Chem 2
Chapter # 17 Equilibrium.
CHEMICAL EQUILIBRIUM
• The state where the concentrations of all reactants and products remain constant with
them.
• All reactions are reversible.
• Forward and reverse rates are equal at equilibrium.
• Any chemical reaction carried out in a close vessel reaches equilibrium.
• On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly
dynamic situation.
The Equilibrium condition:
H2 O (g) + CO (g) __________ H2 (g) +
CO2(g)
X moles
x moles
y moles
y moles
Same concentrations.
At start reactants conc decreases, products conc increases. Conc of R and P don’t change when
equilibrium is reached. Although equilibrium position lies to far right reactants never go to zero
conc. (they are negligible).
If we add more H2 O collisions increase , this will form more products causing reverse reaction to
occur. Both reactions take place till them become equal again at equilibrium.
Equ position of areaction depends on:
1) initial conc 2) Relative energies of R and P 3) Relative degree of organization of R and P.
4) minimum energy, maximum disorder.
THE EQUILIBRIUM CONSTANT:
The Law of Mass Action
For jA + kB  lC + mD
The law of mass action is represented by the equilibrium expression:
C l D m
K
A j Bk
LAW OF MASS ACTION: At a given temperature a given reaction reaches a state when the ratio
of conc of P and R is constant.
K=equ constant.
Equilibrium constant= Ratio of equilibrium concentrations of product and reactant at a particular
temperature.
Magnitude of K indicates how far a reaction proceeds towards the given product.
Write equ expression for:
4NH3(g) + 7O2(g)  4NO2(g) + 6H2O(g)
The Extent of a Reaction:
1) Small K: If a reaction yields very little product before going to equilibrium.
NO REACTION.
K < 1, equ system consists only of R, equ lies to left.
2) Large K: If a reaction reaches equilibrium with very little reactant left.
GOES TO COMPLETION.
K > 1, equ system consists only of P, equ lies to right.
3) Intermediate K: Both R and P present in significant amount at equilibrium.
REACTION QUOTIENT: It helps to determine the direction of the move toward equilibrium.
The law of mass action is applied with initial concentrations.
1) Q=K, system is equ. No shift will occur.
2) Q> K, system shifts to left.
3) Q< K, system shifts to right.
For jA + kB  lC + mD
Q=
Write reaction quotient for:
1) H2(g) + F2(g)  2HF(g)
2) N2O5(g)  NO2(g) + O2(g)
3) C3H8(g) + O2(g)  CO2(g) + H2O(g)
Form of Q for an Overall Reaction:
K and Q have no units.
Qoverall= Q1x Q2x Q3x …………………
Koverall=K1x K2x K3x …………………..
(a) Show that the Qc for the overall reaction sequence is the same as the product of the Qcs
of the individual reactions.
(b) Calculate the Kc for the overall reaction
Qc(forward)= 1/Qc (reverse)
Kc(forward)= 1/Kc (reverse)
n ( jA + kB  lC + mD)
Q’ =Qn
and K’ =Kn
For


Ratio of initial concs varies widely but always gives same ratio of equ concs.
The individual equilibrium concs are different , but ratio of these equ concs is constant.
Equilibrium expression:
1)The Equilibrium Expression for a reaction is the reciprocal of that for the reaction written
in reverse.
2)When the equation for a reaction is multiplied by n, EEnew = (EEoriginal)n
2) K has no units.
Equ position: Each set of equ. Constants is called equ position.
Equ constant : It is always the same and does not depend on the concentrations.
Table 17.2, Pg. 732.
For the ammonia formation reaction
N2(g) + 3H2(g) ↔ 2NH3(g)
the equilibrium constant, Kc, is 2.4x10-3 at 1000K. If we change the coefficients of the equation,
which we’ll call the reference (ref) equation, what are the values of Kc for the following balanced
equations?
a) 1/3N2(g) + H2(g) ↔
2/3NH3(g)
(b) NH3(g)
↔ 1/2N2(g) + 3/2H2(g)
Heterogeneous equilibria: that involve more than one phase.
CaCO3(s)  CaO(s) + CO2(g)
K = [CO2]
The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or
liquids present.
EQU EXPRESSIONS INVOLVING PRESSURE:
PV=nRT
P=n/V(RT)= CRT, C=molar concentration of gas.
For
jA + kB  lC + mD
Kp = Kc(RT)n
n = sum of coefficients of gaseous products minus sum of coefficients of gaseous reactants.
Calculate Kp for the following reaction:
PCl3(g) + Cl2(g) ↔ PCl3(g) where Kc= 1.67(500 K)
Do Follow up Problem 17.5 Pg.735.
Solving Equilibrium Problems
1. Balance the equation.
2.Write the equilibrium expression.
3.List the initial concentrations.
4.Calculate Q and determine the shift to equilibrium.
5. Define equilibrium concentrations.
6.Substitute equilibrium concentrations into equilibrium expression and solve.
7.Check calculated concentrations by calculating K.
1) 2HI (g) ↔ H2 (g) + I2 (g)
A 2.00 L flask is filled with 0.200 mol of HI gas , at equilibrium [HI]= 0.078 M . Calculate Kc.
2) CH4 (g) + H2O (g) ↔ CO (g) + 3H2(g)
A 0.32 L flask is filled with methane and water at 1200 K. At equilibrium, the flask contains
0.26 mol of CO, 0.091 mol of H2, 0.01 mol of CH4. What is the [H2O] at equilibrium? Kc=0.26.
3) CO (g) + H2O (g) ↔CO2 (g) + H2(g)
At 700 K the equilibrium constant for this reaction is 5.10. Calculate the equilibrium
concentrations of all species if 1.000 mol of each component is mixed in 1.000 L flask.
4)H2(g) + F2(g) ↔2HF(g)
If 3.000 mol of H2 and 6.000 mol of F2 are mixed together in a 3.000 L flask the equilibrium
constant is 1.15 x 102. Calculate the equilibrium concentration of each component.
5)COCl2 ↔CO (g) + Cl2 (g) , Kc=8.3 x 10-4
5.00 mol of phosgene was decomposed in 10.0 L flak. Calculate the concentration of all the
species present at equilibrium.
6) In one experiment, 1.00mol of CH4, 1.00mol of CS2, 2.00mol of H2S, and 2.00mol of H2 are
mixed in a 250-mL vessel at 9600C. At this temperature, Kc = 0.036
In which direction will the reaction proceed to reach equilibrium?
If [CH4] = 5.56M at equilibrium, what are the equilibrium concentrations of the other
substances?
LE CHÂTELIER’S PRINCIPLE
If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a
direction that tends to reduce that change.
Effects of Changes on the System
1.Concentration: The system will shift away from the added component.
2.Temperature: K will change depending upon the temperature (treat the energy
change as a reactant).
3.Pressure:
a. Addition of inert gas does not affect the equilibrium position.
b. Decreasing the volume shifts the equilibrium toward the side with fewer moles.
1)How does an increase in temperature affect the concentration of the underlined substance and
Kc for the following reactions?
CaO(s) + H2O(l)
↔ Ca(OH)2(aq) DH0 = -82kJ
CaCO3(s) ↔
CaO(s) + CO2(g) DH0 = 178kJ
SO2(g) ↔
S(s) + O2(g) DH0 = 297kJ
2) 2H2S(g) + O2(g) ↔ 2S(s) + 2H2O(g)
What happens to
[H2O] if O2 is added?
[H2S] if O2 is added?
[O2] if H2S is removed?
[H2S] if sulfur is added?
The Van’t Hoff Equation: The Effect of T on K: