( g mol 1 )( J mol 1 K 1

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1 Chem 4501 Introduction to Thermodynamics, 3 Credits Kinetics, and Statistical Mechanics Fall Semester 2017 Homework Problem Set Number 11 Solutions 1. McQuarrie and Simon, Paraphrase: If a solution of 1.47 g of dichlorobenzene in 50 g of benzene boils at C and 1 bar, what is the molecular weight of dichlorobenzene if the boiling point of pure benzene is C and the heat of vaporization 32.0 kj mol 1? Well, this problem would be a bit more interesting if we weren t told the name of the solute. It is certainly easier to compute the molecular weight of dichlorobenzene from its formula, C6H4Cl2 (147 g mol 1 )! But, in the spirit of the exercise, we use eqs and to write ΔT vap = M g kg 1 * R( T vap ) 2 Δ vap H m Rearranging to solve for m, plugging in the values given in the problem, and computing the molecular weight of benzene as g mol 1, we have m = ( 0.51K) = mol kg 1 ( 1000 g kg 1 )( 32,000 J mol 1) ( g mol 1 )( J mol 1 K 1 ) K ( ) 2

2 So, the solution is 0.2 molal. Given that it is only 0.05 kg solvent, it must contain 0.01 mol. Thus, the molecular weight is g 0.01 mol = 147 g mol 1, as expected. 2. McQuarrie and Simon, Paraphrase: Sketch a phase diagram showing colligative effects on boiling and melting. Taken from firstyrchemcolligativeindex.htm 3. McQuarrie and Simon, Paraphrase: If a solution of 100 ml of water containing 800 mg of a protein (wow! that s a lot of protein) has an osmotic pressure of 2.06 torr at 25 C, what is the molecular weight of the protein?

3 Another colligative properties problem, but this time a bit more interesting as we don t know the answer more trivially. So, we use eq rearranged (see also example 11-7) to write c = Π RT ( 2.06 torr) 1 atm 760 torr = ( L atm mol 1 K 1 ) K = mol L 1 ( ) ( ) As we re working with 100 ml (0.1 L), the total number of moles in the solution is x So, the molecular weight is 0.8 g x 10 5 mol = 72,210 g mol 1 (or about 72 kd, where kd is kilodaltons, a common unit used for biopolymer molecular weights). 4. McQuarrie and Simon, Paraphrase: If the water activity in a 2 molal sucrose solution is , what external pressure will need to be applied at 25 C to make the activity of the water equal to 1 (i.e., pure water; thus, at this pressure, pure water could be forced through a semiporous membrane). Assume a water density of g ml 1. Now we use eq rearranged to write Π = RT ln a 1 * V 1 ( L atm mol 1 K 1 ) K = $ ( g mol 1 1 ' )( L ml 1 ) % g ml 1 ) ( = 58 atm ( )( ln )

4 which is a lot of pressure! 5. McQuarrie and Simon, Paraphrase: When 40.7 g of the salt HgCl 2 is dissolved in 100 g water, the freezing point depression is observed to be 2.83 K. Based on this, does one expect the solution to readily conduct electricity? The freezing point depression lets us compute the concentration of dissolved things in solution. Using m = ΔTfus Kf and the value for Kf of 1.86 K kg mol 1 found in the textbook, we know that m is 2.83 K 1.86 K kg mol 1 = molal. As there are 100 g of water (0.1 kg), the dissolution of the HgCl2 has resulted in moles of things dissolved. Now, the molecular weight of HgCl2 is g mol 1, so 40.7 g would correspond to moles of solute before any separation of ions might occur. Evidently, then, less than 2% of the HgCl2 dissociates into ions, suggesting that this solution will be at best a poor conductor of electricity. 6. McQuarrie and Simon, Paraphrase: Derive the temperature dependence of activities. Start from the equation for the chemical potential in solution µ = µ * + RT ln a ln a = µ µ* RT If we now take the temperature derivative of the log of the activity we may use the Gibbs-Helmholtz equation to determine

5 T ln a = 1 $ µ µ * ' R T % T ) ( = 1 $ R H T 2 + H *' % T 2 ) ( = H * H RT 2 Note that the denominator, at room temperature, will be J mol 1 K 1 times (300 K) 2, or about 750 kj mol 1 K. So, the difference in molar enthalpies of a solution and its pure liquid must be divided by 750 kj mol 1 to get the per degree variation in the activity. The variation in the molar enthalpy would be expected to be much, much less than 750 kj mol 1, so activities are expected to have rather weak temperature dependence, at least at room temperature. 7. McQuarrie and Simon, Paraphrase: In the gaseous equilibrium between N 2 O 4 and 2 equivalents of NO 2, derive an expression for the extent of reaction ξ at equilibrium and assess the variation in ξ eq as a function of the total pressure P eq. Note that K P is 6.1 (1 bar standard state). If we start with n0 moles of N2O4, then at equilibrium we will have n0 ξ eq moles of N2O4 and 2 ξ eq moles of NO 2. The total pressure P will be equal to the partial pressure of N2O4, plus the partial pressure of NO2. If both gases are taken to be ideal, the individual partial pressures will simply be (nxrt V) for each gas x. Using this in the pressure-dependent equilibrium constant, we have

6 2 P NO2 ( ) 2 RT V ( ) RT V K P = = 2ξ eq P N2 O 4 n 0 ξ eq ( ) 2 ( n 0 ξ eq ) ( ) 2 ( ) = 2ξ eq ( RT V ) The total pressure P at equilibrium will be the total number of moles times (RT V), and that total number is n0 ξ eq + 2 ξ eq = n0 + ξ eq. So, we can rewrite the above equation as ( 2ξ eq) 2 K P = ( n 0 ξ eq ) = 2ξ eq n 0 ξ eq ( RT V ) ( ) 2 P ( ) ( n 0 + ξ eq ) = 4ξ eq 2 2 n 0 2 P ξeq The percentage progress at equilibrium for this reaction is measured as (ξ eq n 0 ), so we can multiply both numerator and denominator of the final expression above by (1 n 0 ) 2 to obtain ( ) 2 ( ) 2 P K P = 4 ξ eq n 0 1 ξ eq n 0 and, if we solve for (ξ eq n 0 ), we have ( ξ eq n 0 ) = K P K P + 4P

7 Note that if we decrease the pressure to near zero (by increasing the volume to infinity, for example), the percentage of reaction goes to 1, i.e., all N2O4 is converted to NO2. If, on the other hand, we make the pressure arbitrarily high (by decreasing the volume), the conversion percentage drops to zero and no NO2 is produced. This reflects Le Châtelier s principle: pressure shifts the equilibrium to the side with fewer particles. 8. McQuarrie and Simon, Paraphrase: Use experimental standard free energies of formation to compute standard free energies of reaction and equilibrium constants at 298 K. The specific cases in this problem are always solved using the general formula ΔG r o = products ( ) n o j ΔG f j o n i ΔG f j j reactants where n is the number of moles of the species in the balanced chemical reaction. The corresponding equilibrium constant is computed as K P = e ΔG r RT i ( ) The answers are (a) ΔG o r = kj mol 1, KP = 0.148; (b) ΔG o r = kj mol 1, KP = 690; (c) ΔG o r = kj mol 1, KP = 481, McQuarrie and Simon, Paraphrase: If K P for the water gas shift reaction ( H 2 + CO 2 D CO + H 2 O ) is 1.59, is the reaction at equilibrium if the partial pressures of the gases from left to right are 0.55, 0.20, 1.25, and

8 0.10 bar, respectively. If not at equilibrium, in which direction will the reaction proceed under these conditions? To answer we compute the reaction quotient Q P = P CO ( )( P H2 O) ( P H2 )( P CO2 ) = ( )( 0.10) ( )( 0.20) =1.14 The free energy of reaction (not the standard-state free energy of reaction!) is computed as (eq ) ΔG r = RT ln Q # P = RT ln 1.14 % ( K P $ 1.59' As the argument of the logarithm is less than 1, the free energy of reaction is negative, and products are formed spontaneously from reactants. 10. McQuarrie and Simon, Paraphrase: Derive the statistical mechanical expression for the equilibrium constant for the reaction 2 HBr D H 2 + Br 2 and compute K P for temperatures 500, 1000, 1500, and 2000 K. Plot ln K against 1 T to determine the standard enthalpy of reaction by the van t Hoff relation. To answer we compute the equilibrium constant as (eq )

9 ( )( q Br2 V ) K P = q H 2 V " o % c RT ( q HBr V ) 2 $ # P o ' ( )( q Br2 V ) = q H 2 V ( q HBr V ) 2 ( ) The general expression for q V for a diatomic is q A V = # 2πm A k B T % ( $ ' h # T # 1 % A (% $ σ A Θ rot ' $ 1 e Θ vib A A T ( ed 0 RT ' So, inserting this expression in as needed, and eliminating a number of constants that drop out, we derive * " K P = m H 2 m % 3 2" 2 % Br2 σ, HBr, $ 2 ' $ # m HBr # σ H2 σ ' Br2, +, * " 1 e Θ vib HBr T %, 2 $ ' #,," 1 e Θ H2 vib T %" $ ' 1 e Θ vib, $ + # # HBr ( Θ rot ) 2 H Θ 2 Br rot Θ 2 rot Br2 T - % '. -. ( ) RT e D H 2 Br 0 +D 2 HBr 0 2D 0 While this equation looks mildly imposing, typing it carefully into a spreadsheet with all the relevant constants permits simple evaluation at different temperatures T (note that only the last two terms depend on T). Note also that McQuarrie and Simon fail to mention that one should not really use natural abundance averaged isotopic

10 masses in this equation. Instead, there will be a different value of KP for every possible choice of isotopes, and measuring such equilibrium isotope effects can be very useful in understanding some chemical reactions. We won t go more deeply into this point, however. For this problem, we ll simply choose to use masses for the H and Br atoms of 1 and 79, respectively, these being the most abundant isotopes. For the temperature-independent first three terms, the net contribution to KP is So, these terms disfavor HBr disproportionating into H2 and Br2. Including the remaining temperature-dependent terms, one obtains the following values for KP: 1.15 x 10 12, 4.31 x 10 7, 3.35 x 10 5, and 2.98 x 10 4 at T = 500, 1000, 1500, and 2000 K, respectively. Plotting ln KP against 1 T gives a slope of K 1. When multiplied by R in order to compute the standard enthalpy of reaction, one has kj mol 1. This is in very good agreement with the experimental value of kj mol 1 ; the deviation is attributable to some breakdown of the rigidrotator-harmonic-oscillator approximation at very high temperatures. Note that the MS solutions manual must have a mistake, as they report rather different numbers for KP and determine a rather bad standard enthalpy of reaction of 98.8 kj mol 1. The source of their error is not obvious to me. However, there is a typo in their final equation where they have positive arguments of the exponentials in the vibrational terms, instead of negative arguments. I get closer to their reported values when I introduce that error, but still not the same. Ah well