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

Option c. ΔH reaction = - 2,512.4 kJ

Explanation:

1) dа ta:

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