1. Keq vs. Q Javell, Shantell, Akosua
EQUILIBRIUM
Keq vs. Q
Javell, Shantell, Akosua

2. Keq Facts Keq is the constant equilibrium.
Keq is a part of the Law of Mass Action which expresses the relative concentrations of reactants and products at equilibrium in terms of a quantity.
Keq is the constant equilibrium.

3. Keq RULES
To find the Keq of a molecule put the products over the reactants.
The Keq or the equilibrium expression is written with symbols of the elements that’s on the periodic table.
Raised powers indicate coefficients, and square brackets are used to denote concentration.
aA+bB ----> cC+dD
[C]^c[D]^d
Keq=
[A]^a[B]^b

4. Example of Keq Equation
What is the equilibrium expression for this reaction ? 2CO (g) + O2(g) 2CO2 (g)
The equilibrium expression is the ratio of the product concentration to the reactant concentration.
Each concentration is raised to a power as indicated by its coefficients in the equation.
Keq= [CO2]^2
[CO]^2 [O2]

5. Now YOU TRY
What is the equilibrium expression for the reaction 2 SO2 (g) + O2  2 SO3 (g)
Answer :Keq= [SO3]^2
[SO2]^2 [O2]

6. Q Facts
The reaction quotient (Q) is used to determine if a reaction is at equilibrium.
It is different from Keq because it uses numbers at the time the concentration measurement is taken.
In order for a reaction to be at equilibrium Q would have to equal Keq because the quotient is the equilibrium expression using concentrations at a given time.

7. Rules for Q
If Q < Keq then the reaction will proceed to the right in the direction of the products because at the time of the reaction the product was too small and the reactants was too large; therefore the reaction will consume reactants and form products to reach equilibrium.
If Q > Keq then the reaction will proceed to the left in the direction of the reactants because the reactants were too small and the products were too big. For the system to reach equilibrium reactants must be formed and products must be consumed.
If Q = Keq the system is at equilibrium. There is no shift.

8. Example of a Q equation COCl2 (g)  CO (g) + Cl2 (g) Keq= 170
If the concentrations of CO and Cl2 are each 0.15 M and the concentration of COCl2 is 1.1 x 10^-3 M, is the reaction at equilibrium? If not, in which direction will it proceed ?
Q= [CO] [Cl2] = (0.15) (0.15) = 20
[COCl2] x 10^-3
Q < Keq which means the reaction is not at equilibrium and it proceeds to the right.

9. Now YOU try
A vial contains M NO2, and M N2O4. Calculate Q for the reaction 2 NO2 (g)  N2O4 (g)
Answer 13.3