Introduction to Introduction to EquilibriumEquilibrium

AP ChemistryAP Chemistry

EquilibriumEquilibrium Defined as the state when the Defined as the state when the

concentrations of all reactants and concentrations of all reactants and products remain constant over time. products remain constant over time.

This occurs when the This occurs when the rate of the rate of the forwardforward = = rate of the reverse rate of the reverse reaction. reaction.

Equilibrium happens because Equilibrium happens because reactants and products never stop reactants and products never stop colliding!colliding!

Equilibrium Graph•After 60 seconds the reaction has reached equilibrium

•The concentrations of the reactants and products are constant, but not necessarily equal!!!

•Rate of forward reaction = rate of reverse reaction

Not all reactions will go to completion. The Not all reactions will go to completion. The equilibrium expression will tell us how much of equilibrium expression will tell us how much of each species is present at equilibriumeach species is present at equilibrium

The Keq is constant at a given temperature and The Keq is constant at a given temperature and depends on the depends on the equilibrium concentrationsequilibrium concentrations of of products and reactants. products and reactants.

In a general reaction:In a general reaction: aAaA ++ bBbB ⇄⇄ cCcC ++ dDdD

The lower case letters represent the stoichiometric The lower case letters represent the stoichiometric coefficients and the upper case letters represent reactants coefficients and the upper case letters represent reactants and products.and products.

The law of Mass Action states that the The law of Mass Action states that the

Equlibrium constant is: Equlibrium constant is: KeqKeq = = [C][C]cc[D][D]dd

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

Example 1Example 1 What is the equilibrium expression What is the equilibrium expression

(Keq) for the following dissociation?(Keq) for the following dissociation? 2N2N22OO55(g) ⇄(g) ⇄ 4NO4NO22(g) + O(g) + O22(g)(g)

[NO2]4[O2]

[N2O5]2

Keq =

Equilibrium ConstantsEquilibrium Constants

When Keq > 1 then products are When Keq > 1 then products are favored in the reactionfavored in the reaction– Which means that they are more Which means that they are more

products than reactants at equlibriumproducts than reactants at equlibrium When Keq < 1 then the reactants are When Keq < 1 then the reactants are

favored in the reactionfavored in the reaction– Which means there are more reactants Which means there are more reactants

then products at equlibrium.then products at equlibrium.

Equilibrium ConstantsEquilibrium Constants When a reaction is multiplied by a When a reaction is multiplied by a

factor of “n”, then the Keq is raised factor of “n”, then the Keq is raised to the ‘nto the ‘nthth’ power.’ power.

2H2H22O O 2H 2H22 + O + O22 Keq = Keq = [H[H22]]22[O[O22]]

[H[H22O]O]22

Multiply by 2 4HMultiply by 2 4H22O O 4H 4H22 + 2O + 2O22(Keq)2 = [H2]4[O2]2

[H2O]4

Equilibrium ConstantsEquilibrium Constants

The equilibrium constant for the The equilibrium constant for the reverse reaction is reverse reaction is

KKeqeq(rev) = 1/K(rev) = 1/Keqeq(fwd)(fwd)

Heterogenous EquilibriaHeterogenous Equilibria Pure solids and pure liquids, and pure Pure solids and pure liquids, and pure

water are not included in equilibrium water are not included in equilibrium expressions because there expressions because there concentrations remain essentially concentrations remain essentially constant over time.constant over time.

Write the equilibrium expression for Write the equilibrium expression for the followingthe following

PClPCl55 (s) ↔ PCl (s) ↔ PCl33 (l) + Cl (l) + Cl22(g)(g)

Keq = [Cl2]

Kp versus KcKp versus Kc In homogeneous In homogeneous gasgas equilibria, partial pressure equilibria, partial pressure

may be used in place of the concentrations—the may be used in place of the concentrations—the format remains the same. format remains the same.

The term Kp is often used to describe the The term Kp is often used to describe the equilibrium constant in terms of partial pressure.equilibrium constant in terms of partial pressure.

Kc may be used when the concentrations of the Kc may be used when the concentrations of the reactants and products are expressed in terms of reactants and products are expressed in terms of molarities. molarities.

Numerical values can be calculated for any of Numerical values can be calculated for any of these. these.

By convention, units are usually omitted for any By convention, units are usually omitted for any equilibrium constant. equilibrium constant.

Water is excluded in aqueous media because the Water is excluded in aqueous media because the concentration of water does not change and/or concentration of water does not change and/or water as a product of a reaction is insignificant in water as a product of a reaction is insignificant in aqueous media.aqueous media.

Example 2Example 2 In the homogeneous gas equilibrium, In the homogeneous gas equilibrium,

dinitrogen tetraoxide decomposes as dinitrogen tetraoxide decomposes as follows:follows:

NN22OO44 (g) (g) ⇄⇄ 2NO2NO22(g)(g)

Kc = Kc = Kp = Kp = [NO2]2

[N2O4]

(PNO2)2

PN2O4

The Relationship Between Kp and The Relationship Between Kp and KcKc

For the general reaction:For the general reaction: aA(g)aA(g) ⇄⇄ bBbB    Using the Ideal Gas Law:Using the Ideal Gas Law: ppAA V V = nRT = nRT and p and pAA = = nnAA RT RT V V Similarly:Similarly: ppBB = = nnBB RT RT

VV

Substituting these terms into the Substituting these terms into the expression for Kp:expression for Kp:

nB RT V nART V

Kp =

b

a

Since n/V is concentration (Molarity):

Since n/V is concentration Since n/V is concentration (Molarity):(Molarity):

Kp = [B]b

[A]a (RT) Δn

Kp = Kc(RT) Δn

Δn = Σ (moles products) – Σ(moles reactants)

Δn = b - a

Sample ProblemSample Problem For the reaction:For the reaction: NN22(g)(g)++ 3H3H22(g)(g) ⇄⇄ 2NH2NH33(g)(g)   The following equilibrium pressures at The following equilibrium pressures at

472472ooC were obtained: HC were obtained: H22 at 7.38atm, N at 7.38atm, N22 at at 2.46 atm and NH2.46 atm and NH33 at 0.166 atm. From at 0.166 atm. From these data, calculate the Kthese data, calculate the Kpp and the K and the Kcc at at

this temperature.this temperature.

Determining Keq :Quantitative Determining Keq :Quantitative EquilibiumEquilibium

What is the equilibrium constant for the What is the equilibrium constant for the following reaction in terms of Kc?following reaction in terms of Kc?

2NO(g)2NO(g) +O+O22(g) ⇄(g) ⇄ 2NO 2NO22(g)(g) The reaction was run at 230The reaction was run at 230oo C and the C and the

following concentrations were obtained:following concentrations were obtained: [NO]= 4.065x 10[NO]= 4.065x 10-2-2 M M [O[O22] =9.525 x 10] =9.525 x 10-2-2MM

[NO[NO22] = 11.625 M] = 11.625 M

Sample Problem 2Sample Problem 2 The Kp for phosgene gas (COClThe Kp for phosgene gas (COCl22) at 800K is 3.3. ) at 800K is 3.3.

At equilibrium, the pressure of CO was found to At equilibrium, the pressure of CO was found to be 0.455 atm and that of Clbe 0.455 atm and that of Cl22 was found to be was found to be 0.356 atm. What is the partial pressure of the 0.356 atm. What is the partial pressure of the phosgene at equilibrium? The overall reaction is:phosgene at equilibrium? The overall reaction is:

COClCOCl22(g)(g) ⇄⇄ CO(g)CO(g)++ ClCl22(g)(g)