The salt cesium perchlorate dissolves in water according to the reaction: CsClO4(s) Cs+(aq) + ClO4-(aq) (a) Calculate the standard enthalpy?

CsClO₄(s) ⇌ Cs⁺(aq) + ClO₄⁻(aq) …… ΔH°

ΔH° = ΔHf°[Cs⁺(aq)] + ΔHf°[ClO₄⁻(aq)] - ΔHf° [HClO₄]
= (-258.3) + (-129.3) - (443.1) kJ
= +55.5 kJ

Molar mass of CsClO₄ = (132.9 +35.5 + 16.0×4) g/mol = 232.4 g/mol
Moles of CsClO₄ dissolved = (46.0 g) / (232.4 g/mol) = 0.1979 mol

Heat gained in dissolving CsClO₄ = Heat lost by the water
(0.1979 mol) × (55.5 × 1000 J/mol) = (196 g) × (4.184 J/g°C) × (26.0 - T)°C
26.0 - T = 13.4
T = 12.6
Temperature reached by solution formed = 12.6°C

a.
products - reactants:
(-258.3 kJ/mol) + (-129.3 kJ/mol) - (-443.1 kJ/mol) = +55.5 kJ

b.
(46.0 g CsClO4) / (232.3561 g CsClO4/mol) x (55500 J/mol CsClO4) / (4.184 J/g·°C) / (196 g) =
13.4°C change

26.0°C - 13.4°C = 12.6°C