Consider the following cases of standard enthalpy of reaction (Δ Hᵣ° in kJ mol⁻¹)
C₂ H₆( g)+7/2 O₂( g) → 2 CO₂( g)+3 H₂ O(l) Δ H₁°=-1550 ; C( graphite )+O₂( g) → CO₂( g) Δ H₂°=-393.5 ; H₂( g)+1/2 O₂( g) → H₂ O(l) Δ H₃°=-286
The magnitude of Δ H_(f C₂ H₆( g))° is ____________ kJ mol⁻¹ (Nearest integer).

Solution

Chemistry ma’am · 4 steps

Burn the elements to make the products, then run the ethane combustion backwards to get ethane.

1. Target reaction. Formation of ethane means making one mole of it from its elements in their standard states: 2C(graphite) + 3H₂(g) → C₂H₆(g).
2. Combine the equations. Take 2 × equation 2 to burn 2 C, and 3 × equation 3 to burn 3 H₂. Together they give 2CO₂ + 3H₂O. Then reverse equation 1 to turn 2CO₂ + 3H₂O back into C₂H₆ and 7/2 O₂. The 7/2 O₂ used up in burning the elements is given back here, so the oxygen cancels.

   Δ H_f° = 2Δ H₂° + 3Δ H₃° - Δ H₁°

3. Put in numbers. Δ H_f° = 2(-393.5) + 3(-286) - (-1550) = -787 - 858 + 1550 = -95 kJ mol⁻¹.
4. So. The enthalpy of formation of ethane is -95 kJ mol⁻¹, so its magnitude is 95.

The common slip: 3195. Adding Δ H₁° instead of reversing it gives -787 - 858 - 1550 = -3195, which is a magnitude of 3195, not 95.

The idea to keep. To get a formation enthalpy from combustion data, use this rule: Δ H_f = (combustion enthalpies of the elements) - (combustion enthalpy of the compound). Always scale each one by its coefficient.

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