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

Answer:  600 kJ

-

Explanation:

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

Δ⁰Hf   kJ/mol       -104             0                                       -393.5         -285.8

Δ⁰Hcomb C₃H₈ = 3(-393.5) + 4 (-285.80) - (-104)  kJ/mol

Δ⁰Hcomb = 2219.70 kJ/mol

n= m /MW   MW c₃H₈ = 44.1 g/mol

n= 12 g/44.1 g/mol = 0.27 mol

then for 12 g the heat released will be

0.27 mol x 2219.70 kJ/mol =  600 KJ