Question

2C2H2 + 5 O2 → 4CO2 + 2H2O

Use the given standard enthalpies of formation to calculate ∆H for this reaction

C2H­2 = 227.4

CO2 = -393.5

H2O = -241.8

1256.2 kJ b. -1256.2 kJ c. -2512.4 kJ d. 2512.4 kJ

Answer 1

• Option c. ΔH reaction = - 2,512.4 kJ

Step-by-step explanation:

1) Data:

a) ΔHf C₂H₂ = 227.4 kJ

b) ΔHf CO₂ = -393.5 kJ

c) ΔHf H₂O = -241.8 kJ

2) Chemical equation:

• 2C₂H₂ + 5 O2 → 4CO₂ + 2H₂O [see the note below about the phases]

3) Necessary assumptions:

• The phases of the reactants and products in the given reaction are the same at which the standard enthalpies of formation are given.

• The reactant O₂ is at its fundamental state (gas) which implies that the correspondant standard enthaly of formation is zero.

• The units of the given standard enthalpies are the same of the units indicated in the choices for the ΔH of the reaction (kJ).

4) Formula:

• ΔH reaction = Δ∑ ΔH products - ∑ ΔH reactants

5) Solution:

• ΔH reaction = 4×ΔHCO₂ + 2×ΔH₂O - 2×ΔH C₂H₂ - 5 ΔH O₂

• ΔH reaction = 4×(-393.5 kJ) + 2×(-241.8 kJ) - 2×(227.4 kJ) - 5×0

• ΔH reaction = - 2,512.4 kJ ← answer