Worksheet Week 4 Solutions

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1 adapted from Rickey Kellow by ADH Chemisty 124 Fall 2018 Problem 1: Worksheet Week 4 Solutions (a) The reaction quotient QQ (written in terms of the partial pressures) for the given reaction is QQ = PP N 2 O 4 2 PP NO2 (b) In last week s worksheet, we saw that the standard change in Gibbs energy for a reaction is related to the reaction s equilibrium constant through GG o = RRRR ln KK. From the given equilibrium constant at 25 o C, the corresponding standard change in Gibbs energy, GG o is therefore GG o = RRRR ln KK = (8.314 J K 1 mol 1 )( K) ln(6.8) 1 kj 10 3 = 4.75 kj mol 1 J (c) Let s make use of the van t Hoff equation, which you also used in lab this week. If we make the approximation that HH o and SS o do not change with temperature, note the linear relationship that arises from this equation. ln KK yy = HHo RR mm 1 TT xx + SSo RR bb This equation implies that for a given set of equilibrium constants and various temperatures, we could take the natural log of the equilibrium constants and plot them versus their corresponding reciprocal absolute (i.e. the temperatures must have units of Kelvin!) temperatures and obtain a straight line with slope mm = HH o /RR and yy-intercept bb = SS o /RR. For only two temperatures and their corresponding equilibrium constants, let s subtract the van t Hoff equation containing the equilibrium constant for temperature TT 1, KK 1, by the van t Hoff equation containing the equilibrium constant for temperature TT 2, KK 2. ln KK 1 = HHo + SSo RRTT 1 RR ln KK 2 = HHo RRTT 2 + SSo RR ln KK 1 ln KK 2 = ln KK 1 = HHo KK 2 RR 1 1 TT 1 TT 2

2 Therefore, from the given data, the standard enthalpy change HH o is calculated to be RR ln KK 1 HH o KK = 2 1 TT 1 1 TT 2 = 57.9 kj mol 1 ln KK 1 = ΔΔHHo KK 2 RR 1 1 TT 1 TT 2 (8.314 J K 1 mol 1 ) 1 kj 10 = 3 J ln (6.8) 1 [( )K] 1 ( ) [( )K] Note that we took the given temperatures to be exact (they don t limit the number of significant figures in the conversion from to Kelvin. It also makes no difference which temperature we denote as TT 1 in the equation, you just must be careful to correctly match the temperatures with their corresponding equilibrium constants. d. Recall also from the last worksheet that we were able to relate the standard change in Gibbs energy GG o, the standard change in entropy SS o, and the standard change in enthalpy HH o through GG o = HH o TT SS o. Therefore, we can solve for the standard change in entropy SS o using our work from (b) as follows. SS o = GGo HH o TT = 178 J K 1 mol 1 GG o = HH o TT SS o ( 4.75 kj mol 1 ) 103 J 1 kj ( 57.9 kj mol 1 ) 103 J 1 kj = ( K) As a good review of previously learned concepts, you can verify that the signs of the calculated standard enthalpy change and standard entropy change make sense. From the given data, we see that the equilibrium constant decreases as the temperature increases. This means that the forward reaction is favored at lower temperatures. In problem 4(c) of the W3 Solutions, we discussed in terms of Le Cha telier s principle why exothermic reactions are favored at lower temperatures and why endothermic reactions are favored at higher temperatures. Thus, from the given data, it makes sense that we calculated a negative value for the standard enthalpy change HH o. What about the standard entropy change SS o? The forward reaction goes from 2 moles of gas to 1 mole of gas, which indicates a decrease in entropy of the system. This is consistent with the fact that we calculated a negative value for SS o. 2

3 Problem 2: (a) The equilibrium constant corresponding to the first ionization, KK a1, of phosphoric acid can be calculated using the relationship between the standard change in Gibbs energy, GG o, and the equilibrium constant KK discussed previously. The standard Gibbs energy for the reaction is calculated using the given information as follows. GG o = nn prod GG f o [products] nn react GG f o [reactants] = = {(1) GG f o [H 2 PO 4 (aaaa)] + (1) GG f o [H + (aaaa)]} {(1) GG f o [H 3 PO 4 (aaaa)]} = {(1)( kj mol 1 ) + (1)(0)} {(1)( kj mol 1 )} = kj mol 1 Using the relationship between the standard change in Gibbs energy, GG o, and the equilibrium constant KK, GG o = RRRR ln KK, we then have 10 (12.26 ln KK a1 = GGo kj mol 1 ) 3 J RRRR = 1 kj (8.314 J K 1 mol 1 ) ( )K = KK a1 = ee ln KK a1 = ee = Make sure that you are careful about tracking your units throughout the calculation! You should not take the anti-log of a number with units that have not canceled. (b) An equilibrium constant KK that is much less than 1 for a reaction indicates a favoring of reactants over products at equilibrium. Therefore, from our work in (a), the acid ionizes to only a small extent and is thus characterized as a weak acid. (c) We are given the standard change in Gibbs energy for the reaction at a new temperature and we calculated the standard change in Gibbs energy for the reaction at 25 o C in (a). We would like to find the standard entropy change SS o, so it is logical to make use of the relation between the standard change in Gibbs energy GG o, the standard change in entropy SS o, and the standard change in enthalpy HH o : GG o = HH o TT SS o. As we did when we derived the van t Hoff equation, let s subtract (in an effort to cancel out the standard enthalpy change term and isolate the standard entropy change) this equation containing one temperature and GG o for that temperature by the equation containing the other temperature and GG o for that temperature. Note that we are again making use of the approximation that HH o and SS o do not change with temperature. 3

4 GG o (TT 1 ) = HH o TT 1 SS o ( GG o (TT 2 ) = HH o TT 2 SS o ) GG o (TT 1 ) GG o (TT 2 ) = SS o (TT 2 TT 1 ) Therefore, we calculate the standard entropy change SS o to be SS o = GGo (TT 1 ) GG o (TT 2 ) (12.26 kj mol 1 ) 103 J 1 kj (15.65 kj mol 1 ) 103 J 1 kj = TT 2 TT 1 [( )K] [( )K] = 67.8 J mol 1 K 1 For additional practice, you should try to solve this problem using the long way mentioned in the short solutions for this worksheet. Use the van t Hoff equation to calculate HH o and use that in GG o = HH o TT SS o. This is good practice for rearranging equations and tracking your units. You should get the same answer using the method above, and you should also note that as we mentioned in problem 1, when we make use of the van t Hoff equation as we are here, we are approximating that HH o and SS o do not change with temperature. Problem 3: (a) We are asked to calculate the change in Gibbs energy for a particular set of concentrations. This implies that we will need to use the relation between the change in Gibbs energy and the reaction quotient QQ: GG = GG o + RRRR ln QQ. Making use of this equation requires that we first solve for the standard change in Gibbs energy, GG o, for the process. To determine this, we need to note that the process of interest is the transfer of 1 mol of K + from blood plasma to the intracellular fluid. We can write this in the form of a balanced equation as K + blood plasma (aq) K + intercellular fluid (aq) The (aq) in the balanced equation above denotes that the ions are dissolved in water in both the plasma and the intracellular fluid. We can write the (aq) above because the intracellular fluid and blood plasma primarily consist of water. (If you didn t know this a quick internet search will tell you that intracellular fluid is all fluid that is contained inside cells. The majority of this fluid is cytosol, which contains mostly water, and water constitutes roughly 70% of the total cell volume! In blood plasma, water constitutes roughly 95% of the total cell volume!). The standard change in Gibbs energy, GG o, for a process can be calculated using o GG rxn = nn prod GG f o [products] nn react GG f o [reactants] 4

5 Using the fact that in blood plasma and intracellular fluid, potassium ions are primarily aqueous, we then have o GG rxn = nn prod GG f o [products] nn react GG f o [reactants] = GG f o [K + (aq)] GG f o [K + (aq)] = 0 Let s think about what this means physically. We have seen that GG o is inherently related to the equilibrium constant KK through GG o = RRRR ln KK in order to have GG o = 0, we must have KK = 1, which for this process would mean we would have equal concentrations of potassium ions in the blood plasma and intracellular fluid at equilibrium. You have seen an example of a process like this before consider a system with two bulbs of the same volume with a stopcock between them that is initially closed and preventing the flow of any contents from one bulb to the other. Suppose one of the bulbs initially contains nothing, and the other bulb contains a gas. After the stopcock is opened and an equilibrium is reached, the gas particles will be evenly distributed between the two bulbs. This process is spontaneous, after opening the stopcock, the system will do this naturally without outside intervention. A similar idea applies to the system in this problem. Because the potassium ions are in the same state (essentially aqueous) in both the blood plasma and the intracellular fluid, at equilibrium, they will be distributed between the blood plasma and intracellular fluid such that their concentrations are the same in both regions. We are told that the potassium ion concentration is greater in the intracellular fluid than it is in the blood plasma, and we want to transfer potassium ions from the blood plasma to the intracellular fluid. This will take us even further away from equilibrium, and as we discussed last week, reactions spontaneously approach equilibrium. Thus, GG should be positive for this process we must perform work on the system if this process is to occur. We can calculate GG for this process as follows. GG = GG o + RRRR ln QQ = (0) + RRRR ln [K+ ] intracellular fluid [K + ] blood plasma = (8.314 J K 1 mol 1 ) 1 kj 10 3 J ( )K ln (0.15 M) ( M) = 8.77 kj mol 1 In other words, this process is not spontaneous under the given conditions and will not occur without outside intervention. (b) We just stated that the process described in this problem will not occur without outside intervention. In this problem, we will derive an important relation between Gibbs energy and 5

6 work. Let s start with the definition of Gibbs energy GG, GG HH TTTT. The change in Gibbs energy is then GG = HH (TTTT) Let s consider a constant temperature process. We then have GG = HH TT SS Let s now use the definition of enthalpy HH, HH EE + PPPP. The change in enthalpy is then HH = EE + (PPPP) Let s consider a constant pressure process. We then have HH = EE + PP VV Substituting our expression above for HH into our expression for the change in Gibbs energy for this process gives GG = HH TT SS = ( EE + PP VV) TT SS Now let s use the first law of thermodynamics, which states the change in internal energy of the system is a sum of the heat absorbed by the system and the work done on the system (i.e. EE = qq + ww). GG = EE + PP VV TT SS = (qq + ww) + PP VV TT SS Recall that the thermodynamic definition of entropy is SS = qq rev /TT. Note that the definition contains a term for heat exchanged reversibly. Therefore, we have TT SS = qq rev. Substituting this into the expression above gives GG = qq + ww + PP VV TT SS = qq + ww + PP VV (qq rev ) Let s now separate the term denoting work ww into a sum of the PPPP work and non-pppp work. GG = qq + ww + PP VV qq rev = qq + (ww PV + ww nonpv ) + PP VV qq rev The amount of PPPP work is given by ww = PP ext VV. For a system at constant pressure such as this one, the external pressure is equal to the system pressure, which we will simply denote as PP. Therefore, GG = qq + ww PV + ww nonpv + PP VV qq rev = qq + ( PP VV) + ww nonpv + PP VV qq rev = qq + ww nonpv qq rev 6

7 Finally, if we suppose that the process under consideration is reversible, then the first qq term becomes qq rev and the ww nonpv term denotes non-pppp work that is reversible. Therefore, we have GG = qq rev + ww nonpv qq rev = ww nonpv Let s note again all of the constraints we placed on the process under consideration. We said that this process is at constant temperature, constant pressure, and it is reversible. In order to make the process in (a) go, we must make the change in Gibbs energy non-positive. So, the minimum amount of (non-pppp) work that must be done on the system is 8.77 kj mol -1, and this is the amount of work that will be done if the process is completely reversible and takes place at constant temperature and pressure. If any of the work done on the system is irreversible, we will have to do more than 8.77 kj mol -1 of work to get the process in (a) to go. Thus, we see that a reversible process serves as an idealized limit. What about processes where GG is negative? These processes are spontaneous and can do work. Thus, if a process has a negative change in Gibbs energy, GG, the magnitude of the change in Gibbs energy represents the maximum amount of work that can be done by the system (if GG is negative, then ww nonpv is negative, so we would say that negative work is being done on the system meaning that work is being done by the system). This is the amount of work that will be done by the system if the process is completely reversible and takes place at constant temperature and pressure. If any of the work done by the system is irreversible, the process will do less work than the magnitude of the change in Gibbs energy. The relation between Gibbs energy and work is one that is very important to understand, and it will soon pop up again in our discussions of electrochemistry in the following week(s)! 7

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