The formation of the blue crystalline solid copper(II) sulfate pentahydrate can be…

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

Problem

The formation of the blue crystalline solid copper(II) sulfate pentahydrate can be represented as CuSO4(s)+5 H2O(l)→CuSO4⋅5H2O(s)\text{CuSO}_4(s) + 5\,\text{H}_2\text{O}(l) \to \text{CuSO}_4\cdot5\text{H}_2\text{O}(s)  ΔH∘=−78.2 kJ\Delta H^\circ = -78.2\,\text{kJ} (per mole of hydrate formed) Tabulated standard molar enthalpies of formation (298\,K) are: • CuSO4(s):−769 kJ⋅mol−1\text{CuSO}_4(s): -769\,\text{kJ}\cdot\text{mol}^{-1} • H2O(l):−285.83 kJ⋅mol−1\text{H}_2\text{O}(l): -285.83\,\text{kJ}\cdot\text{mol}^{-1} • H2O(g):−241.82 kJ⋅mol−1\text{H}_2\text{O}(g): -241.82\,\text{kJ}\cdot\text{mol}^{-1} • CuSO4⋅5H2O(s):?\text{CuSO}_4\cdot5\text{H}_2\text{O}(s): ? Address the items below. (1) Part 1: Identify which value for water (liquid or gaseous) is appropriate for use in the calculation and justify your choice with reference to the reaction conditions. (2) Part 2: Using the correct value from part (a), calculate the standard molar enthalpy of formation, ΔHx∘\Delta H^\circ_x, for CuSO4⋅5H2O(s)\text{CuSO}_4\cdot5\text{H}_2\text{O}(s). Show your work clearly. (3) Part 3: If a student mistakenly used the other water value in the calculation, would the magnitude of the computed ΔHx∘\Delta H^\circ_x for CuSO4⋅5H2O(s)\text{CuSO}_4\cdot5\text{H}_2\text{O}(s) be too large or too small? Explain your reasoning without re-doing the full calculation.

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