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Unformatted text preview: Chemistry 132, Winter 2008 Solutions to Homework No. 1 Problem 1. See Levine problem 11.1. (a) True. The activity of component i in a reaction mixture is defined in terms of an exponential function, which is di mensionless: a i = e ( μ i μ ◦ i ) /RT (Levine eq. 11.1). (b) True. The chemical potential of the standard state is μ i = μ ◦ i , therefore the activity of the standard state is a i = e ( μ ◦ i μ ◦ i ) /RT = e = 1. Problem 4. See Levine problem 11.12. Use m = 0 . 40 mol / kg for the stoichiometric molality of NaCl. Assume that there is no ion pairing, so that m (Na + ) = m (Cl ) = m . Since water is the solvent, the mole fraction scale is used for the activity of water, so that a (H 2 O) = γ x (H 2 O) x (H 2 O) (Levine p. 321). Since water is uncharged, γ x (H 2 O) can be assumed to be 1. Then a (H 2 O) = x (H 2 O), i.e., the activity of water is just the mole fraction of water, which can be calculated as follows: x (H 2 O) = m (H 2 O) m (H 2 O) + m (Na + ) + m (Cl ) = m w m w + 2 m . The stoichiometric molality of water, m w , is the reciprocal of the molecular mass of water: m w = 1000 g / kg 18 . 01528 g / mol = 55 . 5084 mol / kg . Therefore a (H 2 O) = x (H 2 O) = 55 . 5084 55 . 5084 + 2(0 . 40) = 0 . 986 . Problem 7. See Levine problem 11.25. The solvation reaction for this problem is CaF 2 ( s ) * Ca 2+ ( aq ) + 2F ( aq ) . Thus ν + = 1, ν = 2, z + = 2, z = 1. Let x be the amount of dissolved CaF 2 , so that m + = m (Ca 2+ ) = x and m = m (F ) = 2 x . Then the equilibrium relation (Levine eq. 11.27) can be written as: K sp = ( γ ± ) ν + + ν ( m + ) ν + ( m ) ν = γ 3 ± x 1 (2 x ) 2 = 4 x 3 γ 3 ± ⇒ x = 1 γ ± K sp 4 1 / 3 . To calculate x : 1) assume initially that γ ± = 1; 2) cal culate x using the above equation; 3) calculate the ionic strength I m (Levine eq. 10.62); 4) calculate γ ± using the Davies equation (Levine eq. 10.71); 5) repeat the above steps until the change in x becomes sufficiently small. Fig ure 2 shows an Excel spreadsheet that was used to calculate x = m (Ca 2+ ) = 1 . 118 × 10 4 mol / kg according to the fore going procedure. Figure 1: Excel spreadsheet for Levine problem 11.25 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 A B C D E F z + 2 ν + 1 z −  1 ν − 2 K sp 3.20E11 ν 3 iteration γ ± m (Ca 2+ ) m (F − ) I m 1 1.00000 2.000E04 4.000E04 6.000E04 2 0.94579 2.115E04 4.229E04 6.344E04 3 0.94435 2.118E04 4.236E04 6.354E04 4 0.94431 2.118E04 4.236E04 6.354E04 cell formula C7 =($B$3/4)^(1/3)/B7 D7 =2*C7 E7 =0.5*($B$1^2*C7+$B$2^2*D7) B8 =10^(0.51*$B$1*$B$2*(1/(1+SQRT(1/E7))0.3*E7)) Problem 10. See Levine problem 11.32. Use Levine eq 11.25 and assume that the gas behaves ideally: K ◦ ≈ P [CO 2 ] P ◦ = RTn [CO 2 ] V P ◦ = RT V P ◦ ( n [CO 2 ] + ξ ) = 1 P ◦ P [CO 2 ] + RTξ V ⇒ ξ = P ◦ K ◦ P [CO 2 ] RT/V , where ξ is the extent of reaction and P [CO 2 ] is the initial pressure of carbon dioxide. A positive value of ξ indicates...
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This note was uploaded on 05/22/2008 for the course CHEM 132 taught by Professor Lindenberg during the Winter '08 term at UCSD.
 Winter '08
 Lindenberg
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