Phosphorus pentachloride is produced from white phosphorus and chlorine gas via phosphorus trichloride as shown in these equations:
P4(s)+6Cl2(g)→4PCl3(g)ΔH=−1148 kJ mol−1 \text{P}_4(\text{s}) + 6\text{Cl}_2(\text{g}) \rightarrow 4\text{PCl}_3(\text{g}) \quad \Delta H = -1148\text{ kJ mol}^{-1} P4(s)+6Cl2(g)→4PCl3(g)ΔH=−1148 kJ mol−1 PCl3(g)+Cl2(g)→PCl5(g)ΔH=−93 kJ mol−1 \text{PCl}_3(\text{g}) + \text{Cl}_2(\text{g}) \rightarrow \text{PCl}_5(\text{g}) \quad \Delta H = -93\text{ kJ mol}^{-1} PCl3(g)+Cl2(g)→PCl5(g)ΔH=−93 kJ mol−1What is the enthalpy change (in kJ mol−1\text{kJ mol}^{-1}kJ mol−1) for the direct preparation of phosphorus pentachloride from white phosphorus?
P4(s)+10Cl2(g)→4PCl5(g) \text{P}_4(\text{s}) + 10\text{Cl}_2(\text{g}) \rightarrow 4\text{PCl}_5(\text{g}) P4(s)+10Cl2(g)→4PCl5(g)−776-776−776
−1520-1520−1520
−1241-1241−1241
−1334-1334−1334
34 exam-style questions on AQA A Level Chemistry Energetics, covering Enthalpy change, Calorimetry, Applications of Hess's law, and Bond enthalpies. Each one has a worked solution and a mark scheme showing where the marks go.