CHEM 1062 Lecture Notes - Lecture 27: Recap (Software), Vant (Band)

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8 Apr 2015

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Chem 1062: lecture #27 free energy and chemical equilibrium: free energy and chemical equilibrium, temperature dependence of equilibrium constant, free energy and useful work. If, qp > kp, the reaction goes to the left. Qp < kp, the reaction goes to the right. Qp = kp, the reaction is at equilibrium. G < 0, the reaction goes to the right. G > 0, the reaction goes to the left. G = 0, the system is at equilibrium. Ts is equivalent to f(v) = f(nrt/p) G = e + pv - ts. Pv = nrt, therefore pv is a f(t) Go is at 1 atm of pressure. When p = 1 atm, g=go ie. a (g) 2b (g) G = 2gb - ga: + rt*ln(pa, + rt*ln(pb) At equilibrium g = 0, q = k. If q > k, g > 0, the reaction goes left. If q < k, g < 0, the reaction goes right.

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Diamond
a. At 298 K, what is the Gibbs free energy change ïG for the following reaction? Cgraphite -> ï Cdiamond

b. Is the diamond thermodynamically stable relative to graphite at 298 K?

c. What is the change of Gibbs free energy of diamond when it is compressed isothermally from 1 atm to 1000 atm at 298 K?

d. Assuming that graphite and diamond are incompressible, calculate the pressure at which the two exist in equilibrium at 298 K.

e. What is the Gibbs free energy of diamond relative to graphite at 900 K? Assume the heat capacities of the two materials are equal.

f. Diamond is synthesized from graphite at high pressures and high temperatures. The need for high pressure should be obvious from your calculations, but why is the process carried out at high temperature?

DATA
Density of graphite is 2.25 g/cm3 Density of diamond is 3.51 g/cm3

	Delta Hf(298K)	S^o(298K)
Diamond	1.897 kJ/mol	2.38 J/(K mol)
Graphite	0	5.73 J/(K mol)

ïHf(298 K)S0

298 K

Diamond 1.897 kJ/mol 2.38 J/(K mol)

Graphite 0 5.73 J/(K mol)

Consider the reaction: N₂ + 3 H₂ Â« 2 NH₃ .

For this reaction 1 bar of N₂ and 3 bars of H₂ are introduced into a 10.0 L vessel at 455 K. From ICE diagrams, we see that at equilibrium, we have:

N₂ + 3 H₂ Â« 2 NH₃ .

(1 â x) 3Ã(1 â x) 2x

Let us maintain this system at a constant pressure of P_T = 4 bar.

a.) From the above ICE diagram, produce an expression in terms of the extent of reaction, x, that gives the total amount of material in the system at any point.

b.) From the above ICE diagram and your answer in a.), produce expressions for the mole fractions of each species in terms of the extent of reaction, x.

c.) From your result in b.), produce expressions for the partial pressures of each species using Daltonâs Law.

d.) From your results above, produce an expression for the equilibrium constant, K_p in terms of x, P_Tot and the standard pressure, P_o

The Gibbs Free Energy for this system is plotted below vs âxâ the âextent of reactionâ defined above.

e.) Using the graphic above, calculate the partial pressures of all species at equilibrium.

f.) Calculate the equilibrium constant from your results at 455 K.

g.) From standard tables, calculate the standard Gibbs Free Energy for this reaction and use the result to calculate K_p at 298.15 K. DG_f(NH₃) = -16.45 kJ/mol.

h.) From standard tables, calculate the reaction enthalpy for this reaction at 298.15 K. DH_f(NH₃) = -46.11 kJ/mol.

i.) Using your results in g.) and h.), calculate K_p at 455 K and compare with your estimated result that you obtained from the graphic provided. Since the temperature is elevated and involve gases, use the extended equation that gives the proper temperature dependence on the enthalpy

CHEM 1062 Lecture Notes - Lecture 27: Recap (Software), Vant (Band)

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