This question is about enthalpy changes of reactions involving organic compounds.
A student determines the enthalpy change of combustion, ΔcH\Delta_{\text{c}}HΔcH, of pentan-1-ol, C5H11OH\text{C}_5\text{H}_{11}\text{OH}C5H11OH, using the following method:
The temperature of the water increased by 18.2 ∘C18.2\text{ }^\circ\text{C}18.2 ∘C.
The spirit burner decreased in mass by 0.320 g0.320\text{ g}0.320 g.
Use the student's results to determine the enthalpy change of combustion of pentan-1-ol, ΔcH(C5H11OH)\Delta_{\text{c}}H(\text{C}_5\text{H}_{11}\text{OH})ΔcH(C5H11OH), in kJ mol−1\text{kJ mol}^{-1}kJ mol−1. Assume the specific heat capacity of water is 4.18 J g−1 K−14.18\text{ J g}^{-1}\text{ K}^{-1}4.18 J g−1 K−1.
Heptane, C7H16(g)\text{C}_7\text{H}_{16}(\text{g})C7H16(g), can be broken down by heat to form butane, C4H10(g)\text{C}_4\text{H}_{10}(\text{g})C4H10(g), and propene, C3H6(g)\text{C}_3\text{H}_6(\text{g})C3H6(g):
C7H16(g)→C4H10(g)+C3H6(g)ΔH=+81 kJ mol−1Reaction 1 \text{C}_7\text{H}_{16}(\text{g}) \to \text{C}_4\text{H}_{10}(\text{g}) + \text{C}_3\text{H}_6(\text{g}) \quad \Delta H = +81\text{ kJ mol}^{-1} \quad \text{Reaction 1} C7H16(g)→C4H10(g)+C3H6(g)ΔH=+81 kJ mol−1Reaction 1The enthalpy changes of combustion of C7H16(g)\text{C}_7\text{H}_{16}(\text{g})C7H16(g) and C3H6(g)\text{C}_3\text{H}_6(\text{g})C3H6(g) are shown in the table below:
| Hydrocarbon | ΔcH/kJ mol−1\Delta_{\text{c}}H / \text{kJ mol}^{-1}ΔcH/kJ mol−1 |
|---|---|
| C7H16(g)\text{C}_7\text{H}_{16}(\text{g})C7H16(g) | −4817-4817−4817 |
| C3H6(g)\text{C}_3\text{H}_{6}(\text{g})C3H6(g) | −2058-2058−2058 |
Use ΔH\Delta HΔH in Reaction 1 and the enthalpy changes of combustion in the table to determine the enthalpy change of combustion of C4H10(g)\text{C}_4\text{H}_{10}(\text{g})C4H10(g).
