KCSE 2025 Chemistry P2 Q3 — Energetics and Hess's Law
Published
The Question
“(a) Define the standard enthalpy of formation of a compound. (b) The standard enthalpies of formation of carbon dioxide and water are −393 kJ mol⁻¹ and −286 kJ mol⁻¹ respectively, and the standard enthalpy of formation of propane, C₃H₈, is ΔH₁ = −104 kJ mol⁻¹. In the energy cycle the elements 3C(s) + 4H₂(g) can reach the combustion products 3CO₂(g) + 4H₂O(l) either directly (ΔH₂) or by first forming propane (ΔH₁) which is then burned (ΔH₃). (i) Write the equations of formation for CO₂ and H₂O. (ii) Calculate ΔH₂ and hence use Hess's law to find ΔH₃, the standard enthalpy of combustion of propane. (c) In an experiment, burning butane, C₄H₁₀, raised the temperature of 500 g of water by 35 °C. Given the molar enthalpy of combustion of butane is 2880 kJ mol⁻¹, the specific heat capacity of water is 4.2 J g⁻¹ K⁻¹ and the molar mass of butane is 58 g mol⁻¹, calculate the mass of butane burned (Q = mcΔT).”
(a) Standard enthalpy of formation
The standard enthalpy of formation of a compound is the heat change when one mole of the compound is formed from its elements, with everything in its standard state and measured under standard conditions (298 K and 1 atmosphere pressure). The two details that earn the marks are 'one mole of the compound' and 'from its elements in their standard states'.
(b)(i) Formation equation for carbon dioxide
To form one mole of carbon dioxide from its elements, solid carbon reacts with oxygen gas. The equation already gives exactly one mole of CO₂ and is balanced as written, and from the data this releases 393 kJ, so the enthalpy of formation is −393 kJ mol⁻¹.
(b)(i) Formation equation for water
For water the key is to keep the product to exactly one mole, so only half a mole of oxygen can be used. Hydrogen gas plus half a mole of oxygen gives one mole of liquid water, with an enthalpy of formation of −286 kJ mol⁻¹.
(b)(ii) Setting up the Hess cycle
The elements 3C(s) + 4H₂(g) can reach the same products, 3CO₂(g) + 4H₂O(l), by two routes. The direct route, ΔH₂, forms the oxides straight from the elements. The indirect route first forms propane (ΔH₁) and then burns it (ΔH₃). Hess's law states that the total enthalpy change is the same by either route, so ΔH₁ + ΔH₃ = ΔH₂.
(b)(ii) ΔH₁ — forming propane
Making propane directly from its elements is simply the standard enthalpy of formation of propane, which the data give as −104 kJ mol⁻¹.
(b)(ii) ΔH₂ — the direct route
The direct route forms three moles of CO₂ and four moles of water from the elements, so ΔH₂ is three times the enthalpy of formation of carbon dioxide plus four times that of water.
(b)(ii) ΔH₃ — the enthalpy of combustion of propane
Applying Hess's law, ΔH₁ + ΔH₃ = ΔH₂. Substituting the known values and making ΔH₃ the subject gives the standard enthalpy of combustion of propane, −2219 kJ mol⁻¹.
(c) Heat released to the water
Now switch to the butane experiment. Burning the butane raises the temperature of 500 g of water by 35 °C. The heat absorbed by the water is found from Q = mcΔT, using the specific heat capacity of water, 4.2 J g⁻¹ K⁻¹.
(c) Mass of butane burned
That heat came from burning butane, whose molar enthalpy of combustion is 2880 kJ mol⁻¹, so dividing the heat released by 2880 gives the number of moles burned. Multiplying by the molar mass of butane, 58 g mol⁻¹, gives the mass of butane burned.
Final Result
(a) The standard enthalpy of formation is the heat change when one mole of a compound is formed from its elements in their standard states at 298 K and 1 atm. (b) C(s) + O₂(g) → CO₂(g), ΔHf = −393 kJ mol⁻¹; H₂(g) + ½O₂(g) → H₂O(l), ΔHf = −286 kJ mol⁻¹. ΔH₂ = 3(−393) + 4(−286) = −2323 kJ, and by Hess's law ΔH₃ = ΔH₂ − ΔH₁ = −2323 − (−104) = −2219 kJ mol⁻¹. (c) Q = mcΔT = 500 × 4.2 × 35 = 73.5 kJ; moles of butane = 73.5 ÷ 2880 = 0.02552 mol, so mass = 0.02552 × 58 = 1.48 g.
Why this method works
Hess's law works because enthalpy is a state function: the heat change depends only on the starting and finishing states, not on the path taken between them. That is why the elements can be imagined reaching the combustion products either directly or through propane, and both routes must release the same total energy. This lets you find an enthalpy that is hard to measure directly — the combustion of propane, ΔH₃ — from ones that are easier, by treating the cycle as a simple equation, ΔH₁ + ΔH₃ = ΔH₂. Watch the signs: every value here is negative because these are exothermic reactions, and the rearrangement must respect those signs. The calorimetry part links energy to amount of substance: the water can only tell you how much heat was released, so you convert that heat to moles using the molar enthalpy of combustion, and only then to a mass using the molar mass. Keeping propane (the cycle) and butane (the calorimetry) apart is essential — they are different fuels with different molar masses and combustion enthalpies.
Rearranged directly, ΔH₃ = ΔH₂ − ΔH₁ = −2323 − (−104) = −2219 kJ mol⁻¹, matching the cycle. For the mass, 0.02552 mol × 58 g mol⁻¹ ≈ 1.48 g, and multiplying back, 0.02552 × 2880 ≈ 73.5 kJ, recovers the heat released.