Answer:
a) 2.541 mol/MJ;
b) 1.124 mol/MJ;
c) 0.4354 mol/MJ;
d) 0.1835 mol/MJ
Explanation:
The enthalpy of formation (ΔH°f) is the enthalpy of a reaction to form a compound by its constituents. For CO₂, ΔH°f = - 393.5 kJ/mol.
The enthalpy of a reaction is the sum of the enthalpy of the products (each one multiplied by the number of moles) less the sum of the enthalpy of the reactants (each one multiplied by the number of moles). The ΔH°f for simple substances (with one atom) is 0. The combustion is the reaction between the fuel and the oxygen.
a) The combution reaction is:
C(s) + O₂(g) → CO₂(g)
ΔH°rxn = -393.5 kJ/mol = -393.5x10⁻³ MJ/mol
Number of moles per MJ released: 1/|ΔH°rxn|
n = 1/(393.5x10⁻³) = 2.541 mol/MJ
b) The combustion reaction is:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
H₂O is in the liquid state because it's at 1 atm and 25ºC.
ΔH°f, H₂O(l) = -285.3 kJ/mol
ΔH°f, O₂(g) = 0
ΔH°f, CH₄(g) = -74.8 kJ/mol
ΔH°rxn = [2*(-285.3 ) + 1*(-393.5)] - [1*(-74.8)]
ΔH°rxn = -889.3 kJ/mol = -889.3x10⁻³ MJ/mol
n = 1/889.3x10⁻³ = 1.124 mol/MJ
c) C₃H₈(g) + 10O₂(g) → 3CO₂(g) + 4H₂O(l)
ΔH°f,C₃H₈(g) = -25.2 kJ/mol
ΔH°rxn = [4*(-285.3) + 3*(-393.5)] - [1*(-25.2)]
ΔH°rxn = -2,296.5 kJ/mol = -2.2965 MJ/mol
n = 1/2.2965 = 0.4354 mol/MJ
d) C₈H₁₈(l) + (25/2)O₂(g) → 8CO₂(g) + 9H₂O(l)
ΔH°f, C₈H₁₈(l) = -250.1 kJ/mol
ΔH°rxn = [9*(-283.5) + 8*(-393.5)] - [1*(-250.1)]
ΔH°rxn = -5,449.4 kJ/mol = -5.4494 MJ/mol
n = 1/5.4494 = 0.1835 mol/MJ
