Exam 3 Solutions

1 b.) Continued2The equilibrium constant expression isK eq =K eq =2P FP !"$#"$#P F2P !%'&%'&2" x F P %$# P ! '&" x F2P %$# P ! '&.The equilibrium constant can be written terms of the extent of reaction,€K eq ==K eq =2" x F P %$# P ! '&" x F2P %$# P ! '&2" 2ξ % " P %$ ' $# 1+ξ & # P ! '&" 1−ξ %$# 1+ξ &' " P %$# P ! '&4ξ 21+ξ( )( 1−ξ)2" P %$ ' .# &P !Substituting the value for the equilibrium constant, equilibrium pressure (P=1.3 bar), and€K eq =0.120 =4ξ 21+ξ( )( 1−ξ)4ξ 21+ξ( )( 1−ξ)$ P '& )% (P !$ 1.3bar '& ).% 1.0 bar (€P ! = 1bar yieldsSolving for the extent of reaction,€0.120 =0.0924 =4ξ 21+ξ( )( 1−ξ)4ξ 21+ξ0.0924 ( 1+ξ) ( 1−ξ) = 4ξ 20.0924 1−ξ 2( ) = 4ξ 20.0924 − 0.0924ξ 2 = 4ξ 2( )( 1−ξ)0.0924 = 4.0924ξ 20.02258 = ξ 2ξ = 0.150.$ 1.3bar '& )% 1.0 bar (€

3.) (14 points) Sketch a typical phase diagram for a one-component system. Make sure to carefully label allthe appropriate sections and points on the phase diagram. Explain the Gibbs phase rule and the role thatit plays in determining the degrees of freedom for each region or feature of the phase diagram.4A typical phase diagram for a one-component system is shown below.Critical PointPSLCoexistence curvesVTriple PointTNote that all phase diagrams of this type are graphs in which pressure is plotted on the y-axis with temperatureon the x-axis.The Gibbs Phase Rule is F = C – Φ + 2, where F is the number of degrees of freedom, C is the number ofcomponents, and Φ is the number of phases. For a one-component system, C = 1, so the phase rule reduces to F= 1 – Φ + 2, or F = 3 – Φ.The regions labeled S, L, and V correspond to regions in which only one phase is present, solid, liquid, orvapor, respectively. In these regions, the Gibbs phase rule tells us that F = 3 – 1 = 2. This means that we arefree to select both P and T in those regions.The regions on the phase diagram corresponding to the coexistence curves are lines along which two phases arein equilibrium, either S-L, S-V, or L-V. The Gibbs phase rule tells us in this case that F = 3 – 2 = 1. Thismeans that if we select the temperature, then the pressure is fixed because it must lie on the coexistence curve.(The same number of degrees of freedom holds for the critical point as well since it lies on the liquid-vaporcoexistence curve.)Finally, at the triple point, all three phases are in equilibrium. The Gibbs phase rule in this case gives F = 3 – 3= 0. This means that neither temperature nor pressure can be selected for a three-phase equilibrium. Both thetemperature and pressure are fixed at the triple point.

4.) (15 points) True/false, short answer, multiple choice.5a.) True or False : The chemical potential provides a measure of the progress of a chemical reaction fromreactants to products.b.) True or False: The equilibrium constant of a particular reaction is determined to be 7.8×10 -4 at 300 K.The magnitude of this equilibrium constant implies that reactants are favored.c.) Short answerThe notion that a chemical system in equilibrium will shift in order to relieve an applied stress is known as_______LeChatelier's Principle___________ .d.) Short answerIn a one-component system, the ________triple point__________ corresponds to solid, liquid, and vaporall in equilibrium.e.) Multiple Choice: For the two graphs shown below, circle the one that exhibits the correct behavior.SVµLµLVST(a)T(b)

5.) (14 points) Nitrogen tetroxide decomposes according to the reaction6N 2 O 4( g)2 NO 2 ( g) .At 55˚C and 1 atm, the average molecular weight of N 2 O 4 and NO 2 present at equilibrium is 61.2 g/mol.Determine the equilibrium € constant. [Note: atomic € mass N = 14.007 g/mol; atomic mass O = 15.999g/mol.]We can use the average molecular weight to determine the equilibrium mole fractions. The equation for theaverage molecular weight M isM = x NO2 M NO2 + x N2 O 4M N2 O 4,€where x is the gas phase mole fraction and M is the molecular weight. Since the sum of the mole fractions mustequal 1, we have€x NO2 + x N2 O 4= 1 ,or€x NO2 = 1 − x N2 O 4.Substituting into the expression for the average molecular weight yieldsThen,€M = ( 1− x N2 O 4)M NO2 + x N2 O 4M N2 O 461.2 g/mol = 1− x N2 O 4( ) 46.005g/mol61.2 = 46.005 x N2 O 4+ 46.00515.195 = 46.005 x N2 O 4x N2 O 4= 0.330.( ) + x N2 O 4( 92.010 g/mol)€x NO2 = 1 − x N2 O 4= 1 − 0.330x NO2 = 0.670.Knowing the mole fractions, we can now determine the equilibrium constant. Treating the gases as ideal gases,the equilibrium constant can be determined as€K eq ="$#"$#P NO2P !P N2 O 4P !2%'&.%'&€

5.) Continued7In terms of mole fractions, the equilibrium constant becomes€K eq ===2" x NO2 P %$# P ! '&" x N2 O 4P %$# P ! '&2x NO2 Px N2 O 4P !K eq = 1.38.( 0.670) 2 1.013bar0.330( )( )( 1bar)Here, we have used the equilibrium pressure, P =1 atm = 1.013 bar.

6.) (14 points) The melting point of mercury is –38.9˚C at 1 bar and –19.9˚C at 3540 bar. The density ofliquid mercury is 13.69 g/mL and the density of solid mercury is 14.19 g/mL. Determine the molarenthalpy of fusion. [The atmoic mass of Hg is 200.59 g/mol.]8The Clapeyron equation for solid-liquid phase equilibrium isdPdT= ΔH fus,mT fus ΔV m.We can approximate the left side of the equation as€dPdT≈ ΔPΔT .Substituting,€ΔPΔT= ΔH fus,mT fus ΔV m.We can now solve for the desired molar enthalpy of fusion,€ΔH fus,m = ΔPΔT ⋅T fus ΔV m .The molar volume of each phase can be calculated from the molecular weight M and the density D. For solidmercury,€V s,m = M D ⋅# 1L &% ($ 1000 mL '# 200.59 g mol −1 & # 1L &= %$ 14.19 g mL −1 ( ⋅%(' $ 1000 mL 'V s,m = 0.01414 L/mol.For liquid mercury, the molar volume is€V l,m = M D ⋅# 1L &% ($ 1000 mL '# 200.59 g mol −1 & # 1L &= %$ 13.69 g mL −1 ( ⋅%(' $ 1000 mL 'V l,m = 0.01465L/mol.Substituting the information for the solid to liquid phase transition,€# ΔP &ΔH fus,m = % ( T fus ΔV m$ ΔT '# 3540 −1 bar &= %( 234.25 K$ 253.25− 234.25 K '# 100 J &= 22.53 Lbar/mol % ($ 1 Lbar 'ΔH fus,m = 2253 J/mol .( )( 0.01465− 0.01414 L/mol)€

7.) (15 points) True/false, short answer, multiple choice.9a.)True or False: For two phases to be in equilibrium, their chemical potentials must be equal.b.) True or False : The reaction CO (g) + 3 H 2 (g)to the right if carbon monoxide is removed from the system.CH 4 (g) + H 2 O (g) is expected to shiftc.) Short answerThe _______Clausius-Clapeyron_____________ Equation gives an expression for the vapor pressure ofa solid or liquid in a one-component system.d.) Short answerThe ________activity___________ of a solid or liquid takes a value of approximately unity in theequilibrium constant expression for a heterogeneous reaction.e.) Multiple Choice: The equilibrium constant for the reaction( )2ClO gCl 2 O 2 g( ) is2.48 ×10 11 at200 K. The standard molar enthalpy of reaction is –72.4 kJ/mol. If the temperature is increased to 500 K,the equilibrium constant is expected to:€€ €(a) increase.(b) decrease.(c) remain the same.