Solutions - Heat of Solution

[Mitchtwitchita]

When LiNO3 dissolves in water, the enthalpy of solution is -2.5 KJ/mol. The standard enthalpy of formation of solid LiNO3 is -483.1 KJ/mol.
(a)When LiNO3 dissolves in water, is the dissolution endothermic or exothermic?
(b)Will the solubility of LiNO3 increase or decrease as the temperature increases?
(c)Calculate the standard enthalpy of formation (KJ/mol) for aqueous LiNO3.

(a)The process would be exothermic because the enthalpy of solution is -2.5 KJ/mol

(b)The solubility would increase because the solubility of every nitrate salt I've seen increases with an increase in temperature. Also, the solubility of most salts in general increase with an increase in temperature.

(c)I'm pretty lost on this one.

Li+ + NO3- ---> LiNO3 delta Hrxn = -483.1 KJ/mol
LiNO3 + H2O ---> Li+(aq) + NO3-(aq) delta Hsoln = -2.5 KJ/mol

= -485.6 KJ/mol?

Any help with this one would be appreciated!

 

[TeethWhitener]

(a)The process would be exothermic because the enthalpy of solution is -2.5 KJ/mol

Yes.

(b)The solubility would increase because the solubility of every nitrate salt I've seen increases with an increase in temperature. Also, the solubility of most salts in general increase with an increase in temperature.

No. This is an illustration of Le Chatelier's principle. Your equilibrium equation goes like this:
$$\mathrm{LiNO_{3(s)}} \rightleftharpoons \mathrm{Li^+_{(aq)}} + \mathrm{NO_{3(aq)}^{-}} + \mathrm{heat}$$
Increasing the temperature adds heat which, following Le Chatelier, will push the equilibrium toward solid LiNO₃. So increasing temperature decreases solubility of this salt.

(c)I'm pretty lost on this one.

Li+ + NO3- ---> LiNO3 delta Hrxn = -483.1 KJ/mol
LiNO3 + H2O ---> Li+(aq) + NO3-(aq) delta Hsoln = -2.5 KJ/mol

= -485.6 KJ/mol?

This looks fine.