Hess cycles from combustion data and Core Practical 2

Using combustion data

Formation of most organic compounds cannot be carried out directly, but every substance involved can be burned. Draw the cycle with the elements and the compound at the top and the combustion products below:

ΔrH = sum of ΔcH(reactants) - sum of ΔcH(products)

Note this is the reverse order from the formation-data rule.

Core Practical 2

Pearson's Core Practical 2 uses Hess's law to find an enthalpy change that cannot be measured directly. A typical version finds ΔH for the thermal decomposition 2KHCO3(s) -> K2CO3(s) + H2O(l) + CO2(g):

• Reaction 1: KHCO3 + HCl -> KCl + H2O + CO2 (temperature falls)
• Reaction 2: K2CO3 + 2HCl -> 2KCl + H2O + CO2 (temperature rises)

Measure each with Q = mcΔT, then ΔH(decomposition) = 2ΔH1 - ΔH2.

Building any cycle

Write the target reaction across the top. Link both sides to a common set of substances using reactions whose ΔH you know. Going along an arrow, add ΔH; going against an arrow, subtract it. Multiply by the number of moles in the target equation.

Worked example

Find ΔfH of ethane. ΔcH: C(s) -394, H2(g) -286, C2H6(g) -1560 kJ/mol.

1. Target: 2C(s) + 3H2(g) -> C2H6(g)
2. ΔfH = [2(-394) + 3(-286)] - (-1560)
3. = -788 - 858 + 1560

Answer: ΔfH = -86 kJ/mol

Check yourself

1. ΔcH: C(s) -394, H2(g) -286, C3H8(g) -2220 kJ/mol. Find ΔfH of propane.
   Show answer

   -106 kJ/mol. Target 3C + 4H2 -> C3H8. ΔfH = [3(-394) + 4(-286)] - (-2220) = -1182 - 1144 + 2220 = -106 kJ/mol.

2. In Core Practical 2, ΔH1 (KHCO3 + HCl) = +29.0 kJ/mol and ΔH2 (K2CO3 + 2HCl) = -33.0 kJ/mol. Find ΔH for 2KHCO3 -> K2CO3 + H2O + CO2.
   Show answer

   +91.0 kJ/mol. ΔH = 2ΔH1 - ΔH2 = 58.0 + 33.0 = +91.0 kJ/mol.

3. 3.00 g of KHCO3 (M = 100.1 g/mol) is added to 30.0 cm³ of excess HCl and the temperature falls by 6.5 °C. Find ΔH1 in kJ/mol. (c = 4.18 J/g/K)
   Show answer

   +27 kJ/mol. Q = 30.0 x 4.18 x 6.5 = 815 J; n = 0.0300 mol; temperature fell, so ΔH = +0.815 / 0.0300 = +27 kJ/mol (2 s.f.).

4. Why can ΔH for the thermal decomposition of KHCO3 not be measured directly by calorimetry?
   • It is too exothermic
   • KHCO3 does not decompose
   • The products are all gases
   • It needs continuous strong heating, so the heat change cannot be measured with a thermometer in a cup
   Show answer

   It needs continuous strong heating, so the heat change cannot be measured with a thermometer in a cup. You cannot measure the heat absorbed while heating a solid strongly over time.

5. Using combustion data, ΔrH equals
   • sum ΔcH(products) - sum ΔcH(reactants)
   • sum ΔcH(reactants) - sum ΔcH(products)
   • sum ΔcH(products) + sum ΔcH(reactants)
   • ΔcH of the largest molecule
   Show answer

   sum ΔcH(reactants) - sum ΔcH(products). Combustion arrows point down from both sides, so reactants minus products.

6. A -> B has ΔH = -40 kJ/mol and C -> B has ΔH = +25 kJ/mol. Find ΔH for A -> C.
   Show answer

   -65 kJ/mol. A -> B -> C: -40 + (-25) = -65 kJ/mol (reverse C -> B, so change its sign).

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Sources

• Pearson Edexcel IAL Chemistry (2018) scheme of work
• Pearson Edexcel International Advanced Level Chemistry (2018) qualification page
• Pearson IAL Chemistry teacher practical guide

Last updated 7 November 2026. Spotted a mistake? Tell us.