Chapter 13
Chemical Equilibrium
AP Chemistry
Ms. Vorak
David Choi
#16625
►Equilibrium constant Concepts
● The Equilibrium condition is highly dynamic situation. The equilibrium is analogous to the flow of cars
across a bridge connecting two island cities.
● We use concentrations in order to calculate chemical equilibrium rate.
● When the system is a t chemical equilibrium, the forward and reverse reactions are so slow that the
system moves toward equilibrium at a rate that can’t be detected
● The law of mass action is based on experimental observation.
Important concepts of Equilibrium expression
1. The equilibrium expression for a reaction is the reciprocal of that for the reaction written in
reverse
2. When the balanced equation for a reaction is multiplied by a factor n, the equilibrium expression
for the new reaction is the original expression raised to the nth power. Thus Knew = (Koriginal)^n.
3. K values are customarily written without units.
● For the reaction jA + kB = lC + mD, K = ([c]^l*[d]^m)/([a]^j*[b]^k)
►Equilibrium constant using pressure
● Remember that since we are using concentrations for K value we can replace the concentrations using the
equation PV = nRT.
Kp = K(RT)^(n)
Also one can just simply use pressures to find the Kp = (Preactants)^n / (Pproducts)^m
►Heterogeneous Equilibria
When writing out a heterogeneous equilibrium constant remember that the constant, K, doesn’t
depend on the amounts of pure solids or liquids present. Thus, one should take out pure solids or liquids
when obtaining the value of K.
►The extent of a reaction
● The inherent tendency for a reaction to occur is indicated by the magnitude of the equilibrium constant. A
value of K much larger than 1 means that at equilibrium the reaction system will consist of mostly products.
The equilibrium lies to the right.
● If the K value is very small. The system at equilibrium will consist mostly of reactants.
►Reaction Quotient
Q = [reactant]^n / [product]^m
This is obtained by applying the law of mass action using initial concentrations instead of equilibrium
concentrations.
● If Q is equal to K the system is at equilibrium
● If Q is greater than K the ratio of initial concentrations of products to initial concentrations of reactants is
too large. This system shifts to the left.
● If Q is less than K. The ratio of initial concentrations of products to initial concentrations of reactants is
too small. The system must shift to the right, consuming reactants and forming products to attain
equilibrium.
►Le Chatelier’s principle
● If a component is added to a reaction system at equilibrium at constant T and P or constant T and V the
equilibrium position will shift in the direction that lowers the concentration of that component. If a
component is removed, the opposite effect occurs.
- Effect of a change in pressure
Three ways to change the pressure
1. Add or remove a gaseous reactant or product
2. Add an inert gas (one not involved in the reaction) Also it only increases the pressure
3. Change the volume of the container
Lowering the volume means lowering the molecules. Thus, the system will try to lower the number of
molecules in the system. In order to predict the shift, one should count the molecules of gas first.
- Effect of a change in temperature
● Value of K changes by temperature unlike volume.
● Shift will be in the direction that consumes the energy. The easy way to find it is treat energy as a
reactant or product.
►Conclusive summary of Le Chatelier’s principle
Le Chatelier’s principle allows us to predict qualitatively the effects of changes in concentration,
pressure, and temperature on a system at equilibrium. This principle states that if a change is imposed, the
equilibrium position will shift in a direction that tends to compensate for the imposed change.
►5 multiple choice questions
1. non-calculation
2. non-calculation
3. non-calculation
4. calculation
The reaction for the formation of nitrosyle chloride
2NO(g) + Cl2(g) 2NOCl(g)
was studied at 25 ˚C. The pressures at equilibrium were found to be
P(NOCl) = 1.2 atm
P(NO) = 5.0*10^(-2) atm
P(Cl2) = 3.0 * 10^-1 atm
Calculate the value of Kp for this reaction at 25 ˚C
(A) 111.2
(B) 122412.12
(C) 1.9 * 10^3
(D) 1.3 *10^3
5.Using the value of Kp from number 4 calculate the value of K at 25 ˚C for the reaction.
2NO(g) + Cl2(g)  2NOCl(g)
Answer
1. D
2. C
3. A
4. C
For this reation,
Kp = (P(NOCl)^2) / ((P(NO2))^2(PCl2)) = (1.2)^2/((5.0*10^-2)^2(3.0*10^-1))
= 1.9* 10^3
5. Using Kp = K(RT) ^(difference of moles)
where T = 298 K and the difference of moles is -1
Thus Kp – K/RT and K = Kp(RT) = (1.9*10^3)(0.08206)(298)
= 4.6 * 10^4
►1 multipart free response
Calculating the Values of K
The following equilibrium concentration were observed for the Haber proess at 127 ˚C:
[NH3] = 3.1 * 10^-2 mol/L
[N2] = 8.5 * 10^-1 mol/L
[H2] = 3.1 *10^-3 mol/L
a.
Calculate the value of K at 127 ˚C for this reaction
b.
Calculate the value of the equilibrium constant at 127 ˚C for the reaction
2NH3(g)  N2(g) + 3H2(g)
Calculate the value of the equilibrium constant at 127 ˚C for the reaction given by the
equation
1/2N2(g) + 3/2H2(g)  NH3(g)
c.
Answer
a.
the balanced equation for the Haber process is
N2(g) + 3H2(g)  2NH3(g)
Thus K = [NH3]^2 / ([N2][H2]^3)  (3.1*10^-2)^2 / ((8.5*10^-1)(3.1*10^-3)^3)
Note that K is written without units
b. The reaction is written in the reverse order from the equation given in part a. This leads to the
equilibrium expression
K’ = ([N2][H2]^3) / [NH3]^2
Which is the reciprocal of the expression K
1/K = 1/(3.8*10^4) = 2.6 *10^-5
c. We use the law of mass action
K” = [NH3] / ([N2]^(1/2)[H2]^(3/2))
If we compare this expression to that obtained in part a, we see that since
[NH3] / ([N2]^(1/2)[H2]^(3/2)) = ([NH3]^2 / ([N2][H2]^3))^(1/2)
K” = K^(1/2)
Thus K” = K ^(1/2) = (3.8*10^4)^(1/2) = 1.9 * 10^2