Chem 1310: Introduction to physical chemistry Part 3: Equilibria Peter H.M. Budzelaar

Kinetics and Equilibria Kinetics: how fast does a reaction go (initially) Equilibrium: what will be the final composition of the reaction mixture?

What is "equilibrium" ? A system is at (dynamic) equilibrium when its composition doesn't change any more over time. There are equal numbers of molecules going to the left as to the right (and this number is not zero!). A mixture with a composition that does not appear to change over time is not necessarily at equilibrium. The reaction could just be very slow!

How do you know that equilibrium has been established? The reaction started, but then did not proceed any further. This is not an infallible criterion! The equilibrium was approached from both the product and the reactant side (in two separate experiments), and both resulted in the same composition. Molecules have no memory, so the equilibrium composition does not depend on how you got there.

Approaching equilibrium from different sides

How do you know that equilibrium has been established? Heating accelerates approach to equilibrium, but also changes the equilibrium composition. On cooling, the reaction might slow down so much that the new equilibrium will not be established. To make sure you are in a dynamic equilibrium, you could also add one pure component and see how the system responds. If nothing happens, you are probably not at equilibrium.

For a reaction A ⇋ B at equilibrium: For a reaction A + B ⇋ C + 2 D at equilibrium: K C is constant for a given temperature, independent of pressure, volume, etc. The equilibrium constant

Units and equilibrium constants MSJ says (p678) that we should simply omit units from the equilibrium constant, because they are always mol/L. That is not the real reason. We do it because the equilibrium constant is actually something like where [A]° etc are the concentrations at an agreed reference state (which is 1 mol/L for solutes). Because of this division, units cancel and K C is truly dimensionless. Because the reference state is 1 mol/L, we can ignore it when we use mol/L as units.

Kinetics vs equilibria Rate laws can have a complicated dependence on concentrations, unrelated to the reaction stoichiometry. They might depend e.g. on concentrations of catalysts and poisons. Equilibrium constant expressions can be deduced directly from the reaction equation and do not depend on the reaction mechanism, nor on catalysts etc.

Writing an equilibrium constant K C : equilibrium constant expressed in concentrations. Preferably in a single phase, usually gas or solution. NH CH 3 COO - ⇋ NH 3 + CH 3 COOH (in water) Adding NaOH or HCl may change the values of some of the concentrations involved, but K C will indeed remain constant.

Writing an equilibrium constant NH 3 + ¾ O 2 ⇋ ½ N 2 + 1½ H 2 O 4 NH O 2 ⇋ 2 N H 2 O K C (b) = (K C (a)) 4 An equilibrium constant belongs to a chemical equation as written.

Reactions involving solids and liquids Pure solids and liquids are not included in K C. Nearly-pure solvents are also left out. If you have a vapour or solute in equilibrium with the pure solid or liquid, you also leave out that vapour or solute. So if you have ⅛ S 8(s) + O 2 ⇋ SO 2, it doesn't matter whether you also have some S 8(g) present, or whether that is involved in the reaction. You don't need to include S 8(g) as long as S 8(s) remains present.

Why drop pure solids and liquids? This is not simply because their concentrations are constant (MSJ 677). The correct K C would contain not [S 8(s) ] but [S 8(s) ]/[S 8(s) ]°, where [S 8(s) ]° is the concentration in the reference state, which is the pure solid/liquid by convention. Since solids and liquids are not compressible, [S 8(s) ]/[S 8(s) ]° will always be very close to 1, and there is no reason to include it. We are not just dropping a constant, we are dropping a factor 1.

Reactions involving solids and liquids CH 3 COOH + H 2 O ⇋ CH 3 COO - + H 3 O + Hg (l) + Cl 2 ⇋ HgCl 2(s) [Cl 2 ] (eq) = 1/K C Hg (l) + ⅛ S 8(s) ⇋ HgS (s) No equilibrium constant! If product-favoured, reaction will proceed until one or both reactants consumed.

Using equilibrium constants Calculate K C, given equilibrium concentrations. Just plug the concentrations in the formula. Calculate whether a reaction will go forward or backward, given K C and initial concentrations. –Calculate Q C (same formula as K C, but now for non- equilibrium concentrations). –If Q C < K C : reaction will go forward –If Q C > K C : reaction will go backward –If Q C = K C : reaction is at equilibrium

Using equilibrium constants Calculate concentrations at equilibrium, given K C and initial conditions. This may involve serious calculations; use the "x method" of the book (MSJ p ), e.g. ABC initial110 change-x +x equilibrium1-x x

Dissociation of HI H 2 + I 2 ⇋ 2 HI We begin with 1 mol/L of HI, and let this reach equilibrium with H 2 and I 2 at 25°C (K C = 25, see MSJ p685). What will be the final concentrations? H2H2 I2I2 HI initial001 change+ ½ x -x equilibrium½ x 1-x

Dissociation of HI (2) Plug into K C equation: Final concentrations: [HI] = 0.71, [H 2 ] = [I 2 ] = Put into K C expression for final check.

Dissociation of N 2 O 4 N 2 O 4 ⇋ 2 NO 2 We begin with 1 mol/L of N 2 O 4, and let this reach equilibrium with NO 2 at 500K (K C = 46, see MSJ p702). What will be the final concentrations? N2O4N2O4 NO 2 initial10 change-x-x+ 2 x equilibrium1-x2 x

Dissociation of N 2 O 4 (2) Plug into K C equation: Final concentrations: [N 2 O 4 ] = 0.07, [NO 2 ] = Put into K C expression for final check. Quadratic equations

Concentration or pressure dependence Some equilibria are concentration- or pressure-dependent: if all concentrations/pressures change by the same factor (by compressing a gas mixture, or diluting a solution), they are no longer at equilibrium. This happens if the sums of exponents in numerator and in denominator of K C differ. For the case of N 2 O 4 dissociation: numerator 2, denominator: 1. The HI dissociation equilibrium is not pressure dependent (both exponents 2).

Dissociation of N 2 O 4 (3) What will happen if we compress the previous equilibrium mixture ([N 2 O 4 ] = 0.07, [NO 2 ] = 1.85) to 1/10 th of its original volume? Immediately after compression: [N 2 O 4 ] = 0.7, [NO 2 ] = 18.5, new Q C = 489 (should have been 460, but there are rounding errors). So Q C > K C, the equilibrium will shift towards the reactant N 2 O 4.

Dissociation of N 2 O 4 (4) Change table: N2O4N2O4 NO 2 initial change+x+x- 2 x equilibrium0.7+x x

Dissociation of N 2 O 4 (5) Plug into K C equation: (the other solution, x = 27.1, would make [NO 2 ] negative and is unphysical) Final concentrations: [N 2 O 4 ] = 3.56, [NO 2 ] = Put into K C expression for final check.

Le Chatelier's principle The system reacts to counteract an imposed change: A + B ⇋ X + Y Add A (reactant): drive reaction towards X, Y. Add X (product): drive reaction towards A, B. Add something non-reacting (within same volume): no change in concentrations, Q C stays equal to K C, no change. Compress: all concentrations change, but Q C does not (no change in #particles): no change.

Le Chatelier's principle (2) A ⇋ X + Y Add A (reactant): drive reaction towards X, Y. Add X (product): drive reaction towards A. Add something non-reacting (within same volume): no change in concentrations, Q C stays equal to K C, no change. Compress: Q C increases, so reaction is driven towards A (fewer particles), reducing pressure.

Le Chatelier's principle (3) Effect of temperature: heating drives an exothermic reaction (  H 0) towards products. (but not always to completion; the figure on p687 is incorrect!) cooling has the opposite effect.

K P and K C For gases, instead of K C one usually uses K P : where the pressures p are in bar (atm). For any component X, with the ideal-gas law pV = nRT, one has For the above example:

K P and K C Whenever the system is pressure-dependent (sums of exponents in numerator and denominator differ) you will also find RT terms in the conversion between K P and K C. If gases are involved, always make clear whether you are using K C /Q C or K P /Q P !!!!!

Ammonia synthesis