Question

Calculate the standard enthalpy of formation of acetylene (C₂H₂) from its elements:

2C_(graphite) + H₂(g) →C₂H₂(g)

The equations for each step and the corresponding enthalpy changes are

(a) C_(graphite) + O₂(g) →CO₂(g) ∆H°rxn = 2393.5 kJ/mol

(b) H₂(g) + ½ O₂(g →H₂O(l) ∆H°rxn = 2285.8 kJ/mol

(c) 2C₂H₂(g) + 5O₂(g) →4CO₂(g) + 2H₂O(l) ∆H°rxn = 22598.8 kJ/mol

Answer 1

C_graphite + O₂ (g) → CO₂ (g), \(\Delta_{next}=2393.5 KJ/mol\)

-------(1)

H₂(g) + \(\frac12\)O₂ (g) → H₂O (l), \(\Delta H^o_{rea}=2393.5 KJ/mol\)

----(2)

2C₂H₅(g) + 5O₂ + 5O₂ (g) → 4CO₂(g) + 2H₂O(l)-----(3)

\(\Delta H^o_{rxn} = 22598.8 KJ/mol\)

equation (3) can also be written as--

2CO₂ + H₂O(l) → C₂H₂(g) + \(\frac s2\)O₂------(4)

\(\Delta H^o_{rex} = -22598.8 KJ/mole\)

applying (2 x (1) + (2) + (4)) we get

2 C_\(graphite\) + H₂(g) → C₂H₂(g)

therefore, \(\Delta H_{formation}\) = (2 x 2393.5) + 2285.8 + (-2298.8)

 = -15526 KJ/mole.

Hence, the standard enthalpy of formation of acetylene (C₂H₂) will be -15526 KJ/mole.