Chemical Equilibrium

The study of reactions that occur in both directions.

So far with reactions….

Looked at reactions that go to completion

Used stoichiometry for calculation of many quantities

Looked at spontaneity and rates of reactions

Now…

Reactions can be reversible– They reach a state of equilibrium

Examples

Vapor pressure– Evaporation begins– Over time, the

system undergoes evaporation and condensation at the same rate

Examples

Dissolving and Crystallization– A system could

have an equal amount of precipitate (changes states at an equal rate)

System of iron (II) dichromateat equilibrium

Examples

NO2 (g) + NO2 (g) N2O4 (g)

– NO2 = dark brown, N2O4 = colorless

– Ultimately ends up somewhere in between

Equlibrium Change in Action

NO2

N2O4

Equilibrium!!

Equilibrium Defined

Concentration of products and reactants remain constant over time– The reaction is reversible (can go both

directions)– The rate of forward reaction equals the

rate of the reverse reaction– Dynamic!! (looks the same when taking a

snapshot, but constantly moving back and forth)

Demo, then Graph of Equilibrium

Time

[R]

[P]

Explaining the Graph

Time

[R]

[P]

What’s happening to the products?

What’s happening to the reactants?

Which one is favored?

Different reactions have different equilibria…

On a molecular level…

Where does the following reach equilibrium?

Equilibrium Expressions

Idea by Guldberg and Waage (1864)– Called the Law of Mass Action

– Given a reaction:

aA + bB cC + dD

Equilibrium Expressions

aA + bB cC + dD

=[C]c [D]d

[A]a [B]b

Equilibriumconstant

MUST useconcentrations ofproducts over reactants

coefficients of balancedequation becomeexponents

Equilibrium Expressions

Works only for GASES and IONS– No pure solids or liquids included

Example– C3H8 (g) + O2 (g) CO2 (g) + H2O (g)

=[C]c [D]d

[A]a [B]b

=[CO2]3 [H2O]4

[C3H8] [O2]5

Equilibrium Expressions

What if it has solids or liquids?– Called heterogeneous equilibrium– The concentration of solids and liquids is

assumed to always remain constant, so they are not included…

– Example: Ca (s) + O2 (g) CaO (s)

Equilibrium Expressions

More examples

– H2 (g) + I2 (s) 2 HI (g)

– CuSO45 H2O (s) CuSO4 (s) + 5 H2O (g)

– N2 (g) + 3 H2 (g) 2 NH3 (g)

Values of K Equilibrium constant, K, is found by:

[products]

[reactants]

– If K = 1…..• equal ratio of products and reactants

– If K > 1 ….• reaction favors products

– If K < 1 ….• reaction favors reactants

What changes K?

1. Change the temperature.– Equilibrium is temperature dependent.

What changes K?

2. Change the reaction– Look at the original reaction and see the change

that was made to it.• Whatever you do to the reaction, you do to the

power of K

– Example• 2 NO(g) + O2 (g) 2 NO2 (g) K= 4.67 x 1013

• Change the coefficients• NO(g) + 1/2 O2 (g) NO2 (g)

• What is the change?• K’ = K1/2 = (4.67 x 1013)1/2 = 6.83 x 106

What changes K?

2. Change the reaction– Look at the original reaction and see the change

that was made to it.• Whatever you do to the reaction, you do to the

power of K

– Example• 2 NO(g) + O2 (g) 2 NO2 (g) K= 4.67 x 1013

• Reverse the reaction• 2 NO2 (g) 2 NO (g) + O2 (g)

• What is the change?• K’ = 1/K = 1/(4.67 x 1013) = 2.14 x 10-14

Equilibrium Constant

Can also find from reaction mechanism– OVERALL:

• N2O(g) + 3/2 O2 (g) 2 NO2 (g) Kc = ??

– Steps:• N2O(g) + 1/2 O2 (g) 2 NO (g) Kc1 = 1.7 x 10-13

• 2 NO(g) + O2 (g) 2 NO2 (g) Kc2 = 4.67 x 1013

– To find Kc?

• Kc = Kc1 x Kc2

Equilibrium including Gases

Look at partial pressures instead of molarities– aA (g) + bB (g) cC (g) + dD (g)

=(PC)c (PD)d

(PA)a (PB)b

Equilibrium including Gases

If it says Kp, you must use pressures

– Given moles and L, must use temperature and PV = nRT to get pressures of each species

K or Kc is still molarity

Practice Problems

Write equilibrium expressions for:– a) 2 O3(g) 3 O2(g)

– b) 2 NO(g) + Cl2(g) 2 NOCl(g)

– c) BaSO4(s) Ba+2(aq) + SO4-2(aq)

Practice Problems

Changing K by changing the equation– Original equation:

2 NO (g) + O2 (g) 2 NO2 (g) Keq = 0.00103

• NO (g) + ½ O2 (g) NO2

• 2 NO2 (g) 2 NO (g) + O2 (g)