15 Bluffers Guide

Bluffer’s Guide: 15.1
- Chemical Equilibrium- occurs when opposing reactions are proceeding at equal rates
- At equilibrium, the rate at which products are produced from reactants equals the rate at
which reactants are produced from products
- For Example:
Forward Reaction:
A → B Rate = kf[A]
Reverse Reaction:
Rate = kr[B]
Where kf and kr are rate constants for the forward and reverse reactions,
- For gaseous substances, use the ideal-gas equation to convert between
concentration (in molarity, M) and pressure (in atm):
PV = nRT,
M = (n/V) = (P/RT)
So for substances A and B:
[A] = (PA/RT)
[B] = (PB/RT)
- And the rates for the forward and reverse reactions are:
Forward: Rate = kf (PA/RT)
Reverse: Rate = kr (PB/RT)
-At equilibrium, Kf
= Kr B
-By rearranging this equation, a constant is established:
( P B / RT ) P B k f
= constant
( P A / RT ) P A k r
At equilibrium, the ratio of the partial pressures of A and B equals a constant.
- At equilibrium, the partial pressures of A and B don’t change, aka an equilibrium
mixture, but the reactants continue to reactant; this occurs because the forward and
reverse reactions are occurring at the same rate, so that there is no net change in the
reactants’ amounts.
- The equilibrium is dynamic.
-Opposing reactions naturally lead to an equilibrium situation.
- ex: the Haber Process
15.2 The Equilibrium Constant
The equilibrium condition can be reached from either direction
The ratio of the partial pressures in the reaction A(g) B(g) has a constant value
Law of mass action: relationship between concentrations of reactions and
products present in equilibrium
Develops the equilibrium constant expression…
Look at this reaction: aA(g) + bB(g) cC(g) + dD(g)
o A,B,C,D are chemicals, and a,b,c,d are coefficients
o The law of mass action states (when everything is in the gas phase):
o This is the equilibrium constant expression
o Keq is the equilibrium constant
o The expression depends only on the stoichiometry of the reaction,
NOT the mechanism
o When temperature is constant, Keq doesn’t vary with initial amounts of
reactants and products, but only with temperature
The Magnitude of Equilibrium Constants
 Keq>>1 equilibrium lies to the right (toward the product side, products
Keq<<1 the equilibrium lies to the left (towards the reactants, reactants
The Direction of the Chemical Equation and Keq
 Since equilibrium can be approached from either side of a reaction, the direction
that the chemical equation is written is arbitrary
o RXN X: aA(g) + bB(g)
cC(g) + dD(g), then:
o RXN Y: cC(g) + dD(g)
aA(g) + bB(g), then
o Keq of X =
(They are reciprocals of one another)
o So, for a Keq specify reaction and temperature
Other Ways to Manipulate Chemical Equations and Keq Values
o Raise Keq to the power of the factor that the coefficients are multiplied by
 If the coefficients of a reaction are doubled, square the equilibrium
 If coefficients were divided by 2, take the square root of the
original Keq
o If two reactions are added, multiply the equilibrium constants together
Units of Equilibrium Constants
o Keq is unitless BECAUSE…
 The numbers entered into an equilibrium constant expression are
divided by a Reference Pressure (1 atm.) or Reference
Concentration (1 M)
 ALLOWS US to use partial pressures and concentrations in the
SAME expression
 See p. 586 for amazing explanation of unitlessness of Keq
Sample Problems: p.605 #15 - 16
15. HClO2(aq) H+
+ ClO2-(aq) Keq = 1.1
a) If coefficients are divided in half, take the square root of Keq.
b) If coefficients are doubled, square the original Keq
= 1.2
c) If the coefficients are doubled and the reverse reaction occurs, square the
inverse of the
original Keq.
16. Assume all components are aqueous…
RXN 1: A + B  C
RXN 2: C + D  E + A
Using Hess’s Law, the reaction simplifies to:
RXN 3: B + D  E
(A is a catalyst and C is an intermediate, so they drop out)
Keq for RXN 3 = (Keq RXN 1)(Keq RXN 2)
Keq =
Keq = 0.16
15.3 Heterogeneous Equilibria
Homogeneous equilibria- involve substances that are all in the same phase
Heterogeneous equilibria- involve substances that are in different phases
Pure solids, pure liquids, and solvents are not used in the equilibrium-constant
Only partial pressures of gases and molar concentrations of substances in solution
are included in the expression
Pure solids and liquids which are involved in the reaction are present at
equilibrium, even though they do not show up in the equilibrium-constant
Section 15.4
To find the equilibrium concentration and Keq value complete the following:
o Tabulate the known initial and equilibrium concentrations of all the species
in the equilibrium-constant expressions.
o For those species for which both the initial and equilibrium concentrations
are known, calculate the change in concentration that occurs as the system
reaches equilibrium.
o Use the stoichiometry of the reaction to calculate the changes in
concentration for all the other species in the equilibrium.
o From the initial concentrations and the changes in concentration, calculate
the equilibrium concentrations. These are used to evaluate the equilibrium
Example Problem.
A 1.000-L flask is filled with 1.000 mol of H2 and 2.000 mol of I2 at 448 degrees Celsius.
The value of the equilibrium constant for the reaction at this temperature is 50.5. What
are the partial pressures of H2, I2, and HI in the flask?
H2(g) + I2 (g)
First, calculate the initial pressures of the reactants.
P =( nRT) / V
For H2:
For I2:
59.19 atm
-x atm
59.19-x atm
118.4 atm
-x atm
118.4-x atm
0 atm
+2x atm
2x atm
Keq = (PHI)2 / (PH2) (PI2)
Reject 137.6. Thus, x=55.3
Partial Pressures:
H2= 59.19 – 55.3 = 3.9 atm
I2 = 118.4 – 55.3 = 63.1 atm
HI = 2(55.3) = 110.6 atm
Section 15.5
The equilibrium constant allows us to…
o Predict the direction in which the reaction mixture will proceed to achieve
o Calculate the concentrations of reactants and products when equilibrium
has been reached
-When the reactant and products partial pressures or concentrations are substituted into
the equilibrium-constant expression, the result is the reaction quotient.
o The letter Q represents the reaction quotient.
o ***The reaction quotient is equal to the Keq value ONLY at equilibrium.
o When Q > Keq substances on the right side of the chemical equation will
react to form substances on the left; the reaction moves right to left to
approach equilibrium.
o When Q < Keq substances on the left side of the chemical equation will
react to form substances on the right; the reaction moves left to right to
approach equilibrium.
Bluffer’s Guide for 15.6: Le Chatelier’s Principle
-Le Chatelier’s principle- if a system at equilibrium is disturbed by a change in
temperature, pressure, or the concentration of one of the components, then the system
will shift its equilibrium position so as to counteract the effect of the disturbance
Change in Reactant or Product Concentrations
-The addition of either a reactant or a product to a chemical system will shift to
reestablish equilibrium by consuming part of the added substance
-Removing either a reactant or product will cause the reaction to move in the direction
that forms more of that substance
-increase concentration of reactants, equilibrium shifts to right
-increase concentration of products, equilibrium shifts to left
-decrease concentration of reactants, equilibrium shifts to left
-decrease concentration of products, equilibrium shifts to right
Effects of Volume and Pressure Changes
-At constant temperature, reducing the volume of a gaseous equilibrium mixture causes
the system to shift in the direction that reduces the number of moles of gas
-Pressure-volume changes do not change the value of Keq, but change the concentrations
of the substances
-Consider the reaction: A(g)  2B(g)
-A decrease in volume (increase in pressure) shifts the reaction in the direction
that produces a smaller number of moles of gas (shifts to left in above reaction)
-An increase in volume (decrease in pressure) shifts reaction in direction that
produces a larger number of moles of gas (shifts to right in above reaction)
-If there’s no change in the number of moles of gases in a reaction, a volumepressure change does not affect the position of equilibrium
Effect of Temperature Changes
-When temperature is increased, it’s as if a reactant or product has been added to the
system at equilibrium. The equilibrium shifts in the direction that consumes the excess
reactant or product, namely heat
-Endothermic: reactants + heat  products; increasing the heat in an endothermic
reaction results in an increase in Keq
-Exothermic: Reactants  products + heat; increasing the heat in an exothermic
reaction results in a decrease in Keq
-In an endothermic reaction, heat is absorbed as reactants are converted to
products, thus increasing the temperature will cause equilibrium to shift to the right, and
Keq increases. Opposite for exothermic reactions
-Cooling reactions has opposite effect of heating them: cooling an endothermic
reaction shifts the equilibrium to the left, decreasing Keq. Opposite for exothermic
The Effect of Catalysts
-A catalyst increases the rate at which equilibrium is achieved, but does not change the
composition of the equilibrium mixture