1. Chemical Equilibrium…demo time…
Chapter 13
Chemical Equilibrium…demo time…

2. Chemical Equilibrium
The state where the *concentrations* of all of reactants and products remain constant with time
Any chemical reaction carried out in a closed container will reach equilibrium
Reactions can reach equilibrium far to the right (products)…or the left (reactants)
Yet to come: one can calculate the concentration of reactant(s) and product(s) at equilibrium

3. The Equilibrium Condition
At equilibrium, no NET change is occurring. This does NOT mean that the reaction has stopped. Rate of forward reaction = rate of reverse reaction
Cars on bridge analogy, demo
H2O + CO <-> H2 + CO2

4. Equilibrium Position (R, L, Center…)
Determined by…
Initial concentrations
Relative energies of reactants/products
Relative degree of “organization” of reactants/products
*Nature wants minimum energy and maximum disorder (we will revisit this later)

5. Equilibrium Expression (Law of Mass Action)
For reaction: jA + kB <-> lC + mD
K = [C]l[D]m
[A]j[B]k
Liquids and solids aren’t included
K = equilibrium constant (no unit)
If [product] > [reactant], then K>1. If [reactant] > [product], then K<1.

6. 4NH3(g) + 7O2(g) <-> 4NO2(g) + 6H2O(g)
Example
Write the equilibrium expression for the following reaction:
4NH3(g) + 7O2(g) <-> 4NO2(g) + 6H2O(g)
You can use the equilibrium expression to determine the value for K

7. Example
Using the expression from the last example and [NH3] = 3.1 X 10-2mol/L, [N2] = 8.5 X 10-1mol/L, [H2] = 3.1 X 10-3mol/L
Calculate the value of K
Calculate the value of the equilibrium constant for the reaction 2NH3(g) <-> N2(g) + 3H2(g)
Calculate the value of the equilibrium constant for the reaction given by the equation /2N2(g) + 3/2H2(g) <-> NH3(g)

8. Changing the EE You can write equilibrium expressions in reverse
**example with last reaction
You can multiply all of the coefficients in a reaction by a constant which will also change the equilibrium expression

9. K Summary… EE in reverse reaction -> K’=1/K
When exponents * by n -> K’’ = Kn
K has no unit

10. Equilibrium Position
The equilibrium constant for a reaction system at a given temperature never changes
Equilibrium Positions (sets of equilibrium concentrations) have infinite possibilities
See page 600 Table 13.1

11. Example
Determine the equilibrium constant for both sets of data. Reaction: sulfur dioxide and oxygen are added to form gaseous sulfur trioxide
Experiment 1
Experiment 2
Initial
Equilibrium
[SO2]0 =2.00 M
[SO2] = 1.50 M
[SO2]0=0.500M
[SO2] =0.590 M
[O2]0 = 1.50 M
[O2] = 1.25 M
[O2]0 = 0
[O2] = M
[SO3]0 =3.00 M
[SO3] = 3.50 M
[SO3]0=0.350M
[SO3] =0.260 M

12. Let’s Add Some Pressure
Rearranging the Ideal Gas Law (PV = nRT), “C” can be substituted for n/V (P = CRT)
C = moles per volume = molar concentration of the gas
Equilibrium constant can be written in terms of partial pressures (Kp)
Kp = PNH32
(PN2)(PH23)

13. Example Calculate Kp for the following reaction
CH3OH(g) <-> CO(g) + 2H2(g)
given the equilibrium pressures as follows:
PCH3OH = 6.10 X 10-4 atm PCO = atm PH2 = 1.34 atm
Answer: 1.14 X 103 atm2

14. Relating K to Kp
When the sum of the coefficients on either side of a balanced equation are equal, K = Kp.
In general,
Kp = K(RT)∆n
∆n = sum of coefficients of the gaseous products - sum of coefficients of gaseous reactants
R = kgm2/s2Kmol…J/Kmol OR Latm/Kmol
T = Kelvin (use if 273 doesn’t work online)

15. Example
Calculate K for the reaction at 25°C using the previous value for Kp.
Answer: 1.90 mol2L-2

16. Heterogeneous Equilibria
The equations we’ve seen so far have gases as all of the reactants and products (Homogeneous Equilibria)
Heterogeneous Equilibria: more than one phase in a reaction
Concentrations of solids and liquids will not change, so do not need to be included in the equilibrium expression. GASES and SOLUTIONS (aq) are included
2H2O(l) <-> 2H2(g) + O2(g)

17. Examples Write the expressions for K and Kp for the following:
a. Solid phosphorus pentachlorde decomposes to liquid phosphorus trichloride and chlorine gas.
b. Deep blue solid copper (II) sulfate pentahydrate is heated to drive off water vapor to form white solid copper (II) sulfate.

18. What Does It Mean??
Values can be substituted in for concentrations, pressures, and/or K to determine missing variables
Whether or not a reaction is at equilibrium can be determined through a calculation similar to this (given concentrations, solve for K to compare)