The salt copper(II) sulfate dissolves in water according to the reaction: CuSO4(s) Cu2+(aq) + SO42-(aq) (a) Calculate the standard enthalpy change ΔH° for this reaction, using the following data: CuSO4(s) = -771.4 kJ mol-1 Cu2+(aq) = 64.8 kJ mol-1 SO42-(aq) = -909.3 kJ mol-1 kJ (b) Calculate the temperature reached by the solution formed when 13.0 g of CuSO4 is dissolved in 0.109 L of water at 25.0 °C. Approximate the heat capacity of the solution by the heat capacity of 109 g of pure water (specific heat capacity = 4.18 J g-1 °C-1), ignoring the mass of the salt. °C (c) Heats of reaction find practical application in hot packs or cold packs. Would this dissolution reaction be appropriate for the preparation of a hot pack or a cold pack?

(a) In this case, given the formation enthalpies for copper (II) sulfate, copper (II) ion and sulfate ion, we can compute the enthalpy change for such process as follows:

[tex]\Delta H=\Delta H_{Cu^{2+}}+\Delta H_{SO_4^{-2}}-\Delta H_{CuSO_4}\\\\\Delta H=64.8-909.3-(-771.4)\\\\\Delta H=-73.1\frac{kJ}{mol}[/tex]

(b) In this case, since 13.0 g of copper (II) sulfate (molar mass: 159.55 g/mol) we can compute the gained heat by water via:

[tex]Q_{water}=-n_{CuSO_4}\Delta H=-13.0g*\frac{1mol}{159.55g}* -73.1\frac{kJ}{mol}\\ \\Q_{water}=5.96kJ=5960J[/tex]

As the heat lost by the reaction is gained by the water, therefore, the final temperature is:

[tex]Q_{water}=m_{water}Cp_{water}(T_2-T_1)\\\\T_2=T_1+\frac{Q_{water}}{m_{water}Cp_{water}}\\ \\T_2=25.0\°C+\frac{5960J}{0.109L*\frac{1000g}{1L}*4.18\frac{J}{g\°C} } \\\\T_2=38.1\°C[/tex]

(c) In this case, since this dissolution reaction is exothermic as it increases the temperature when undergone, we can infer that yes, this dissolution would be appropriate for the preparation of a hot pack, because a cold pack would be with an endothermic dissolution reaction.

Best regards.