Question

If enthalpies of formation of C₂H₄ (g) , CO₂(g)
and H₂O(₁) at 25°C and 1atm pressure are 52,
– 394 and – 286 kJ/mol respectively, the change
in ethalpy is equal to
(a) – 141.2 kJ/mol (b) – 1412 kJ/mol
(c) + 14.2 kJ/mol (d) + 1412 kJ/mol

Answer 1

The change  in enthalpy is equal to (a) – 141.2 kJ/mol

Explanation:

The chemical equation of the reaction given in the question is:

2C (s) + 2H₂ → C₂H₄ (g)

⇒ ΔH₁ = 53 kJ/mole → (equation 1)

C (s) + O₂ (g) → CO₂ (g)

⇒ ΔH₂ = - 394 kJ/mole → (equation 2)

H₂ (g) + 1/2O₂ (g) → H₂O (l)

⇒ ΔH₃ = - 286 kJ/mole → (equation 3)

The change in enthalpy is given as:

C₂H₄ (g) → 2C (s) + 2H₂ ⇒ ΔH₁ = 53 kJ/mole

C (s) + O₂ (g) → CO₂ (g) ⇒ ΔH₂ = - 2 × 394

H₂ (g) + 1/2O₂ (g) → H₂O (l) ⇒ ΔH₃ = - 2 × 286

⇒ ΔH = 53 - (2 × 394) - (2 × 286)

∴ ΔH = - 1412 kJ/mol