Nitrogen monoxide is produced from ammonia and air as shown in these equations:
4NH3(g)+3O2(g)→2N2(g)+6H2O(g)ΔH=−1268 kJ mol−1 4\text{NH}_3(\text{g}) + 3\text{O}_2(\text{g}) \rightarrow 2\text{N}_2(\text{g}) + 6\text{H}_2\text{O}(\text{g}) \quad \Delta H = -1268\text{ kJ mol}^{-1} 4NH3(g)+3O2(g)→2N2(g)+6H2O(g)ΔH=−1268 kJ mol−1 N2(g)+O2(g)→2NO(g)ΔH=+180 kJ mol−1 \text{N}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{NO}(\text{g}) \quad \Delta H = +180\text{ kJ mol}^{-1} N2(g)+O2(g)→2NO(g)ΔH=+180 kJ mol−1What is the enthalpy change (in kJ mol−1\text{kJ mol}^{-1}kJ mol−1) for the following reaction?
4NH3(g)+5O2(g)→4NO(g)+6H2O(g) 4\text{NH}_3(\text{g}) + 5\text{O}_2(\text{g}) \rightarrow 4\text{NO}(\text{g}) + 6\text{H}_2\text{O}(\text{g}) 4NH3(g)+5O2(g)→4NO(g)+6H2O(g)−1088-1088−1088
−1448-1448−1448
−908-908−908
−1628-1628−1628
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.