General Chemistry revisited

Transcription

1 General Chemistry revisited A(g) + B(g) C(g) + D(g) We said that G = H TS where, eg, H = f H(C) + f H(D) - f H(A) - f H(B) G < 0 implied spontaneous to right G > 0 implied spontaneous to left In a very unconnected way, we discussed the equilibrium A(g) + B(g) C(g) + D(g) with equilibrium constant K eq PP C D PP A B We talked about the reaction proceeding in both directions to establish equilibrium. ie, proceeding both when ΔG < 0 and when ΔG > 0!!! What s going on??

2 General Chemistry revisited A(g) + B(g) C(g) + D(g) We said that ΔG = ΔH T ΔS where, eg, ΔH = Δ f H(C) + Δ f H(D) - Δ f H(A) - Δ f H(B) ΔG < 0 implied spontaneous to right ΔG > 0 implied spontaneous n to left So where is equilibrium? What s wrong? The description above basically treats A,B, C and D as pure components that never mix. That is to say, as though the composition is constant. This has to be made more sophisticated if we are to couple this concept with that of equilibrium. We now know about chemical potentials and molar Gibbs energies, so that we can follow G as the composition changes, gaining a much more sophisticated viewpoint. We now do just that!

3 No Mixing The importance of free energy of mixing G(A) A B A B G = (1-n A )G B n A G A < 0 With Mixing G(B) P A, P B A o P B G RT lnk RT ln r P A

4 Gibbs Energy Minimum The composition that makes the chemical potentials equal is the equilibrium composition.

5 ConcepTest 1 2 NO 2 (g) N 2 O 4 (g) 300K A. For the above reaction, find K C if K P = 6.9 bar 1 RT= bar dm 3 K -1 mol -1 x 300 K = 24.9 bar dm 3 mol -1 A 172 dm 3 /mol B 0.28 dm 3 /mol K C = K P /(RT) Δn C 0.28 mol/dm 3 D 172 mol/dm 3

6 ConcepTest 1 2 NO 2 (g) N 2 O 4 (g) 300K A. For the above reaction, find K C if K P = 6.9 bar 1 RT= bar dm 3 K -1 mol -1 x 300 K = 24.9 bar dm 3 mol -1 K C = K P /(RT) Δn Here Δn = -1 A 172 dm 3 /mol C 0.28 mol/dm 3 B 0.28 dm 3 /mol D 172 mol/dm 3

7 Sample Problem If 1.00 mol NO 3 2 is injected into a 1 dm vessel at 300 K, find [N 2 O 4 ] at equilibrium. K C = 172 dm 3 /mol and K P = 6.9 bar 1 At equilibrium, x 2 NO 2 (g) N 2 O 4 (g) 300K (1-x)/2 moles of each Assuming gases are ideal, then P NO2 n NO2 and P N2 O4 n N 2 O4 We use K C but could have used K P 3 dm 172 mol 1 x mol 3 2 dm 2 2 mol x 3 dm 2 Simplifying, 344 x 1 x or 2 344x x 1 0 Using my quadratic equation solver, I find x = and x = Two solutions!! Answer: an additional constraint: 0< x < 1 Are both right? x= x NO2 = mol/dm 3 NO Are both wrong? 2 Did I make a mistake? x N2 O 4 = (1-x)/2 = mol/dm 3 N 2 O 4

8 Effects of Changes on Equilibria The Le Chatelier Principle results from consideration of the effects of temperature, pressure or quantity changes on an equilibrium constant. E.g., 2NO 2 (g) N 2 O 4 (g) The reaction shifts to the left when the # moles of N 2 O 4 is increased. The reaction shifts to the right when the total pressure is increased. We now consider in a bit more detail pressure and temperature effects on equilibria. (Reasons that we might do this include having tests for equilibrium, and to have ways to measure H or S.)

9 Effects of Pressure on Equilibria We know G = RT ln K P ΔG is defined at P = P = 1 bar, and so it is independent of pressure. (ideal gases assumed) For the gas-phase reaction aa + bb yy + zz y z P P o Y Z K P a b P P Obviously, K P must also be pressure independent. A There are three ways that one could imagine changing the pressure: a) Add an inert gas to the reaction mixture b) Change the volume of the reaction mixture c) Change the quantity of one of the components (eg, A,B, Y or Z in above) Take a brief look at each of them. B

10 Effect of adding an ideal inert gas, M Add gas M G = RT ln K P aa + bb yy + zz K y P P P P z o Y Z P a b mm+ aa + bb yy + zz+ mm A B If there is no change in volume, then the partial pressures of each of the ideal gas components remains unchanged by the addition of M. No change P y P z P m in any partial o Y Z M pressures. KP a b m K is unchanged! If the reaction were P P P A B M taking place in an y z However, M might isolated system, M would P P Y Z well affect the almost surely also affect a b rate at which the the final temperature P P A B process occurs.

11 Change the volume Changing the volume can have an effect if the number of moles of gas on the left and right sides of the equation are different. N 2 O 4 2 NO 2 N(1-) 2N N in moles Let be the fraction of the dimer that has dissociated. Total number of moles of gas = N(1 - ) ) + 2N = N(1 + ) ) We now write the mole fractions of each component: NO 1 1 N 2 2 NO 1 1 N 1 1 N N 2 4 2

12 Change the volume (2) We use the mole fractions to obtain the partial pressures of each component: 1 2 P P P P NO 2 4 NO2 1 1 Use these to write the pressure equilibrium constant NO NO P 2 K P P P Solve for : P P 2 4 K P K 4 P p P The degree of dissociation decreases when the pressure increases. Another manifestation of the Le Chatelier Principle

13 Change quantity of one component 2 NO 2 (g) N 2 O 4 (g) P NO 2 4 K P P 2 NO The system will shift in a direction i away from the component that was increased. This is a method that is often used to determine whether or not equilibrium has actually been attained. (Rather than simply having an equilibration time >> observation time) 2

14 Effect of Temperature on Equilibrium We know G = RT ln K P We couple this ΔG relationship with the Gibbs-Helmholtz equation derived in Chapter 3, G 0 H 0 (3.169) 2 T T T P to obtain an expression that gives the temperature dependence of K P d 0 ln K 0 P H the van't Hoff equation 2 dt RT 0 d ln K P 1 dt H Since d, we can rewrite, giving 2 T T d 1/ T R This last form shows that a plot of ln K P vs 1/T has slope -H/R 0

15 Effect of Temperature on Equilibrium This last form shows that a plot of ln K P vs 1/T has slope -H/R, if H does not depend strongly on temperature. Note that the positive slope means that H is negative, an exothermic reaction. Here HH has a large T dependence. Can the slope be negative? What would a negative slope imply? The Y intercept of this line is also interesting!

16 Effect of Temperature on Equilibrium The Y intercept is also interesting! We know that t So ln K G = RT ln K P G H 1 T S P The y-axis is located at RT R T RT 1 0 T So the value of the intercept is S/R These plots can be a very useful way to determine basic thermochemical properties of molecules and reactions. S/R

17 Sample Problem Consider the reaction 4 NH 3 (g) + 3 O 2 (g) 2 N 2 (g) + 6 H 2 O(g) with the dimensionless equilibrium constant K po = at 298 K. A reaction flask is filled to a pressure of 1.0 bar with gases at the following partial pressures: P(NH 3 ) = 0.10 bar, P(O 2 ) = 0.10 bar, P(H 2 O) = 0.60 bar, and P(N 2 ) = 0.20 bar. The gases are well mixed. You may further assume that the total pressure of the reaction flask is maintained constant, the gases behave ideally, and standard state corresponds to partial pressures of 1 bar of all gases. Compared with the initial composition, will the flask contain fewer reactants or fewer products when it reaches equilibrium? P P HO 2 N2 Q P NH P O Q < K so by Le Chatelier s Principle products will be produced Q < K, so by Le Chatelier s Principle, products will be produced as the reaction proceeds to equilibrium. Therefore, fewer reactants will be present at equilibrium. 4

18 Sample Problem Consider the reaction 4 NH 3 (g) + 3 O 2 (g) 2 N 2 (g) + 6 H 2 O(g) with the dimensionless equilibrium constant K po = at 298 K. A reaction flask is filled to a pressure of 1.0 bar with gases at the following partial pressures: P(NH 3 ) = 0.10 bar, P(O 2 ) = 0.10 bar, P(H 2 O) = 0.60 bar, and P(N 2 ) = 0.20 bar. The gases are well mixed. You may further assume that the total pressure of the reaction flask is maintained constant, the gases behave ideally, and standard state corresponds to partial pressures of 1 bar of all gases. Calculate G for this reaction for the concentrations and temperature given above. What does this value imply about the spontaneity of this point in the reaction? o o kj 4 ln ln kj G RT K K 25.4 p mol K mol o G G RT lnq kj kj K ln mol mol K kj 1.04 mol Since G < 0, this reaction is spontaneous at the given concentrations and temperature

19 Sample Problem Consider the reaction 4 NH 3 (g) + 3 O 2 (g) 2 N 2 (g) + 6 H 2 O(g) with the dimensionless equilibrium constant K po = at 298 K. A reaction flask is filled to a pressure of 1.0 bar with gases at the following partial pressures: P(NH 3 ) = 0.10 bar, P(O 2 ) = 0.10 bar, P(H 2 O) = 0.60 bar, and P(N 2 ) = 0.20 bar. The gases are well mixed. You may further assume that the total pressure of the reaction flask is maintained constant, the gases behave ideally, and standard state corresponds to partial pressures of 1 bar of all gases. What will be the value of G at equilibrium? Give one sentence of justification. G = 0 at equilibrium G = 0 at equilibrium At equilibrium, the reaction is neither spontaneous (G < 0) nor non-spontaneous (G > 0).

20 Sample Problem (2) Use the thermochemical data below to estimate the boiling temperature of hydrazoic acid, HN 3. Thermodynamic data (Standard state: 298 K, 1 bar) H fo HN 3 (liquid) = kj mol -1 S fo HN 3 (liquid) = J K -1 mol -1 H fo HN 3 (gas) = kj mol -1 S fo HN 3 (gas) = J K -1 mol -1 Density: (HN = -3 3 (liquid)) 1.09 kg dm HN HN g The vaporization reaction 3 3 The boiling temperature is when the above Rxn is in equilibrium at 1 bar. G rxn 0 at boiling temperature o o H T S 0 rxn rxn H H T S T SS rxn o o o rxn rxn rxn o o o o kj kj H H HN g H HN rxn f 3 f 3 mol mol o o o J J S S HN g S HN rxn f f molk molk T boil kj mol 305K kj Why is this only an estimate? K mol