Write the balanced reaction equation for this process. qcal = (110.0 g)(4.18 J g-1°C-1)(20.6°C) = 9470 J DHsol = -9470 J/(10.0 g/42.4 g mol-1) = -40,200 J mol-1 = -40.2 kJ/mol Write the balanced reaction equation for this process. LiCl(s)  Li+(aq) + Cl-(aq)

b.) Na+ = 1s22s22p6 c.) The electronic configuration of Li+ is 1s2 . A whole extra energy level is filled in the Na+ ion means that the ionic radius is significantly greater for Na. d.) LiCl. Given lattice energy is governed by Coulomb’s law, the smaller the value of r (the distance between the ions) and the larger the force of attraction between them, the stronger the bond. Since Li+ is smaller than Na+ , Li+ and Cl- can get closer together than Na+ and Cl-

H2O (δ- oxygen) to Li+ , and H2O (δ+ hydrogen) to Cl- H2O (δ- oxygen) to Li+ , and H2O (δ+ hydrogen) to Cl- . Each can be described as dipole (H2O) to ion (Li+ , Cl- ) interactions. Why is it that the overall heat of solution can be determined by the addition of the various steps in the solvation process? Hess’s Law…the first law of thermodynamics.

DG° = -40.2 kJ/mol – (298 K)(0.1345 kJ mol-1K-1) DG° = -80.3 kJ/mol If the absolute entropy of Li+ and Cl- are 138.7 and 55.1 J mol-1 K-1respectively at 298 K and the concentrations of these ions in solution are already 1.00 M, would dissolving additional LiCl take place if solid LiCl were present in the container housing the materials or would additional solid LiCl precipitate out? (The absolute entropy of LiCl solid at that temperature is 59.3 J mol-1K-1) DS° = 193.8 J K-1 - 59.3 J K-1 = 134.5 J K-1 To determine the spontaneity of a process under standard conditions, DG° should be found. This can be done using the Gibb’s-Helmholtz Equation: DG° = DH° - T DS° DG° = -40.2 kJ/mol – (298 K)(0.1345 kJ mol-1K-1) DG° = -80.3 kJ/mol Therefore, more dissolving will take place!!!