Question

Acetylene burns in air according to the following equation:

C2H2(g) + 5 2 O2(g) → 2 CO2(g) + H2O(g) ΔH o rxn = −1255.8 kJ

Given ΔH o f of CO2(g) = −393.5 kJ/mol and

ΔH o f of H2O(g) = −241.8 kJ/mol, find ΔH o f of C₂H₂(g).

Answer 1

A-->E = A-->B-->C-->D-->E (the overall energy change remains the same, independently of the path chosen)





(h1) = formation of 2 moles of C2H2 from elements [we're looking for STANDARD formation enthalpy - meaning, the energy required to form 1 mole of C2H2 = (h1)/2]

(h2) = formation of 4 moles of CO2 and 2 moles of H2O = 4x(-393.5 kJ) + 2x(-285.8 kJ) = -2145.6 kJ

(h3) = -2598.8 kJ



now, (h1) = (h2) - (h3)

[we take the other path: we go from the elements, to the products of combustion and then to the reagents; since we reverse the arrow of (h3), we "reverse" the sign as well]

(h1) = -2145.6 kJ - (-2598.8 kJ)
(h1) = -2145.6 kJ + 2598.8 kJ
(h1) = 453.2 kJ

now, this is the energy required to form 2 moles of C2H2, so we divide by 2 to get the formation enthalpy of 1 mole

Standard formation enthalpy (C2H2) = 226.6 kJ/mol