Mid-term III

Mid-term III April 14, 2023 Name______________________ Chem. 450 1 liter=0.001 m3 1 L•bar = 100 J 1 L = 1000 cm3 R=8.314 J/K*mol; R=8.314 x 10-2 L*bar/K*mol; R=8.206 x 10-2 L*atm/K*mol k=1.38 x 10-23 J/K Na=6.022 x 1023/mol mixG = nRT ( Aln( A) + Bln( B) + A B) regular solution theory nAd A + nBd B + nCd C + … = 0 at constant T & p Gibbs Duhem Eqn. mixS = mix p, n(a), n(b) pV=NkT pV=nRT U q w H U pV dwrev=-pdV dqrev=TdS w Cp CV Vf Vi p ex dV H T p U T V ln(1- B) = (1/T - 1/T* vapH/R) Tf=Ti(Vi/Vf)1/c (adiabatic change) c=cv,m/R F=C–P+2 pV =constant whereby =(Cp,m/CV,m) S=qrev/T Cp,m-Cv,m=R – A= U-T S dG = Vdp -SdT dS-dq/T 0 (Clausius Inequality) G , Sm , Vm n T,p T p p T mixS = -nR(x1ln(x1) + x2ln(x2)) and mixG = nRT(x1ln(x1) + x2ln(x2)) both for ideal mixing S=RTln(Vf/Vi) S=nCp,mln(Tf/Ti) or S=nCv,mln(Tf/Ti) * j = j + RTln(xj) for an ideal solution n l = n l (lever rule) d[ln(p)]/dT= vapH/RT2 (Clausius Clapeyron) dp/dT= transS/ transV (Clapeyron Eqn.) and trnsS= trnsH/Ttrans * + (T-T* fusH/T* fusV) (solid liquid phase boundary) T = kf b, T = Kb b, and PB = bmKb p = p*e (1) (12 points) The pressure-temperature phase diagram for CO2 is fairly typical in that the solid/liquid phase boundary has a positive slope. By contrast, water is anomalous because the solid/liquid phase boundary has a negative slope. Using the appropriate equation as part of your justification, explain the reason one can obtain both a positive and negative slope. One can have both a positive and negative slope due to increasing and decreasing pressure. Increasing pressure leads to the positive slope that CO2 has and decreasing pressure leads to the negative slope has, so depending on whether the pressure is increasing or decreasing can determine the slope for the solid-liquid boundary. dp ¢transH = dT T ¢transV This equation determines the slope of the line by taking the change of pressure over the change in temperature to determine the slope of the line. (2) (12 points) Water in the liquid phase is heated at 1 bar from 45 °C to 75 °C. By how much does its chemical potential change? You may assume that the standard molar entropy of liquid water is 69.9 J/mol K over the entire temperature range. Δµ ΔH TOS ΔH 9 ΔH MCΔT Δ 1144184 30K Δµ 125.4 298 74.15 Δµ 21.971k 5 50 Rin 69.9 8.3141n 69.9 4.25 5 74.15J (3) Short answer (6 points each, total of 18 points) (3a) What are the maximum number of phases that can be in equilibrium in a four component system? F C P Z F 4 P Z F 6 p A total of 6 phases can be in equilibrium (3b) The wider significance of the chemical potential, µ, is that it: (circle the correct answer) (i) Provides a framework for understanding free energy changes as a function of p, T. (ii) Allows scientists to calculate p and T from the free energy of the system. (iii) Shows how the extensive thermodynamic properties U, H, A, and G depend on the composition of the system with different constraints. (iv) Can be used to predict dU = dq + dw (3c) Draw a pressure – area phase diagram for pure water. Please label the solid, liquid, and gas regions of the diagram. Next, a small quantity of a non-volatile additive is introduced that does not partition into the solid phase upon freezing. On the same diagram, depict the change in areas of the solid, liquid, and gas phase regions after the additive has been introduced. change induced liquid pressure solid gas temperature (4) (15 points) Consider an experimental system that consists of two compartments. Pure water is placed into the first compartment. Water with a small amount of protein is placed into the second compartment. A porous membrane separates the two compartments such that water can pass through the membrane, but the protein cannot. (a) Which way does water flow if the pressure in the two compartments is identical? Justify your answer using the appropriate equation. The water ows towards the compartment with pure water because for ideal mixing it would need a 1 to 1 ratio which could be obtained in pure water but not in the water with the protein because the mole fraction for that one would be di erent. (b) Consider the compartment that contains the protein. Is the chemical potential of the water in this compartment higher, lower, or the same if the protein forms dimers compared to conditions where it does not? Explain your answer. It would be lower because dimers would create a higher mole fraction which in turn leads to a lower delta G value and since the partial fraction of delta G equals the chemical potential then the lower the delta G the lower the chemical potential. (5a) (8 points) Consider a two component system that contains 2 moles of A and 10 moles of B. The introduction of an infinitesimal additional amount of A into this system at constant temperature and pressure is associated with d A = 23 J/mol. What is the value of d B under these circumstances? duana AM dMBhB dMAha MB 2311m01 2mol 10mol dm 4.6 J mol (5b) (8 points) The Henry’s law constant for CO2(g) in water at 298 K is KB = 30.1 bar Kg/mol. Find the number of moles of CO2(g) that will dissolve in 1 Kg of water if the pressure of CO2(g) is set to 3.25 bar over the water. s s 30.1barm.gg 325bar 5 9 26 91m01 (6) (13 points) Consider the mixing of two gases A & B. Initially, 2 moles of A are placed in a first compartment that is 20.5 L in volume. Also, 2 moles of B are placed in a second compartment that is also 20.5 L in volume. Entropically, the gases mix ideally, but mixH = 31.6 kJ. Find mixS and mixG under these overall nonideal mixing conditions if the temperature of the system is 298 K. Will the system undergo spontaneous mixing? ΔmixS NR x In x 2 8.314 0.5In 0.5 Δmix5 5.76J ΔmixG ΔmixH T ΔmixS 31.600 J 298 5.760 31,600 1716.48 ΔmixG 29.9Ks non spontaneous mixing (7) (14 points) Calculate the Gibbs energy, entropy, and enthalpy of mixing when 1.00 mole of n-hexane is mixed with 3.00 moles of n-heptane at 298 K. You may assume that the mixing is ideal. 3 0 1.00 hex mix ΔmixS 0.25 sina.es 2.88J ΔmixH 0 ΔmixG ΔmixH TΔmixs 0 ΔmixG 298712.88 858J Xnep 3.08 00 0.75 Δ's ΔmixS 5.38 Δmix 0 sina.is ΔmixG ΔmixH TΔmxs 0 298 5.38 ΔmixG 1.6KJ