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122ch13a - 1 Chapter 13 Homework Solutions Use the...

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1 Chapter 13 - Homework Solutions Use the following diagrams for questions concerning solution process. ) H soln > 0 ( endothermic ); product (solution) has a higher energy than reactants (solute & solvent). Attractive forces between solute and solvent (unlike particles) are not as strong as solute-solute and solvent-solvent AF (like particles). More energy required to separate solute-solute and solvent-solvent particles than is released when form solute-solvent AF. ENDO thermic ( ) H soln > 0) A solution process with ) H soln > 0 (+) tends to not be spontaneous (but can be). It also depends on the entropy of solution, ) S soln . See page 3.
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2 ) H soln < 0 ( exothermic ); product (solution) has a lower energy than reactants (solute & solvent). Attractive forces between solute and solvent (unlike particles) are stronger than solute-solute and solvent-solvent AF (like particles). Less energy required to separate solute-solute and solvent-solvent particles than is released when form solute-solvent AF. EXO thermic ( ) H soln < 0) A solution process with ) H soln < 0 (-) tends to be spontaneous (but may not be). It also depends on the entropy of solution, ) S soln . See page 3.
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3 ) H soln = 0 ( ideal ); product (solution) has same energy as reactants (solute & solvent). Attractive forces between solute and solvent (unlike particles) are similar to solute-solute and solvent-solvent AF (like particles). Energy required to separate solute-solute and solvent-solvent particles is about the same as that released when form solute-solvent AF. Generally, particles with only LF are the most likely to form ideal solutions. IDEAL ( ) H soln = 0) For ideal solutions ( ) H soln = 0) and endothermic ( ) H soln > 0) to form the change in entropy (disorder) must be positive ( ) S soln > 0, disorder must inc .). Generally, ) S soln > 0 for mixing. Remember, as given in class: ) H soln = ) H 1 + ) H 2 + ) H 3 or ) H soln = ) H sep solute + ) H sep solvent + ) H solvation ) G = ) H - T C ) S and ) G < 0 (negative) for a spontaneous process ) G > 0 (positive) for a nonspontaneous process If, ) H > 0 (+) or ) H = 0 then, MUST have ) S > 0 (+) (an increase in disorder) in order to have ) G < 0 (-), at some temperature, in order for a solution to form. A ) S > 0 (+) tends to make a process spontaneous. If ) H < 0 (-), ) S can be negative (-) and still have ) G < 0 (-), so a solution can form at some temp even with a decrease in disorder (inc. in order).
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4 13.2) When an ionic compound dissolves in water the cation becomes surrounded by the H 2 O molecules (see figure below). Courtesy of Prentice Hall (BLB 10 th ed.) This process is called hydration (specific term for solvation when the solute is an ion and the solvent is H 2 O). This is an Ion-Dipole AF between the ion and the polar solvent (H 2 O). This AF is quite strong. The attractive forces we’ve discussed have the relative order in terms of increasing strength: LF < DD < H-bonds < Ion-Dipole < Ionic & Covalent In terms of the energy diagrams depicted on pages 1-3 above (Figure 13.4 in the text, p. 532) to which step does this interaction correspond? This corresponds to step 3 in the diagrams. This is ) H 3 ( ) H solvation ). For this specific case of an ion in H 2 O this is usually referred to as the heat of hydration, ) H hyd This step always
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