- Joined
- Feb 24, 2010
- Messages
- 183
- Reaction score
- 0
If a compound becomes less soluble as the temperature of the solution is increased, then which of the following statements about the entropy and enthalpy changes associated with its solvation is correct?
A. ∆S is probably negative; ∆H must be negative.
B. ∆S is probably positive; ∆H must be negative.
C. ∆S is probably negative; ∆H cannot be determined without more information.
D. ∆S is probably positive; ∆H cannot be determined without more information.
Solution:
B is the best answer. Regardless of whether the compound becomes more or less soluble with increasing temperature, the change in entropy is positive for any salt dissociating into ions. This is because the solution is more random than the lattice structure, so choices A and C are eliminated. As heat is added to a solution, we observe that the reverse reaction is favored, corresponding to a decrease in solubility. This implies that the reaction is exothermic. A solid forms, removing extra heat from the system. Choice B is thus a better answer than choice D. The best answer is B.
I understand that a solution is more random than the lattice structure and why entropy should be positive. I also get that they figured out the sign for enthalpy based on Le Chatelier's principle (since heat is a product, and the addition of heat heat would drive the reaction towards the reactant side).
What confuses me is the contradiction that comes up if I try to solve this problem with the free energy equation delta G = delta H - T(delta S), because if you increase temperature with a positive delta S and negative delta H, delta G should become more negative and the reaction (solvation) should be more favorable.
Can anyone explain this?
A. ∆S is probably negative; ∆H must be negative.
B. ∆S is probably positive; ∆H must be negative.
C. ∆S is probably negative; ∆H cannot be determined without more information.
D. ∆S is probably positive; ∆H cannot be determined without more information.
Solution:
B is the best answer. Regardless of whether the compound becomes more or less soluble with increasing temperature, the change in entropy is positive for any salt dissociating into ions. This is because the solution is more random than the lattice structure, so choices A and C are eliminated. As heat is added to a solution, we observe that the reverse reaction is favored, corresponding to a decrease in solubility. This implies that the reaction is exothermic. A solid forms, removing extra heat from the system. Choice B is thus a better answer than choice D. The best answer is B.
I understand that a solution is more random than the lattice structure and why entropy should be positive. I also get that they figured out the sign for enthalpy based on Le Chatelier's principle (since heat is a product, and the addition of heat heat would drive the reaction towards the reactant side).
What confuses me is the contradiction that comes up if I try to solve this problem with the free energy equation delta G = delta H - T(delta S), because if you increase temperature with a positive delta S and negative delta H, delta G should become more negative and the reaction (solvation) should be more favorable.
Can anyone explain this?
