Intro chem - Chem 5 Winter, 2005 Practice Exam 3 (the...

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Chem 5 Winter, 2005 Practice Exam 3 (the actual third exam last year) Page 1 of 7 1. (7 + 7 points) Benzene, C 6 H 6 , is a volatile liquid at room temperature, meaning it has an appreciable vapor pressure. Listed below are thermodynamic data on both the liquid and gaseous forms of benzene at 298.15 K. C 6 H 6 ( l )C 6 H 6 ( g ) H f o /kJ mol –1 49.03 82.93 S o /J mol –1 K –1 172.8 269.2 G f o /kJ mol –1 124.50 129.66 Use these data (assuming they are independent of temperature) to answer the following questions. ( Big hint: consider C 6 H 6 ( l ) C 6 H 6 ( g ) to be a “reaction” at equilibrium.) (a) What is the vapor pressure of benzene at 298.15 K, the temperature at which the data above hold? If we think of this vaporization as a “reaction,” we can write, where P C 6 H 6 is the vapor pressure, K = P C 6 H 6 = e G r o / RT We can find G r o most directly from the G f o values: G r o = G f o ( g ) G f o ( l ) = 129.66 kJ mol –1 – 124.50 kJ mol –1 = 5.16 kJ mol –1 and then find the vapor pressure: P C 6 H 6 = e G r o / = e –(5160 J mol –1 )/(8.314 J mol –1 K –1 )(298.15 K) = 0.125 atm (b) What is the normal boiling point of benzene, i.e., the temperature at which C 6 H 6 ( g ) has a vapor pressure of 1 atm? There are two different ways to think about and solve this problem. First, we recall that at 1 atm, we are at the standard pressure (the pressure to which our tabulated thermodynamic data refer) and when the “reaction” (the vaporization) is in equilibrium, that is because the total free energy of the reactants equals the total free energy of the products, making G = 0. Thus, we can write, with T vap o being the normal boiling point (i.e., the “standard vaporization temperature”) G = 0 = G r o = H r o T vap o S r o or T vap o = H r o S r o We find H r o from the enthalpy of formation values, and we find S r o from the absolute entropies: H r o = (82.93 – 49.03) kJ mo l –1 = 33.90 kJ mol –1 S r o = (269.2 172.8) J mol –1 K –1 = 96.4 J mol –1 K –1 Finally, we calculate T vap o = (33,900 J mol –1 )/(96.4 J mol –1 K –1 ) = 351.7 K = 78.5 °C. The second way to approach this problem considers the temperature and vapor pressure from part (a) as one pair of T , K values and seeks a second pair where T = T vap o and K = 1 (the value of the equilibrium constant for a partial pressure of 1 atm). These pairs of T , K values are related through the expression
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Chem 5 Winter, 2005 Practice Exam 3 (the actual third exam last year) Page 2 of 7 ln K 2 K 1 = H r o R 1 T 2 1 T 1 We know H r o , and if we let the values from part (a) represent T 1 and K 1 , then we can write ln 1 0.125 = 33,900 J mo l –1 8.315 J mol –1 K –1 1 T vap o 1 298.15 K We solve this expression for T vap o and find T vap o = 351.6 K = 78.4 °C, essentially the same value we found using the other approach. 2. (15 points) Magnesium metal, Mg( s ), forms a dipositive cation in solution with the following standard reduction potential: Mg 2+ ( aq ) + 2e Mg( s ) E red o = –2.363 V The standard reduction potential for solid Mg(OH) 2 is also known: Mg(OH) 2 ( s ) + 2e Mg( s ) + 2 OH ( aq ) E o = –2.690 V
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This note was uploaded on 05/01/2008 for the course CHEM 5/6 taught by Professor Gribble during the Spring '08 term at Dartmouth.

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Intro chem - Chem 5 Winter, 2005 Practice Exam 3 (the...

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