Question

Pentane (C5H12) is a component of gasoline that burns according to the following balanced equation:
C5H12(l)+8O2(g)→5CO2(g)+6H2O(g)
Calculate ΔH∘rxn for this reaction using standard enthalpies of formation. (The standard enthalpy of formation of liquid pentane is -146.8 kJ/mol.)

Answer 1

Enthalpy change of Pentane (C5H12)  that burns is -3,270.5  kJ

Enthalpy change is defined as the difference in enthalpy of total products and reactants, each multiplied by the respective moles. It is expressed as ΔH°rxn

The formula for calculating the enthalpy change of a reaction is:

ΔH°rxn =∑[n x ΔH°f(product) - ∑[n x ΔH°f(reactant)]

The equilibrium reactions are:

C₅H₁₂(l) + 8O₂(g) ⇒ 5CO₂(g) +6H₂O(g)

ΔH°f(CO₂)(g) = -393.5 kJ/mol

ΔH°f(H₂O)(g) = - 241.8 kJ/mol

ΔH°f(C₅H₁₂)(l) = -146.8 kJ/mol

ΔH°f(O₂)(g) = 0 kJ/mol

The equation for the enthalpy change for the above reaction is:

ΔH°rxn =∑[n x ΔH°f(product) - ∑[n x ΔH°f(reactant)]

ΔH°rxn =∑[(n(CO₂) x ΔH°f(CO₂)) + (n(H₂O) x ΔH°f(H₂O)) ]- ∑[(n x ΔH°f(C₅H₁₂)) + (n x ΔH°f(O₂))]

ΔH°rxn =[( 5 x -393.5) + (6 x - 241.8 ) ]- [(1 x -147.8) + (8 x O₂)]

ΔH°rxn = -3,418.3 -(-147.8)

ΔH°rxn = -3,270.5  kJ

Therefore, the enthalpy of burning 1 mole of pentane is -3,270.5  kJ

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