Hydrogen peroxide undergoes disproportionation according to 2 ,H_2O_2(l) 2 ,H_2O(l) +…

CHEMISTRY · Multi-part · A Level · Chemistry · Energetics · Energetics

Problem

Hydrogen peroxide undergoes disproportionation according to 2 H2O2(l)⟶2 H2O(l)+O2(g)\mathrm{2\,H_2O_2(l) \longrightarrow 2\,H_2O(l) + O_2(g)} The following auxiliary thermochemical information is available: (1) H2(g)+12O2(g)→H2O(l)        ΔH1=−286 kJ mol−1\mathrm{H_2(g) + \tfrac12 O_2(g) \to H_2O(l)} \;\;\;\; \Delta H_1 = -286\,\text{kJ mol}^{-1} (2) 2 H2(g)+O2(g)→2 H2O2(l)    ΔH2=+187 kJ mol−1\mathrm{2\,H_2(g) + O_2(g) \to 2\,H_2O_2(l)} \;\; \Delta H_2 = +187\,\text{kJ mol}^{-1} No other enthalpy data are to be used. (a) Using only reactions (1) and (2), determine the standard enthalpy change for the disproportionation of hydrogen peroxide written above. Show your work, including any algebraic manipulation of the given reactions. (b) Indicate whether the disproportionation is exothermic or endothermic and explain how the signs of the individual steps in your calculation support that classification. (c) If the disproportionation reaction were written in the reverse direction (starting from water and oxygen to form hydrogen peroxide), what would be the numerical value and sign of the enthalpy change? Briefly justify your answer.

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