ThermochemistryHard

Question

Use the following data to calculate the enthalpy of hydration for caesium iodide and caesium hydroxide, respectively.

Compound Lattice energy (kJ/mol) ΔHSolution (kJ/mol)

CsI +604 +33

CsOH +724 −72

Options

A.−571 kJ/mol and −796 kJ/mol
B.637 kJ/mol and 652 kJ/mol
C.−637 kJ/mol and −652 kJ/mol
D.571 kJ/mol and 796 kJ/mol

Solution

$\Delta_{\text{Lattice}}H + \Delta_{\text{Hydration}}H = \Delta_{\text{solution}}H$

$\text{For }CSl:\Delta_{\text{Hyd}}H = 33 - 604 = - 571\text{ kJ} $$$\text{For }CsOH:\Delta_{\text{Hyd}}H = ( - 72) - 724 = - 796\text{ kJ}$$

Create a free account to view solution

View Solution Free
Topic: Thermochemistry·Practice all Thermochemistry questions

More Thermochemistry Questions

Heat of neutralisation of CsOH with all strong acid is 13.4 kcal mol-1. The heat released on neutralisation of CsOH with...Enthalpies of combustion of CH4, C2H6 and C3H8 are −210, −370 and −526 kcal/mol, respectively. Enthalpy of combustion of...The bond enthalpies of C–C, C=C and C≡C bonds are 348, 610 and 835 kJ/mol, respectively at 298 K and 1 bar. The enthalpy...For the equilibrium H2O(l) ⇋ H2O(g) at 1atm and 298 K :...If the bond dissociation energies of XY, X2 and Y2 (all diatomic molecules) are in the ratio of 1:1:0.5 and ᐃfH fo...