Question

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.

Answer 1

Answer:

-604 kJ

Explanation:

Let's consider the complete combustion of propane.

C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(l)

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)

$\frac{-2220kJ}{mol} .\frac{1mol}{44.1g}.12.0g =-604 kJ$

Answer 2

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:

$\mathbf{C_3H_{8(g)}+5O_{2(g)} \to 3CO_2{(g)} +4 H_2O_(l) }$

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.

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

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

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

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

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

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:

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

where;

q = quantity of heat produced in 12.0g of propane

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

$\mathbf{q = -605.41 kJ }$

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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