HW Sol#6

HW Sol#6 - Chemistry 131 Fall 2007 Solutions to Homework No...

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Unformatted text preview: Chemistry 131, Fall 2007 Solutions to Homework No. 6 Problem 1. See Levine, problem 7.2. Use the phase rule for systems with no reactions, f = 2- p + c (Levine eq. 7.7). (a) H 2 O( l ) + sucrose( aq ): p = 1 (aqueous solution), c = 2 (water, sucrose), f = 2- 1 + 2 = 3. Can use T , P , and x sucrose as independent variables. (b) H 2 O( l ) + sucrose( aq ) + ribose( aq ): p = 1 (aqueous solu- tion), c = 3 (water, sucrose, ribose), f = 2- 1 + 3 = 4. Can use T , P , x sucrose , x ribose , as independent variables. (c) H 2 O( l ) + sucrose( aq ) + ribose( aq ) + sucrose( s ): p = 2 (aqueous solution, solid sucrose), c = 3 (water, sucrose, ribose), f = 2- 2 + 3 = 3. Can use T , P , and x ribose as independent variables. (d) H 2 O( l ) + sucrose( aq ) + ribose( aq ) + sucrose( s ) + ri- bose( s ): p = 2 (aqueous solution, solid sucrose, solid ribose), c = 3 (water, sucrose, ribose), f = 2- 3 + 3 = 2. Can use T and P as independent variables. (e) H 2 O( l )+H 2 O( g ): p = 2 (liquid water, water vapor), c = 1 (water), f = 2- 2+1 = 1. Can use T or P as the independent variable. (f) sucrose( aq ) + H 2 O( g ): p = 2 (aqueous solution, water vapor), c = 2 (water, sucrose), f = 2- 2 + 2 = 2. Can use T and x sucrose or P and x sucrose as independent variables. (g) sucrose( s ) + sucrose( aq ) + H 2 O( g ): p = 3 (solid sucrose, aqueous solution, water vapor), c = 2 (sucrose, water), f = 2- 3 + 2 = 1. Can use T or P as the independent variable. (h) H 2 O( l ) + C 6 H 6 ( l ) + H 2 O( g ) + C 6 H 6 ( g ): p = 3 (liquid water, liquid benzene, gas mixture of water and benzene), c = 2 (water, benzene), f = 2- 3 + 2 = 1. Can use T or P as the independent variable. Problem 3. See Levine, problem 7.8. Use c ind = c- r- a (Levine eq. 7.10) and f = c ind- p + 2 (Levine eq. 7.11). (a) The reactions are H 2 O * H + + OH- and 2H 2 O * (H 2 O) 2 . Also, charge neutrality requires that x (H + ) = x (OH- ). Therefore, p = 1 (liquid), c = 4 (water, hydronium, hydroxide, water dimers), r = 2 (water dissociation, dimer formation), a = 1 (charge neutrality), c ind = 4- 2- 1 = 1, f = 1- 1 + 2 = 2. (b) In this case there is an additional species, (H 2 O) 3 , so c in- creases by one, and an additional reaction, 3H 2 O * (H 2 O) 3 , so r also increases by one. Therefore, c ind and f remain un- changed. This means that the presence of water multimers does not affect the number of independent intensive variables needed to specify the thermodynamic state of pure liquid wa- ter....
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HW Sol#6 - Chemistry 131 Fall 2007 Solutions to Homework No...

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