Determine the standard enthalpy change,

CHEMISTRY · Subjective · A Level · Chemistry · Energetics · Energetics

Problem

Determine the standard enthalpy change, ΔH298∘\Delta H^\circ_{298}, for the reaction 2N2O5(l)→4NO2(g)+O2(g)2\mathrm{N}_2\mathrm{O}_5(l)\rightarrow4\mathrm{NO}_2(g)+\mathrm{O}_2(g) in two independent ways, and state the numerical difference in kJ between the results. Route A: Combine the following reactions: (a) 2NO2(g)+12O2(g)→N2O5(l)2\mathrm{NO}_2(g)+\frac12\mathrm{O}_2(g)\rightarrow\mathrm{N}_2\mathrm{O}_5(l), ΔH∘=−56.5 kJ\Delta H^\circ=-56.5\text{ kJ}. (b) NO(g)+12O2(g)→NO2(g)\mathrm{NO}(g)+\frac12\mathrm{O}_2(g)\rightarrow\mathrm{NO}_2(g), ΔH∘=−57.2 kJ\Delta H^\circ=-57.2\text{ kJ}. (c) N2O5(l)+NO(g)→32N2(g)+O2(g)\mathrm{N}_2\mathrm{O}_5(l)+\mathrm{NO}(g)\rightarrow\frac32\mathrm{N}_2(g)+\mathrm{O}_2(g), ΔH∘=−348.5 kJ\Delta H^\circ=-348.5\text{ kJ}. Route B: Use the standard enthalpies of formation ΔHf∘(N2O5,l)=−364.0 kJ mol−1\Delta H_f^\circ(\mathrm{N}_2\mathrm{O}_5,l)=-364.0\text{ kJ mol}^{-1}, ΔHf∘(NO2,g)=+33.2 kJ mol−1\Delta H_f^\circ(\mathrm{NO}_2,g)=+33.2\text{ kJ mol}^{-1}, ΔHf∘(N2,g)=0\Delta H_f^\circ(\mathrm{N}_2,g)=0, and ΔHf∘(O2,g)=0\Delta H_f^\circ(\mathrm{O}_2,g)=0.

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