E QUILIBRIUM

R EACTIONS ARE REVERSIBLE A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the forward reaction is possible As C and D build up, the reverse reaction speeds up while the forward reaction slows down. Eventually the rates are equal

Reaction Rate Time Forward Reaction Reverse reaction Equilibrium

W HAT IS EQUAL AT E QUILIBRIUM ? Rates are equal Concentrations are not. Rates are determined by concentrations and activation energy. The concentrations do not change at equilibrium. or if the reaction is very slow.

L AW OF M ASS A CTION For any reaction j A + k B l C + m D K = [C] l [D] m PRODUCTS power [A] j [B] k REACTANTS power K is called the equilibrium constant. is how we indicate a reversible reaction

P LAYING WITH K If we write the reaction in reverse. l C + m D j A + k B Then the new equilibrium constant is K ’ = [A] j [B] k = 1/K [C] l [D] m

P LAYING WITH K If we multiply the equation by a constant nj A + nk B nl C + nm D Then the equilibrium constant is K’ = [A] nj [B] nk = ([A] j [B] k ) n = K n [C] nl [D] nm ( [C] l [D] m ) n

T HE UNITS FOR K Are determined by the various powers and units of concentrations. They depend on the reaction. Are usually dropped, unlike rate constant units which are specifically asked for.

K IS CONSTANT At any temperature. Temperature affects rate. The equilibrium concentrations don’t have to be the same, only K. Equilibrium position is a set of concentrations at equilibrium. There are an unlimited number.

E QUILIBRIUM C ONSTANT UNIQUE FOR EACH TEMPERATURE

C ALCULATE K N 2 + 3H 2 2NH 3 Initial At Equilibrium [N 2 ] 0 =1.000 M [N 2 ] = 0.921M [H 2 ] 0 =1.000 M [H 2 ] = 0.763M [NH 3 ] 0 =0 M [NH 3 ] = 0.157M

C ALCULATE K N 2 + 3H 2 2NH 3 Initial At Equilibrium [N 2 ] 0 = 0 M [N 2 ] = M [H 2 ] 0 = 0 M [H 2 ] = M [NH 3 ] 0 = M [NH 3 ] = 0.157M K is the same no matter what the amount of starting materials

E QUILIBRIUM AND P RESSURE Some reactions are gaseous PV = nRT P = (n/V)RT P = CRT C is a concentration in moles/Liter C = P/RT

E QUILIBRIUM AND P RESSURE 2SO 2 (g) + O 2 (g) 2SO 3 (g) Kp = (P SO3 ) 2 (P SO2 ) 2 (P O2 ) K = [SO 3 ] 2 [SO 2 ] 2 [O 2 ]

E QUILIBRIUM AND P RESSURE K = (P SO3 /RT) 2 (P SO2 /RT) 2 (P O2 /RT) K = (P SO3 ) 2 (1/RT) 2 (P SO2 ) 2 (P O2 ) (1/RT) 3 K = Kp (1/RT) 2 = Kp RT (1/RT) 3

G ENERAL E QUATION j A + k B l C + m D K p = (P C ) l (P D ) m = (C C xRT ) l (C D xRT ) m (P A ) j (P B ) k (C A xRT ) j (C B xRT ) k K p = (C C ) l (C D ) m x(RT) l+m (C A ) j (C B ) k x(RT) j+k K p = K (RT) ( l+m)-(j+k) = K c (RT)  n  n= (l+m)-(j+k) = change in moles of gas

H OMOGENEOUS E QUILIBRIA So far every example dealt with reactants and products where all were in the same phase. We can use K in terms of either concentration or pressure. Units depend on reaction.

H ETEROGENEOUS E QUILIBRIA If the reaction involves pure solids or pure liquids the concentration of the solid or the liquid doesn’t change. As long as they are not used up we can leave them out of the equilibrium expression.

F OR E XAMPLE H 2 (g) + I 2 (s) 2HI(g) K = [HI] 2 [ H 2 ][ I 2 ] But the concentration of I 2 does not change. K [ I 2 ] = [HI] 2 = K’ [H 2 ]

S OLVING E QUILIBRIUM P ROBLEMS Type 1

T HE R EACTION Q UOTIENT Tells you the directing the reaction will go to reach equilibrium Calculated the same as the equilibrium constant, but for a system not at equilibrium Q = [Products] coefficient [Reactants] coefficient Compare Q value to the equilibrium constant

W HAT Q TELLS US If Q Q Not enough products Shifts right If Q>K K < Q Too many products Shifts left If Q=K system is at equilibrium

E XAMPLE For the reaction 2NOCl(g) 2NO(g) + Cl 2 (g) K = 1.55 x M at 35ºC In an experiment 10.0 mol NOCl, 1.0 mol NO(g) and.10 mol Cl 2 are mixed in 2.0 L flask. Which direction will the reaction proceed to reach equilibrium?

S OLVING E QUILIBRIUM P ROBLEMS Given the starting concentrations and one equilibrium concentration. Use stoichiometry to figure out other concentrations and K. Learn to create a table of initial and final conditions.

Consider the following reaction at 600ºC 2SO 2 (g) + O 2 (g) 2SO 3 (g) In a certain experiment 2.00 mol of SO 2, 1.50 mol of O 2 and 3.00 mol of SO 3 were placed in a 1.00 L flask. At equilibrium 3.50 mol SO 3 were present. Calculate the equilibrium concentrations of O 2 and SO 2, K and K P Initial Change Equilibrium

P ROBLEMS WITH SMALL K K<.01

P ROCESS I S LIKE LAST YEAR ’ S ACID / BASE PROBLEMS Set up table of initial, change, and final concentrations. For a small K the product concentration is small.

F OR EXAMPLE nitrogen + oxygen yields nitrogen monoxide K= 4.10 x If.50 moles of nitrogen and.86 moles of oxygen are mixed in a 2.0 L container, what are the equilibrium concentrations of all 3 substance ? Q =

K = Because K is small, assume x is negligible compared to.25 and.43 N2O2O2 2NO Initial Change Equilibrium

Simplified K expression: K = Always check the assumption. X must be less than 5% of the original concentration.

LARGE K K >100

W HAT IF YOU ’ RE NOT GIVEN EQUILIBRIUM CONCENTRATIONS ? The size of K will determine what approach to take. First let’s look at the case of a LARGE value of K ( >100). Allows us to make simplifying assumptions.

E XAMPLE H 2 (g) + I 2 (g) 2HI(g) K = 7.1 x 10 2 at 25ºC Calculate the equilibrium concentrations if a 5.00 L container initially contains 15.9 g of H g I 2. [ H 2 ] 0 = (15.7g/2.02)/5.00 L = 1.56 M [I 2 ] 0 = (294g/253.8)/5.00L = M [HI] 0 = 0

Q= 0<K so more product will be formed. Assumption since K is large reaction will go to completion. Stoichiometry tells us I 2 is LR, it will be smallest at equilibrium let it be x Set up table of initial, change in and equilibrium concentrations.

Choose X so it is small. For I 2 the change in X must be X-.232 M Final must = initial + change H 2 (g) I 2 (g) 2HI(g) initial 1.56 M0.232 M 0 M change eq X

Initial + change = final, Change = final – initial Using to stoichiometry we can find: Change in H 2 = = x M Change in HI = (2(.232-x) = (.464-2x) H 2 (g) I 2 (g) 2HI(g) initial 1.56 M0.232 M 0 M change X eq X

Now we can determine the final concentrations by adding. Now plug these values into the equilibrium expression K = ( X) 2 = 7.1 x 10 2 (1.328+X)(X) Now you can see why we made sure X was small. H 2 (g) I 2 (g) HI(g) initial 1.56 M0.232 M 0 M change X X X eq x X.464 – 2x

W HY WE CHOSE X K = ( X) 2 = 7.1 x 10 2 (1.328+X)(X) Since X is going to be small, we can ignore it in relation to and Avoid algebra nightmare and quadratic equation! So we can rewrite the equation 7.1 x 10 2 = (0.464) 2 (1.328)(X) X = 2.8 x back to ICE table, substitute into equilibrium line

C HECKING THE ASSUMPTION The rule of thumb is that if the value of X is less than 5% of all the other concentrations, our assumption was valid. If not we would have had to use the quadratic equation. More on this later………..anyone have it programmed already?????? Our assumption was valid.

P RACTICE For the reaction Cl 2 + O 2 2ClO(g ) K = 156 In an experiment mol ClO, 1.00 mol O 2 and mol Cl 2 are mixed in a 4.00 L flask. If the reaction is not at equilibrium, which way will it shift? Calculate the equilibrium concentrations of all species.

Cl 2 O2O2 2 ClO Initial Change equilibrium

M ID - RANGE K’ S.01<K<100

QUADRATIC EQUATION TI-82 PROGRAM Hit PRGM key Using right arrow, highlight NEW and hit ENTER Give the program a name (i.e. QUAD) and hit ENTER Hit PRGM key Using right arrow, highlight I/O Highlight Input and hit ENTER Hit ALPHA key and then hit the letter A, then hit ENTER Repeat for B and C Hit PRGM key Using right arrow, highlight I/O Using down arrow, highlight Disp and hit ENTER Type in the following formula: (-B+√(B^2-4*A*C))/(2*A) and hit ENTER Repeat for the following: (-B-√(B^2-4*A*C))/(2*A) and hit ENTER To execute the program: Hit 2 nd and Quit to exit the editing of the program Hit PRGM key Highlight EXEC and use down arrow to highlight QUAD and hit ENTER PRGMQUAD should show up at the top of the screen –hit ENTER ENTER first coefficient after ? (i.e. 1) ENTER second coefficient after ? (i.e. -5) ENTER constant term after ? (i.e. 6) The answers of 3 and 2 DONE should show up in the lower right corner of the screen.

N O S IMPLIFICATION Choose X to be small. Can’t simplify so we will have to solve the quadratic H 2 (g) + I 2 (g) HI(g) K=38.6 What are the equilibrium concentrations if mol H 2, mol I 2 and mol HI are mixed in a L container?

H2H2 I2I2 2 HI Initial Change equilibrium

P ROBLEMS I NVOLVING P RESSURE Solved exactly the same, with same rules for choosing X depending on K P For the reaction N 2 O 4 (g) 2NO 2 (g) K P =.131 atm What are the equilibrium pressures if a flask initially contains atm N 2 O 4 ?

N2O42 NO 2 Initial Change equilibrium

L E C HATELIER ’ S P RINCIPLE If a stress is applied to a system at equilibrium, the position of the equilibrium will shift to reduce the stress. 3 Types of stress 1. Changing amounts of reactants or products 2. Changing pressure 3. Changing temperature

1. C HANGE AMOUNTS OF REACTANTS AND / OR PRODUCTS Adding product makes Q>K Removing reactant makesQ>K Adding reactant makesQ<K Removing product makes Q<K Determining the effect on Q will tell you the direction of shift

2. C HANGE P RESSURE By changing volume System will move in the direction that has the least moles of gas. Because partial pressures (and concentrations) change a new equilibrium must be reached. System tries to minimize the moles of gas.

C HANGE IN P RESSURE By adding an inert gas Partial pressures of reactants and product are not changed No effect on equilibrium position

3. C HANGE IN T EMPERATURE Affects the rates of both the forward and reverse reactions. Doesn’t just change the equilibrium position, changes the equilibrium constant. The direction of the shift depends on whether it is exothermic or endothermic Think about ∆G and its relation to K and ∆H.

E XOTHERMIC D H<0 Releases heat Think of heat as a product Raising temperature pushes toward reactants. Shifts to left.

E NDOTHERMIC D H>0 requires heat Think of heat as a reactant Raising temperature pushes toward products. Shifts to right.