The standard heat of formation ( Δ _fH₂₉₈⁰ ) of ethane (in kj/mol), if the heat of combustion of ethane, hydrogen and graphite are - 1560, -286 and -393.5 Kj/mol, respectively is :

Solution

Chemistry ma’am · 4 steps

Burn the reactants, then un-burn ethane: heat of formation is reactants' combustion minus product's.

1. Target reaction. Ethane forms from its elements: 2C(graphite) + 3H₂(g) → C₂H₆(g). Every substance here burns to CO₂ and H₂O, so combustion data links them all.
2. Hess's law route. Burning 2 C and 3 H₂ gives the same 2CO₂ + 3H₂O as burning one C₂H₆. So the reactants' heats of combustion minus the product's heat of combustion is the heat of formation.

   Δ_f H° = 2 Δ_c H(C) + 3 Δ_c H(H₂) - Δ_c H(C₂H₆)

3. Put in numbers. 2(-393.5) + 3(-286) - (-1560) = -787 - 858 + 1560.
4. So. Δ_f H° = -1645 + 1560 = -85 kJ mol⁻¹, so the answer is -85.

The common slip: 85. Using products minus reactants, as you would with formation data, flips the sign and gives +85.

The idea to keep. With combustion data, Δ H = Σ Δ_c H(reactants) - Σ Δ_c H(products). That is the opposite order to formation data, and each value is multiplied by its stoichiometric coefficient.

Answer from the NTA final answer key. Solution by PrepWiser.