Exam 3 Solutions. ClO g. At 200 K and a total pressure of 1.0 bar, the partial pressure ratio for the chlorine-containing compounds is p ClO2

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1 Chemistry 360 Dr. Jean M. Standard Fall 2016 Name KEY Exam 3 Solutions 1.) (14 points) Consider the gas phase decomposition of chlorine dioxide, ClO 2, ClO 2 ( g) ClO ( g) + O ( g). At 200 K and a total pressure of 1.0 bar, the partial pressure ratio for the chlorine-containing compounds is p ClO2 / p ClO Determine the extent of reaction and equilibrium constant for the reaction at 200 K, assuming that there is 1 mole of ClO 2 initially present in the system. We first need to determine the mole fractions at equilibrium and the equilibrium constant expression. Starting from 1 mole of ClO 2 initially, we have ( ) ( ) + O ( g). ClO 2 g ClO g molesinit molesreacted molesequil. 1 x i The equilibrium constant is P ClO P P O P P ClO2 P P x P O P P 2 P P x O 2 P P. Substituting the mole fractions expressed in terms of the extent of reaction yields x O 2 P P P 1 P. 1+ Simplifying the expression above (not a required step but helpful later) leads to the result 2 ( 1+) 1 ( ) P P P P. The information in the problem does not include the value of the equilibrium constant; however, the ratio p ClO2 / p ClO is provided.

2 1. Continued 2 Expressing the ratio p ClO2 / p ClO in terms of the extent of reaction will allow us to determine, Solving for, p ClO2 p ClO p ClO2 p ClO 2 P P p ClO2 p ClO Finally, the equilibrium constant may be calculated by substitution, 2 ( 1+) P P ( ) P P bar 2 1 bar P P.

3 3 2.) (14 points) Sketch a temperature-composition (T-x) diagram for a binary liquid-vapor system with two components, A and B, that form ideal solutions. Make sure to label the axes as well as the various phases present in each region, along with the bubble and dew point lines. In addition, you should label the boiling points of the pure components, T A and TB. Assume that component A is the more volatile component. For any temperature T lying between the two pure boiling points, draw a tie line and describe its purpose. Discuss whether or not you expect the vapor phase mole fraction to be enriched or depleted in component A relative to the liquid phase. A sketch of a T-x diagram is shown below. The vapor phase lies in the upper part of the diagram, at higher temperatures. The liquid phase lies in the lower part of the diagram, at lower temperatures. The liquid-vapor equilibrium region is in between. The boundaries of the liquid-vapor equilibrium region are the dew point and bubble point lines. The points labeled T A and T B are the boiling points (vaporization temperatures) of pure components A and B, respectively. The more volatile component in this case, A, is the lower boiling component. V T B Dew Point L + V Tie Line Bubble Point T A L x A y A 0 1 x A, y A A tie line in the region of liquid-vapor equilibrium is shown in red. The end points label the compositions of the liquid and vapor phases. The composition of the liquid phase, x A, comes from the intersection of the tie line with the bubble point line, shown in this case as the dashed line on the left of the diagram. The composition of the liquid phase, y A, comes from the intersection of the tie line with the dew point line, shown in this case as the dashed line on the right of the diagram. The vapor phase will be enriched in the more volatile component. Thus, since A is more volatile than B in this case, we would expect to find more A in the vapor phase, such that y A > x A, as seen above.

4 3.) (14 points) The melting point of mercury is 38.9 C at 1 bar and 19.9 C at 3540 bar. The density of liquid mercury is g/ml and the density of solid mercury is g/ml. Determine the molar enthalpy of fusion. [The atomic mass of Hg is g/mol.] 4 The Clapeyron equation for solid-liquid phase equilibrium is dp dt ΔH fus,m T fus ΔV m. We can approximate the left side of the equation as dp dt ΔP ΔT. Substituting, ΔP ΔT ΔH fus,m T fus ΔV m. We can now solve for the desired molar enthalpy of fusion, ΔH fus,m ΔP ΔT T fus ΔV m. The molar volume of each phase can be calculated from the molecular weight M and the density D. For solid mercury, For liquid mercury, the molar volume is V s,m M D # 1L & % ( $ 1000 ml ' # g mol 1 & # 1L & % $ g ml 1 ( % ( ' $ 1000 ml ' V s,m L/mol. V l,m M D # 1L & % ( $ 1000 ml ' # g mol 1 & # 1L & % $ g ml 1 ( % ( ' $ 1000 ml ' V l,m L/mol. Substituting the information for the solid to liquid phase transition, # ΔP & ΔH fus,m % ( T fus ΔV m $ ΔT ' # bar & % ( K $ K ' # 100 J & Lbar/mol % ( $ 1 Lbar ' ΔH fus,m 2253 J/mol. ( )( L/mol)

5 4.) (15 points) True/false, short answer, multiple choice. 5 a.) True or False : The Clapeyron equation is valid only for the solid-liquid phase coexistence curve in one-component systems. b.) True or False: The reaction CO (g) + 3 H 2 (g) CH 4 (g) + H 2 O (g) is expected to shift to the left if hydrogen gas is removed from the system. c.) Short answer A maximum or minimum boiling solution for a binary liquid-vapor system is referred to as an azeotrope. d.) Short answer The van't Hoff Equation gives the temperature dependence of the equilibrium constant. e.) Multiple Choice For the three graphs shown below, circle the one that exhibits the correct behavior for a typical reaction which reaches chemical equilibrium. G G G (a) (b) (c)

6 6 5.) (14 points) Mixtures of benzene and cyclohexane exhibit ideal behavior. A solution was created containing 1.5 moles of liquid benzene and 2.5 moles of liquid cyclohexane, and at 50ºC the total vapor pressure of the solution was measured to be 340 torr. Another solution was created containing 1.5 moles of liquid benzene and 3.5 moles of liquid cyclohexane, and at 50ºC the measured total vapor pressure was 370 torr. Calculate the vapor pressures of pure benzene and pure cyclohexane at 50ºC. The ideal solution can be described by Raoult's Law, P i x i P i. The total pressure of the mixture is given by the sum of partial pressures, P P b + P c or P x b P b + x c P c. Here b refers to benzene and c refers to cyclohexane. We are given the total pressure for two different liquid phase compositions, and are asked to determine the pure vapor pressures. In the first case, the mole fractions are x b x b mol 1.5mol + 2.5mol and x c 1 x b x c In the second case, the mole fractions are x b x b mol 1.5mol + 3.5mol and x c 1 x c x c The total pressure, given by P x b P b + x c P c, in the first case is 340 torr 0.375P b P c. The total pressure in the second case is 370 torr P b P c.

7 5.) Continued 7 There are two equations and two unknowns. One way to solve this is to solve the first equation for P b and substitute into the second equation. So, solving the first equation for P b yields P b or P b P c P c. Substituting this result into the second equation leads to the vapor pressure of pure cyclohexane, P b P c ( ) P c 370 ( 0.300) P c P c P c P c P c 490 torr. From the first equation, we had P b P c. Substituting the vapor pressure of pure cyclohexane allows us to determine the vapor pressure of pure benzene, P b P b P b P c ( 490 torr) 90 torr.

8 6.) (14 points) Chlorine monoxide, ClO, a key component in the destruction of the ozone layer in the polar stratospheric regions, dimerizes to form Cl 2 O 2 according to the reaction 8 2ClO( g) Cl 2 O 2 ( g). At 200 K and a total pressure of 1 bar, ΔH R kj/mol and ΔS R 144 J mol 1 K 1. (a.) Determine the standard molar Gibbs free energy ΔG R bar. and equilibrium constant at 200 K and 1 We can use the equation ΔG R Substituting, ΔH R TΔS R. to calculate the standard molar Gibbs free energy. ΔG R ΔG R ΔH R TΔS R Jmol 1 ( 200 K) ( 144 Jmol 1 K 1 ) Jmol 1, or kjmol 1. Next, the following expression may be used to determine the equilibrium constant, ΔG R RT ln, # or exp ΔG & R $ ' % RT (. Substituting, exp ΔG R RT exp e ( J/mol) 8.314Jmol 1 K 1 ( ) 200 K ( )

9 6.) Continued 9 (b.) Discuss whether increasing the total pressure (at constant temperature) will increase or decrease the extent of reaction. Explain your answer. The effect of pressure on the extent of reaction depends on whether there are more moles of gas on the product side or on the reactant side. In this case, there is one mole of gas on the product side and there are two moles of gas on the reactant side. Therefore, increasing the total pressure will put stress on the reactant side, and the system will shift to the right to alleviate this stress. As a result the extent of reaction will increase if the pressure is increased. An alternate explanation is based upon the expression for the equilibrium constant. For the reaction 2ClO( g) Cl 2 O 2 ( g). the equilibrium constant expressed in terms of mole fractions is # " # " P Cl2 O 2 P P ClO P $ & % $ & % 2 x Cl2 O 2 P $ # " P & % P $ # " P & % 2. Simplifying, we have x Cl2 O 2 P 2 P. Notice that the equilibrium constant expression for this reaction includes the total pressure in the denominator. Therefore, if the total pressure increases, the P P term will decrease. In order for the numerical value of the equilibrium constant to remain the same, that means that the other term, x Cl 2 O 2, 2 must increase. For this to occur, x Cl2 O 2 (the mole fraction of the product) must increase and (the mole fraction of the reactant) must decrease; this has the effect of increasing the extent of reaction.

10 6.) Continued 10 (c.) Discuss whether increasing the temperature (at constant pressure) will increase or decrease the equilibrium constant. Explain your answer. The temperature dependence of the equilibrium constant depends on the sign of the molar enthalpy of reaction. We are given above that ΔH R kj/mol, so it is an exothermic reaction. One way to write the reaction including the heat released is 2ClO( g) Cl 2 O 2 ( g) + heat. Increasing the temperature puts stress on the product side (since that is where the heat resides). To alleviate this stress, the reaction shifts to the left, and the equilibrium constant decreases. An alternative explanation can be given using the van't Hoff plot, shown below. Exothermic case ln slope -ΔH o /R > 0 T high 1/T T low Since the enthalpy of reaction is negative, the slope of the van't Hoff plot is positive. When the temperature increases, 1/T decreases. We can see that this leads to a smaller value of ln and hence a smaller value of ; thus, the equilibrium constant decreases as the temperature increases.

11 7.) (15 points) True/false, short answer, multiple choice. 11 a.) True or False : For the liquid-to-vapor phase transition in a one-component system, the Gibbs free energy change is negative ( ΔG < 0 ). b.) True or False: A chemical equilibrium is an example of a dynamic equilibrium. c.) Short answer The triple point is the place on the phase diagram of a one-component system where the number of thermodynamic degrees of freedom equals zero. d.) Short answer The extent of reaction provides a measure of how far a chemical reaction has progressed from reactants to products. e.) Multiple Choice Henry's Law is generally expected to be valid for which component or components of a solution? 1) the solute 2) the solvent 3) both the solute and the solvent

Problem Set 10 Solutions

Chemistry 360 Dr Jean M Standard Problem Set 10 Solutions 1 Sketch (roughly to scale) a phase diagram for molecular oxygen given the following information: the triple point occurs at 543 K and 114 torr;