Comments re Chapter 6 Problems:

6-2

Some T/F questions about chemical potentials.

6-3

An exercise in calculating equilibrium constants in several forms.  Please remember that the only form of the equilibium constant that can be used in the formula DGo=-RTln Kp° (where DGo is calculated from the DfGo values in standard tables) is Kp°.  The difference between Kp° and Kp is that the latter may have units because the individual factors are pressures rather than pressures divided by the standard pressure (which is 1 bar).  (Kp values are used in General Chemistry courses and are fine for doing equilibrium problems but KP° is the only form that can be used to calculate DrxnGo).   In part (a) it is necessary to convert from mole fraction to pressure using Raoult’s Law (Xi = Pi/Ptot);  the pressures must be calculated in bar if DGo is to be calculated.  In part (c) the value Kc° refers to the equilibrium constant as calculated from concentrations with each concentration being divided by a standard concentration of 1 mol/L.  The concentrations in mol/L of perfect gases are just c = n/V = P/RT.  (Some of this is a review of material from General Chemistry).

6-6

An exercise in determining whether or not a system is at equilibrium with respect to a specific chemical reaction, i.e., the calculation and interpretation of the quantity Qp  (The value QP has the same form as the equilibrium constant but the gas pressures do not necessarily correspond to a set of equilibrium pressures).  If Qp is equal to Kp then the reaction is at equilibrium;  if Qp is less than Kp then the reaction must go forward (more reactants must be converted to products) for the reaction to come to equilibrium.  If Qp is greater than Kp then the reaction must go backwards for the reaction to come to equilibrium.  (Review of material from General Chemistry).
(NB/  All molecules break up to some extent at high temperatures – maybe to their atoms or maybe just to smaller molecules – because breaking up a molecule corresponds to DS > 0 and as the temperature is raised the entropy term in
DG = DH - T DS becomes more and more important).

6-10

A T/F question about equilibrium constants.  (Most of this material was covered in General Chemistry).  In part (d) it is true that Kp° is independent of P because all the pressures used in calculating Kp° = exp(-DrxnGo/RT) are defined to be exactly 1 bar.  (If the gases were not perfect the calculation would be more complicated).

6-34

An exercise that relates the magnitude of the equilibrium constant to the form of the balanced reaction. If the relationship is not obvious write out Kp° for each of the reactions and compare their forms. (Review of material from General Chemistry).

6-37

A demonstration of how a modest error in the DrxnGo value can make a very significant difference in the value of the corresponding Kpo.  To answer this question just do some simple calculations. Write Kp° as DrxtG°±error and remember that exp(a+b) = [exp(a)][(exp(b)].

6-56

Very short question.  (Review of material from General Chemistry).  Remember that the question only applies to equilibria involving perfect gases, for which PiV = niRT.

extra

Calculate DGm = Dm in kJ/mol for water (1.00 g/mL, 18.0 g/mol) when
(a)  the pressure is changed from 1 bar to 5 bar
(b)  the pressure is changed from 1 bar to 0.02 bar = 15 torr
The purpose of this problem is to demonstrate why it is safe to assume that
mP = m° for pure liquids and solids.

 

 

6-14

Use data in Tables to find Kp° at 298 K and then to estimate the value at 400 K.
t­he best way to do this problem is to use the equation taught in General Chemistry:
ln[(Kpo at T2)/(Kpo at T1)] = -[(DHo at T1)/R](1/T2 – 1/T1)
(For this reaction DrxnHo > 0 so that the equilibrium constant decreases as the temperature is raised).

6-16

The value of Kpo is given as a mathematical function of T. Remember that
DGo = - RT ln Kpo and that d(DGo/T)/dT = -DHo/T2 so that d(ln Kpo)/dT =
d(-DGo/RT)/dT = (-1/R)(-DHo/T2).  First find DGo using the standard equation
DGo = -RTln Kop. Then find DHo by using the relationship d(ln Kpo)/dT = DHo/RT2, and DCpo from d(DHo)/dT. Finally, get DSo from DSo = (DHo - DGo)/T. At the end calculate DHo at two temperatures. The two values differ, but only by less than 6%, so the assumption implicit in the usual equation for the variation in Kpo with T that DHo does not vary is reasonable. (The variation in DHo is small because DCpo is usually much smaller than DHo).

Levine's notation "(T/K)", where K means units of Kelvin, can be confusing; it is probably easiest to just ignore all the factors "/K" and to remember that temperature must be given in degrees Kelvin.

6-17

(Calculation only). Shows how the temperature dependence of DHorxn affects the equation for the variation of the equilibrium constant with temperature. This derivation assumes that the Cpo values for all substances are independent of T, but remember that this assumption is less important than the assumption that DHorxn is independent of T.  In any event DCpo changes only very slowly with T.

This problem uses the same chemical reaction and the same temperature as problem 6-16, so comparisons are possible.  Note that Kp values seldom have more than 2 significant digits;  uncertainties of 20% or more are common.

Please do not memorize this equation!

6-23

Short T/F exercise.  For part (a) the equation given just above will be needed.

6-48

A question that can be answered if Le Chatelier's principle is understood. This principle says that if an equilibrium system is “stressed” the system will adjust to (partially) relieve that stress.  So adding a reactant results in production of more product, raising the pressure favors the side of the reaction with the fewer number of gas molecules, and raising the temperature favors the direction of the reaction that is endothermic. (Review of material from General Chemistry).

6-62

For what kind of reaction is the equilibrium position independent of pressure?  (For all other reactions the equilibrium position does depend on the sum of the Pi's even though Kp° itself does not depend on pressure because it is calculated from DrxnGo, where the ° means at 1 bar).
Then, for what kind of reaction is Kpo independent of temperature?  (Again, see the equation given above in the comments on problem 6-14.)

 

 

6-27

Use data in a Table to calculate DrxnGo, then compute Kpo and finally do the equilibrium problem by the method of successive approximations. The reaction itself is interesting.

6-28

An equilibrium problem very similar to the one done in class.  The equilibrium expression for part (a) reduces to a quadratic equation after factoring and cancellation of terms.  It can then be solved using the quadratic formula.  (Two roots will be found but one of them is not physically reasonable).  The equation can also be solved by the method of successive approximations.  No cancellation of terms is possible in part (b).

extra

Calculate the equilibrium pressures for the N2+3H2 « 2NH3 gas-phase reaction at 400 K, where KP° = 36, given that the initial pressures are PN2 = 0.500 bar, PH2 = 0.200 bar, and PNH3 = 2.00 bar.  In doing this problem it is very important to get the limits on x correct.  First calculate whether x is >0 or <0, then work out the limits on x, and then solve the equilibrium problem by the method of successive approximations.

 

 

6-28(a)

An equilibrium problem very similar to the one done in class.  The equilibrium expression for part (a) reduces to a quadratic equation after factoring and cancellation of terms.  It can then be solved using the quadratic formula.  (Two roots will be found but one of them is not physically reasonable).  The equation can also be solved by the method of successive approximations.  Part (b) of this problem will be included in the next assignment.

6-28(b)

Two more equilibrium problems, both based on the same chemical reaction, to be solved by the method of successive approximations.  First, do part (b) of 6-28.  Then do the same problem but starting from 0.500 mol of NH3(g) and 0.0500 mol each of N2(g) and H2(g).

6-63
(not assigned recently)

Another of Levine's thought problems. Very useful preparation for the multiple-choice questions on the hour exams. Most students would get the wrong answer for (d), which deals with a very fine (but interesting) point; part (d) is therefore optional. Parts (m) and (n) are also optional.


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