Final Review Problems Answer Key

# Final Review Problems Answer Key - Chem 10181 Fall 2007...

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Chem 10181, Fall 2007 Final Review Problems SOLUTIONS 1 Chem 10181 Fall 2007 Final Review Problems SOLUTIONS (1) (a) The vapor pressure of liquid ammonia is the equilibrium constant for its vaporization: NH 3 (l) NH 3 (g) K eq = p(NH 3 ) So, since Gº = –RT ln(K eq ) = Hº – T Sº, then ln(P) = vap R ( 1 T ) + vap R . In other words, this is a van’t Hoff plot and from the slope and intercept we can conclude that vap = +23.2 kJ/mol, vap = +96.7 J/mol•K. The sublimation of the solid can be treated similarly, with NH 3 (s) NH 3 (g) K eq = p(NH 3 ) Using the data from the dashed line gives subl = +32.7 kJ/mol, subl = +144.1 J/mol•K. Each van’t Hoff plot is linear because Hº and Sº do not vary significantly with temperature. But since the slopes and intercepts of these lines are determined by their thermodynamic parameters, and the thermodynamic stabilities of the solid and liquid are different, of course they are not going to lie on the same line! (b) At the melting point, the solid and liquid have equal stability, so the vapor pressure of the solid and liquid must be equal. One could use the Hº and Sº values calculated above, or one can just find the intersection of the two lines: ln(P liquid ) = ln(P solid ) 11.63 – 2791( 1 T ) = 17.33 – 3931( 1 T ) 1140( 1 T ) = 5.70 T = 200 K At that temperature, ln(P) = –2.33 P = 0.098 atm (technically this is the triple point, rather than the melting point) (c) From part (a): NH 3 (s) NH 3 (g) Hº = +32.7 kJ/mol Sº = +144.1 J/mol•K NH 3 (g) NH 3 (l) Hº = –23.2 kJ/mol Sº = –96.7 J/mol•K NH 3 (s) NH 3 (l) fus = +9.5 kJ/mol fus = +47.4 J/mol•K (d) H N N N H N N N H N N N 2– major major very minor (e) Two factors are probably most important: (1) Inductive effects. The two additional electronegative nitrogen atoms in HN 3 withdraw electron density from the N-H group, relative to NH 3 . (2) Hybridization. The N in NH 3 is sp 3 hybridized, while the N in HN 3 is sp hybridized; the increased s character in the latter stabilizes the conjugate base. (Resonance may also play a role, as N 3 is resonance-stabilized and NH 2 is not, but it is

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Chem 10181, Fall 2007 Final Review Problems SOLUTIONS 2 unclear how important this is because it is unclear how much more resonance-stabilized N 3 is relative to its conjugate acid, which is what counts.) (2)(a) N O O N O O (b) The N-O bond order in nitrite is 1.5 (two equally important resonance structures, so each bond is average between single and double). The molecular orbital picture for NO indicates that it has the electron configuration ( σ 2s ) 2 ( σ * 2s(2p) ) 2 ( π 2p ) 4 ( σ 2p(2s) ) 2 ( π * 2p ) 1 , so the N-O bond order in NO is 2.5. Thus nitrite would be expected to have a longer N-O bond than nitric oxide (and this is indeed observed). (c) Each atom is sp 2 hybridized (don't forget to distinguish between the shared and the unshared lone pairs!). sp 2 hybridization of the central N, which has one lone pair, implies a bent geometry with a bond angle ~ 120º.
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Final Review Problems Answer Key - Chem 10181 Fall 2007...

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