Chemical Equilibrium……….. Until now, we’ve treated reactions as though they can only go in...

Chemical Equilibrium………..Until now, we’ve treated reactions as

though they can only go in one direction….with all of the reactants turning into products…

Most reactions that we have witnessed appear this way…

Are all reactions like this?

Is it possible to reverse a reaction after it is completed?

Can we turn CO2 and H2O back into gasoline and O2 after it burns?

2 C8H18 + 25 O2 → 16 CO2 + 18 H2O

16 CO2 + 18 H2O → 2 C8H18 + 25 O2 ?

Chemical equilibrium

Most chemical reactions are reversible!Most chemical reactions are reversible! Reactions will continue to occur until they finally Reactions will continue to occur until they finally

maintain a ratio of products versus reactants that maintain a ratio of products versus reactants that will not changewill not change

This ratio is called the Equilibrium Constant This ratio is called the Equilibrium Constant Expression, or KExpression, or Keqeq, or K, or Kcc, and is equal to the final , and is equal to the final product concentration over the final reactant product concentration over the final reactant concentrationconcentration

K = ProductsK = Products ReactantsReactants

Let’s take the following chemical reaction:Let’s take the following chemical reaction:

aA + bB <-------------> cC + dDaA + bB <-------------> cC + dDWhen a chemical reaction reached equilibria, the When a chemical reaction reached equilibria, the

Equilibrium Constant Expression, or KEquilibrium Constant Expression, or Keqeq, is:, is:

KKeqeq = = [C][C]cc[[D]D]dd

[A][A]aa[B][B]bb

Chemical Equilibrium (continued)

A, B, C, and D represent the concentration of each A, B, C, and D represent the concentration of each chemical at the end of the reactionchemical at the end of the reaction

This concentration is measured in Molarity, or MThis concentration is measured in Molarity, or M Once the reaction reaches equilibrium, the concentrations Once the reaction reaches equilibrium, the concentrations

of these molecules WILL NOT CHANGEof these molecules WILL NOT CHANGE

KKeqeq = = [C][C]cc[[D]D]dd

[A][A]aa[B][B]bb

Let’s examine the following reaction……

This reaction is reversibleThis reaction is reversible At equilibrium, the constant, or KAt equilibrium, the constant, or Keqeq, is equal to:, is equal to:

KKeqeq = = [NO[NO22]]22

[N[N22OO44]]11

BUT WHAT DOES THIS TELL US?BUT WHAT DOES THIS TELL US?

NN22OO4(g) 4(g) <-----------> 2<-----------> 2 NO NO2(g)2(g) at 25 at 2500CC

It tells us whether we have more products or reactants at equilibrium!!!!

Looking at the same reaction:Looking at the same reaction:

NN22OO4(g)4(g) <-----------> 2 NO <-----------> 2 NO2(g) 2(g) at 25at 2500CC

If the concentration of NIf the concentration of N22OO4(g)4(g) at equilibrium was .0445 M and the at equilibrium was .0445 M and the

concentration of NOconcentration of NO2(g)2(g) was .0161 M, then: was .0161 M, then:

KKeqeq = = [NO[NO22]]2 2 = = [.0161][.0161]2 2 = .00582 = = .00582 = 582582

[N[N22OO44] [.0445]] [.0445] 100,000 100,000

This is the same as:This is the same as: 5.825.82

10001000

Is this a exothermic or endothermic reaction….and why?

Why is this important in the real Why is this important in the real world?world?

IT IS IMPORTANT BECAUSE:The KThe Keqeq tells us whether we have MORE tells us whether we have MORE

PRODUCTS OR REACTANTS AT PRODUCTS OR REACTANTS AT EQUILIBRIUM!EQUILIBRIUM!Remember, KRemember, Keqeq are temperature specific - at are temperature specific - at

higher or lower temperatures, the ratios will be higher or lower temperatures, the ratios will be different, because temperature affects a reaction!different, because temperature affects a reaction!

Suppose we have this reaction:COCO(g)(g) + H + H22OO(g)(g) <--------> H <--------> H2(g)2(g) + CO + CO2(g)2(g)

The KThe Keqeq = = 5.105.10 at 527 at 52700C.C.

11 This means that products are favored over reactants almost 5 to 1!!This means that products are favored over reactants almost 5 to 1!! Let’s say we are at some point in time, NOT AT EQUILIBRIUM and we measure the concentrations of each of the molecules in the reaction……

Is it possible to predict if a reaction is at equilibrium?

COCO(g)(g) + H + H22OO(g)(g) <--------> H <--------> H2(g)2(g) + CO + CO2(g)2(g)

The KThe Keqeq = 5.10 at 527 = 5.10 at 52700C.C. The concentrations are [CO] = .15 M, The concentrations are [CO] = .15 M,

[H[H22O] = .25 M, [HO] = .25 M, [H22] = .42 M, and ] = .42 M, and

[CO[CO22] = .37 M. ] = .37 M. Can you set up a ratio of concentrations?Can you set up a ratio of concentrations?

DO IT NOW!!!!!DO IT NOW!!!!!

Does your ratio look like this?

Q = [.42]Q = [.42]11[.37][.37]11

[.15][.15]11[.25][.25]11

This is called the REACTION QUOTIENT, OR QThis is called the REACTION QUOTIENT, OR Q It is set up the same way as KIt is set up the same way as Keqeq, but it is not K, but it is not Keqeq – –

Why is it not the same as KWhy is it not the same as Keqeq??

What does Q tell you???

The reaction quotient, or Q, tells you where you The reaction quotient, or Q, tells you where you are in the reaction. You compare it to your Kare in the reaction. You compare it to your Keqeq!!

Q = [.42]Q = [.42]11[.37][.37]1 1 = = 4.144.14

[.15][.15]11[.25][.25]11 11

Q = 4.14, which is less than KQ = 4.14, which is less than Keqeq = 5.10 = 5.10

Q = 4.14, which is less than Keq = 5.10 1 1

If Q < KIf Q < Keqeq, then the reaction will proceed toward the , then the reaction will proceed toward the

products to balance the ratio!products to balance the ratio! If Q > KIf Q > Keqeq, then the reaction will proceed toward the , then the reaction will proceed toward the

reactants to balance the ratio!reactants to balance the ratio! It Q = KIt Q = Keqeq, then the reaction is…….., then the reaction is……..

AT EQUILIBRIUM!!!!!AT EQUILIBRIUM!!!!! That means it is possible for a reaction to reverse itself That means it is possible for a reaction to reverse itself

and turn product back into reactant!and turn product back into reactant!

AT EQUILIBRIUM…- The forward reaction, where molecules A - The forward reaction, where molecules A and B are forming molecules C and Dand B are forming molecules C and D

A + B --------> C + DA + B --------> C + D

- Occurs at the same rate as the reverse - Occurs at the same rate as the reverse reaction, the reaction where molecules C and reaction, the reaction where molecules C and D are forming molecules A and BD are forming molecules A and B

C + D --------> A + BC + D --------> A + B

That is how the ratio at equilibrium never changes!

Chemical Equilibrium does not mean: That there are equal That there are equal

concentrations of concentrations of reactant and product reactant and product when the reaction when the reaction reaches equilibrium!reaches equilibrium!

It simply means that It simply means that there is eventually a there is eventually a ratio reached that will ratio reached that will not change!not change!

Let’s watch a computer simulation of equilibrium in action!

Le Chatelier’s Principle:

If a system at equilibrium is disturbed by a If a system at equilibrium is disturbed by a change in temperature, pressure, or the change in temperature, pressure, or the concentration of one of the molecules, the concentration of one of the molecules, the system will shift its equilibrium position to system will shift its equilibrium position to counteract the disturbance……counteract the disturbance……

WHAT DOES THAT MEAN….?WHAT DOES THAT MEAN….?

Fritz Haber was a German Scientist who utilized LaChatelier’s Principle in making ammonia for explosives in World War I:

NN2(g)2(g) + 3H + 3H2(g) 2(g) ------> 2 NH------> 2 NH3(g)3(g)

KKeqeq = = 9.609.60

11

at 300at 30000C and 200 atmC and 200 atm At equilibrium, products are At equilibrium, products are

favored, but how do we get the favored, but how do we get the reaction to reaction to continuecontinue to make to make NHNH33, or ammonia?, or ammonia?

N2(g) + 3H2(g) <-----> 2 NH3(g)

By adding more NBy adding more N2(g) 2(g) or Hor H2(g)2(g), the reaction , the reaction

would…..?would…..? Shift to the right and make more ammonia - but Shift to the right and make more ammonia - but

this requires more reactant and more money!this requires more reactant and more money! By adding more NHBy adding more NH3(g)3(g), the reaction would….?, the reaction would….?

Shift to the left and form more NShift to the left and form more N2(g)2(g) and H and H2(g)2(g), ,

which is NOT what we want!which is NOT what we want!

N2(g) + 3H2(g) <-----> 2 NH3(g)

By removing NHBy removing NH3(g)…..3(g)…..

The reaction shifts to the right, NThe reaction shifts to the right, N2(g)2(g) and H and H2(g)2(g) hurrying hurrying to create equilibrium again and create more NHto create equilibrium again and create more NH3(g)3(g). .

This causes even more NHThis causes even more NH3(g)3(g) to be produced, driving to be produced, driving the reaction to completion and producing the the reaction to completion and producing the maximum amount of ammonia!maximum amount of ammonia!

The ammonia is removed by liquefying it at -33The ammonia is removed by liquefying it at -3300CC The reaction will always maintain its KThe reaction will always maintain its Keqeq = = 9.609.60

11

Kc, or Keq, only applies to chemicals reacting in the solid, liquid, or gaseous state. It does not apply to aqueous solutions (reactions that occur in water).

Ksp deals with aqueous reactions, or reactions that occur in water!