Chemical Energetics: Question 1

Worked solution

Part (a): Defining the standard enthalpy change of combustion

The standard enthalpy change of combustion, $\Delta H_c^{\ominus}$, is the enthalpy change when one mole of a substance is completely burned in excess oxygen, with all reactants and products in their standard states, measured under standard conditions (298 K, 100 kPa).

Part (b): Heat energy transferred to the water

The volume of water is $150\ \text{cm}^3$, and since the density of water is $1.00\ \text{g cm}^{-3}$, the mass of water is: $m = 150\ \text{cm}^3 \times 1.00\ \text{g cm}^{-3} = 150\ \text{g}$

The temperature rise of the water is: $\Delta T = 36.7 - 19.5 = 17.2\,^\circ\text{C} \ (= 17.2\ \text{K})$

Using $q = mc\Delta T$: $q = 150 \times 4.18 \times 17.2$

Working this in two steps: $150 \times 4.18 = 627.0\ \text{J K}^{-1}$ $627.0 \times 17.2 = 10\,784.4\ \text{J}$

So $q = 10\,784\ \text{J} = 10.8\ \text{kJ}$ (3 s.f.) is transferred to the water.

Check (independent recomputation): $627.0 \times 17.2 = 627.0\times17 + 627.0\times0.2 = 10\,659 + 125.4 = 10\,784.4\ \text{J}$, consistent.

Part (c): Amount of fuel burned and experimental $\Delta H_c$

Mass of propan-1-ol burned: $82.46 - 81.98 = 0.48\ \text{g}$

Amount, in mol, using molar mass $60.0\ \text{g mol}^{-1}$: $n = \frac{0.48}{60.0} = 0.0080\ \text{mol}$

The heat calculated in (b) was released by burning this amount of fuel, so the enthalpy change per mole is: $\Delta H_c = -\frac{q}{n} = -\frac{10.7844\ \text{kJ}}{0.0080\ \text{mol}}$

Dividing: $\frac{10.7844}{0.0080} = 1348.05$

so $\Delta H_c = -1348\ \text{kJ mol}^{-1} \approx -1350\ \text{kJ mol}^{-1} \ (3\ \text{s.f.})$

The value is negative because combustion is exothermic. Heat is released to the water, not absorbed from it.

Check (independent recomputation): $0.48/60.0=0.0080$ exactly; $10\,784.4\ \text{J}\div0.0080\ \text{mol}=1\,348\,050\ \text{J mol}^{-1}=1348.05\ \text{kJ mol}^{-1}$, confirming $\Delta H_c\approx-1350\ \text{kJ mol}^{-1}$.

Part (d): Why the experimental value is smaller in magnitude

Not all of the heat released by the burning fuel reaches the water: some heat warms the metal can, the burner, and the surrounding air, and some is lost by radiation and convection before it can heat the water. Combustion may also be slightly incomplete (producing some soot/carbon monoxide instead of only $\text{CO}_2$ and $\text{H}_2\text{O}$), which further reduces the energy transferred. Both effects make the experimental $\Delta H_c$ less exothermic (smaller in magnitude) than the accepted data-book value.

Final answers

• (a) Enthalpy change when one mole of a substance burns completely in excess oxygen, all species in their standard states, under standard conditions.
• (b) $q = \boxed{10\,784\ \text{J}} = 10.8\ \text{kJ}$
• (c) $n = \boxed{0.0080\ \text{mol}}$; $\Delta H_c = \boxed{-1350\ \text{kJ mol}^{-1}}$ (3 s.f.)
• (d) Heat losses to the surroundings/apparatus (and/or incomplete combustion) mean less energy reaches the water than the true enthalpy of combustion would predict.