Question

The heating value of combustible fuels is evaluated based on the quantities known as the higher heating value (HHV) and the lower heating value (LHV). The HHV has a higher absolute value and is obtained by assuming that the water formed in the combustion reaction is formed in the liquid state. The LHV has a lower absolute value and is obtained by assuming that the water formed in the combustion reaction is formed in the gaseous state. The LHV is therefore the sum of the HHV (which is negative) and the heat of vaporization of water for the number of moles of water formed in the reaction (which is positive). The table below lists the enthalpy of combustion (which is equivalent to the HHV) for several closely related hydrocarbons. Use the information in the table to answer the following questions.

Alkane	ΔHcombΔHcomb (kJ/molkJ/mol)
CH4(g)CH4(g)	−−890.
C2H6(g)C2H6(g)	−−1560
C3H8(g)C3H8(g)	−−2219
C4H10(g)C4H10(g)	−−2877
C5H12(l)C5H12(l)	−−3509
C6H14(l)C6H14(l)	−−4163
C7H16(l)C7H16(l)	−−4817
C8H18(l)C8H18(l)	−−5470

Part A

Given the provided data, complete and balance the combustion equation for the combustion of butane, CH3CH2CH2CH3CH3CH2CH2CH3, using whole-number coefficients for CO2CO2 and H2OH2O, assuming liquid water forms:

CH3CH2CH2CH3(g)+O2(g)→     ––– +      –––CH3CH2CH2CH3(g)+O2(g)→     _ +      _

Express your answer as a balanced chemical equation using the smallest whole-number coefficients including phases.

View Available Hint(s)

Part B

The following table lists heat of formation values for select compounds:

Compound	ΔH∘fΔHf∘ (kJ/molkJ/mol)		Compound	ΔH∘fΔHf∘ (kJ/molkJ/mol)
CO2(g)CO2(g)	−−393.5		C4H10(g)C4H10(g)	−−125.7
H2O(g)H2O(g)	−−241.8		C5H12(g)C5H12(g)	−−146.9
CH4(g)CH4(g)	−−74.6		C6H14(g)C6H14(g)	−−167.4
C2H6(g)C2H6(g)	−−84.68		C7H16(g)C7H16(g)	−−187.9
C3H8(g)C3H8(g)	−−103.85		C8H18(g)C8H18(g)	−−208.7

Part C

Calculate ΔHΔH for the combustion of one mole of butane gas, C4H10C4H10, assuming H2O(l)H2O(l) is produced instead of H2O(g)H2O(g).

Express the change in enthalpy in kilojoules to five significant figures.

Part D

Calculate ΔHΔH for the combustion of one mole of butane gas, C4H10C4H10, assuming H2O(g)H2O(g) is produced instead of H2O(l)H2O(l).

Express the change in enthalpy in kilojoules to five significant figures.

ΔHcombΔHcomb when H2O(g)H2O(g) is produced =	nothing	kJkJ

Answer 1

1-

The combustion reaction of butane is reaction of Butane with Oxygen-

CH3CH2CH2CH3(g) + 13/2 O2(g) → 4CO2(g) + 5H2O(l)

Or we can write it as-

C₄H_10(g) + 13/2 O_2(g) → 4CO_2(g) + 5H₂O_(l)

2-

ΔH for the combustion of one mole of butane gas is the difference between total heat of formation of products and the total heat of formation of reactants.

i.e for the reaction-

C₄H_10(g) + 13/2 O_2(g) → 4CO_2(g) + 5H₂O_(l)

ΔH_comb = ΣΔH^o_f (Prdocuts) - ΣΔH^o_f (Reactants)

= [ΔH^o_f (4CO_2(g)) + ΔH^o_f (5H₂O_(l))] - [ΔH^o_f (C₄H_10(g)) + ΔH^o_f (13/2 O_2(g))]

= [4 * ΔH^o_f (CO_2(g)) + 5 * ΔH^o_f (H₂O_(l))] - [ΔH^o_f (C₄H_10(g)) + 13/2 * ΔH^o_f (O_2(g))]

Now putting the values form a reference table,

ΔH_comb = [4 * ΔH^o_f (CO_2(g)) + 5 * ΔH^o_f (H₂O_(l))] - [ΔH^o_f (C₄H_10(g)) + 13/2 * ΔH^o_f (O_2(g))]

= [4 * (-393.5) + 5 * (-285.83)] - [(−125.7) + 13/2 * (0)]

= [ -1,574 - 1,429.15‬] - [−125.7) 0]

= -1,574 - 1,429.15‬ + 125.7

= -2877.5‬ kJ/mol

3- Similarly, if we form H2O in gaseous state, then

C₄H_10(g) + 13/2 O_2(g) → 4CO_2(g) + 5H₂O_(g)

ΔH_comb = ΣΔH^o_f (Prdocuts) - ΣΔH^o_f (Reactants)

= [ΔH^o_f (4CO_2(g)) + ΔH^o_f (5H₂O_(g))] - [ΔH^o_f (C₄H_10(g)) + ΔH^o_f (13/2 O_2(g))]

= [4 * ΔH^o_f (CO_2(g)) + 5 * ΔH^o_f (H₂O_(gl))] - [ΔH^o_f (C₄H_10(g)) + 13/2 * ΔH^o_f (O_2(g))]

Now putting the values form a reference table,

ΔH_comb = [4 * ΔH^o_f (CO_2(g)) + 5 * ΔH^o_f (H₂O_(l))] - [ΔH^o_f (C₄H_10(g)) + 13/2 * ΔH^o_f (O_2(g))]

= [4 * (-393.5) + 5 * (-241.8)] - [(−125.7) + 13/2 * (0)]

= [ -1,574 - 1,209‬] - [−125.7) 0]

= -1,574 - 1,209‬ + 125.7

= -2657.3‬‬ kJ/mol