = 4.184 kJ, so:
[
-10.93 \, \text{kcal/g} \times 4.184 = -45.74 \, \text{kJ/g}
]
Step 2: Convert the value from kJ per gram to kJ per mole.
Since the molar mass of butane is 58.12 g/mol:
[
-45.74 \, \text{kJ/g} \times 58.12 \, \text{g/mol} = -2654.46 \, \text{kJ/mol}
]
Thus, the standard enthalpy of combustion of butane is approximately -2654.46 kJ/mol.
b) Use this reaction enthalpy to determine the standard enthalpy of formation (ΔHf°) of butane.
The standard enthalpy of combustion is the heat released when one mole of a substance undergoes complete combustion in excess oxygen to form CO₂ and H₂O. The reaction is:
[
\text{C₄H₁₀ (g)} + 13/2 \, \text{O₂ (g)} \rightarrow 4 \, \text{CO₂ (g)} + 5 \, \text{H₂O (g)}
]
The standard enthalpy of combustion is related to the standard enthalpies of formation (ΔHf°) of the reactants and products:
[
\Delta H_{\text{combustion}} = \sum \Delta H_f^\circ \, \text{(products)} – \sum \Delta H_f^\circ \, \text{(reactants)}
]
For the combustion of butane:
[
\Delta H_{\text{combustion}} = \left[4\Delta H_f^\circ (\text{CO₂}) + 5\Delta H_f^\circ (\text{H₂O (g)}) \right] – \left[\Delta H_f^\circ (\text{C₄H₁₀}) + \frac{13}{2} \Delta H_f^\circ (\text{O₂}) \right]
]
Using standard values:
- ΔHf° (CO₂) = -393.5 kJ/mol
- ΔHf° (H₂O (g)) = -241.8 kJ/mol
- ΔHf° (O₂) = 0 kJ/mol (since it is an elemental form)
Thus:
[
\Delta H_{\text{combustion}} = \left[4(-393.5) + 5(-241.8)\right] – \left[\Delta H_f^\circ (\text{C₄H₁₀}) + 0 \right]
]
[
-2654.46 = [-1574 + (-1209)] – \Delta H_f^\circ (\text{C₄H₁₀})
]
[
-2654.46 = -2783 – \Delta H_f^\circ (\text{C₄H₁₀})
]
[
\Delta H_f^\circ (\text{C₄H₁₀}) = -2783 + 2654.46 = -128.54 \, \text{kJ/mol}
]
Thus, the standard enthalpy of formation of butane is approximately -128.54 kJ/mol.
Image of the combustion reaction and calculation:
I will generate an image to visually represent this calculation.
Here is the image representing the combustion reaction of butane along with the standard enthalpy of combustion and formation calculations. Let me know if you need any further clarification!
