Many portable gas heaters and grills use propane, C3H8(g). Using enthalpies of formation, calculate the quantity of heat produced when 12.0 g of propane is completely combusted in air under standard conditions. Assume that liquid water is forming. Express the heat in kilojoules to three significant digits.

Considering the standard enthalpies of formation (ΔH°f) and the moles (n) in the balanced equation, we can calculate the standard enthalpy of the reaction (ΔH°) using the following expression.

ΔH° = 3 mol × ΔH°f(CO₂(g)) + 4 mol × ΔH°f(H₂O(l)) - 1 mol × ΔH°f(C₃H₈(g)) - 5 mol × ΔH°f(O₂(g))

ΔH° = 3 mol × (-393.5 kJ/mol) + 4 mol × (-285.8 kJ/mol) - 1 mol × (-103.8 kJ/mol) - 5 mol × 0

ΔH° = -2220 kJ

2220 kJ of heat are released per mole of propane. The heat released when 12.0 g of propane is combusted is: (molar mass 44.1 g/mol)

[tex]\frac{-2220kJ}{mol} .\frac{1mol}{44.1g}.12.0g =-604 kJ[/tex]

ajeigbeibraheem ajeigbeibraheem · Answer 2 · 2021-10-26T17:42:14+02:00

Many portable gas heaters and grills use propane, C3H8(g). Using enthalpies of formation, the quantity of heat produced when 12.0 g of propane react under combustion is -605.41 kJ

The balanced reaction of the combustion of propane in the presence of oxygen can be expressed as:

[tex]\mathbf{C_3H_{8(g)}+5O_{2(g)} \to 3CO_2{(g)} +4 H_2O_(l) }[/tex]

The standard enthalpy of formation is the standard change that occurs during the formation of a single mole substance under standard conditions.

At standard conditions, the standard enthalpy of formation of the substance participating in the reaction is as follows.

[tex]\mathbf{C_3H_{8(g)}= -103.85 \ kJ/mol} \\ \\ \mathbf{H_2O{(l)}= -285.8 \ kJ/mol} \\ \\ \mathbf{CO_{2(g)}= -393.5 \ kJ/mol}[/tex]

Now, using the formula for calculating the standard enthalpy of formation, we have:

[tex]\mathbf{\Delta \ H^0_{rxn} = \sum m \times \Delta_H^0_{f} (products) - \sum n \times \Delta_H^0_{f} (reactants)}[/tex]

[tex]\mathbf{\Delta H^0_{rxn} = 3 \times (-393.5 \ kJ/mol) +4 \times (-285.8 kJ/mol)- (-103.85 \ kJ/mol) }[/tex]

[tex]\mathbf{\Delta H^0_{rxn} = -2219.85 \ kJ/mol}[/tex]

Hence, one mole of propane on combustion results in 2219.85 kJ/mol of heat.

However,

If 1 mole of propane = 2219.85 kJ/mol

∴

In 12grams of propane, we will have:

[tex]\mathbf{q = \dfrac{mass \ of\ propane}{ molar mass of propane} \times -2219.85 \ kJ/mol }[/tex]

where;

q = quantity of heat produced in 12.0g of propane

[tex]\mathbf{q = \dfrac{12 \ g}{44 \ g/mol } \times -2219.85 \ kJ/mol }[/tex]

[tex]\mathbf{q = -605.41 kJ }[/tex]

Therefore, we can conclude that the quantity of heat produced when 12.0 g of propane react under combustion is -605.41 kJ

Learn more about enthalpies of formation here:

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