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Unformatted text preview: Chemistry 131, Fall 2007 Solutions to Homework No. 5 Problem 1. See Levine, problem 6.2. (a) True. This one of the two properties of an ideal gas mix- ture (Levine, top of p. 176), the other property being that the ideal gas equation applies to the mixture. (b) True. Follows from the expression for the chemical po- tential of a pure ideal gas (Levine eq. 6.2), which is plotted in Levine Fig 6.1. (c) True. Each of the extensive properties U , H , S , G , and C P of an ideal gas mixture can be found by adding the same property calculated for each pure gas at the same tempera- ture and volume as the mixture (Levine, bottom of p. 176). The definition of ideal gas mixture implies that the pressure of each pure gas is equal the partial pressure of the gas in the mixture when the temperature and volume of each pure gas are the same as those of the mixture. So statement (c) is true even though it does not explicitly define the pressure of each pure gas. Problem 3. See Levine, problem 6.9. (a) True. Each factor ( P i, eq /P ) i in the product giving the standard pressure equilibrium constant K P (Levine eq. 6.13) is the quotient of two pressures elevated to some power i . The units of pressure in each factor cancel out so the final product is dimensionless. (b) False. In this case, each factor in the product giving the pressure equilibrium constant K P (Levine eq. 6.19) has di- mensions of pressure, so the whole product is dimensionless only if the sum of the exponents is zero, i.e., if i i = 0 (Levine p. 179), in which case the units of pressure cancel out in the overall product rather than in each factor, as was the case in part (a). (c) False. See part (b). (d) False. In the reverse reaction, the roles of reactacts and products are exchanged, so the signs of the stoichiometric coefficients i (Levine p. 132) are reversed. Reversing the signs of the i s effectively swaps the numerator with the denominator in the product giving K P (Levine eq. 6.13). Therefore, K P for the reverse reaction is the reciprocal of K P for the forward reaction. (e) True. See part (d). (f) False. Each stoichiometric coefficient i appears as the exponent of a ratio of partial pressure to standard pres- sure, so doubling each i effectively raises the standard pressure equilibrium constant K P to the power of 2. For example, for the reaction 2 A * B the equilibrium con- stant is K P = ( P B /P ) / ( P A /P ) 2 , but for 4 A * 2 B the equilibrium constant is K P = ( P B /P ) 2 / ( P A /P ) 4 = [( P B /P ) / ( P A /P ) 2 ] 2 . (g) True. See part (f). (h) True. The ideal-gas equilibrium constant depends on tem- perature only (Levine p. 182). The temperature dependence of K P is described by the vant Hoff equation (Levine eq....
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