Hess's Law Example 7

Worked example: using two measured reactionsThe enthalpy change for hydrating anhydrous copper(II) sulfate cannot be measured directly, because it is hard to add exactly five moles of water to the solid and measure the temperature change accurately.CuSO₄(s) + 5H₂O(l) → CuSO₄·5H₂O(s)Instead, each solid is dissolved separately in water and the two enthalpy changes of solution are measured in a calorimeter. Both give the same solution, CuSO₄(aq).Results:
ΔH₁: CuSO₄(s) + aq → CuSO₄(aq) = -66.5 kJ mol^-1
ΔH₂: CuSO₄·5H₂O(s) + aq → CuSO₄(aq) = +11.7 kJ mol^-1Step 1. Draw the cycle: the target reaction across the top and CuSO₄(aq) at the bottom. Both arrows point down to the solution.Step 2. The route via the solution goes down ΔH₁, then up against ΔH₂, so the sign of ΔH₂ is reversed:Δ_rH = ΔH₁ - ΔH₂Step 3. Substitute:Δ_rH = -66.5 - (+11.7) = -78.2 kJ mol^-1Hydration is exothermic, which is why white anhydrous copper(II) sulfate warms up as it turns blue when water is added.In the real experiment, the mass of each solid is converted to moles and Q = mcΔT is used first to find ΔH₁ and ΔH₂.