Thermodynamic and Equilibria
First Law of Thermodynamics
Second Law of Thermodynamics
Physical Transformation of Pure Substances
Mrs Faraziehan Senusi
Reference: Chemistry: the Molecular Nature of Matter and Change,
6th ed, 2011, Martin S. Silberberg, McGraw-Hill
Kinetics addresses the speed of a reaction, the change in product concentration
(or reactant concentration) per unit time.
Equilibrium addresses the extent of a reaction, the concentration of product that
results given unlimited reaction time.
At equilibrium: rateforward = ratereverse
A system at equilibrium is dynamic on the molecular level; no further net
change in reactant and product concentrations is observed because changes
in one direction are balanced by changes in the opposite direction.
The Concept of Equilibrium
• Consider colorless frozen N2O4. At room
temperature, it decomposes to brown NO2:
• At some time, the color stops changing and we
have a mixture of N2O4 and NO2.
• Chemical equilibrium exists when two
opposing reactions occur simultaneously at
the same rate.
• Using the collision model:
– as the amount of NO2 builds up, there is a chance that
two NO2 molecules will collide to form NO2.
– At the beginning of the reaction, there is no NO2 so
the reverse reaction (2NO2(g)
N2O4(g)) does not
• The point at which the rate of decomposition:
equals the rate of dimerization (a compound formed by combination of two identical molecules) :
is dynamic equilibrium.
• The equilibrium is dynamic because the reaction has not
stopped: the opposing rates are equal.
Consider frozen N2O4: only white solid is present. On the microscopic level,
only N2O4 molecules are present.
As the substance warms it begins to decompose:
A mixture of N2O4 (initially present) and NO2 (initially formed) appears light
When enough NO2 is formed, it can react to form N2O4:
At equilibrium, as much N2O4 reacts to form NO2 as NO2 reacts to re-form N2O4:
The double arrow implies the process is dynamic.
• For an equilibrium we write A
Forward reaction: A B Rate = kf [A]
Reverse reaction: B A Rate = kr [B]
• At equilibrium kf [A] = kr [B].
• As the reaction progresses
Reaction rate constant
– [A] decreases to a constant,
– [B] increases from zero to a constant.
– When [A] and [B] are constant, equilibrium is achieved.
– kf [A] decreases to a constant,
– kr [B] increases from zero to a constant.
– When kf [A] = kr [B] equilibrium is achieved.
Consider a case in which the coefficients in the equation for a reaction are all 1. When substances A
and B react, the rate of the forward reaction decreases as time passes because the concentration of A
and B decrease.
A+ B C + D
As the concentrations of C and D build up, they start to re-form A and B.
As more C and D molecules are formed, more can react, and so the rate of reaction between C and
D increases with time. Eventually, the two reactions occur at the same rate, and the system is at
If a reaction begins with only C and D present, the rate of reaction (2) decreases with time, and the
rate of reaction (1) increases with time until the two rates are equal.
• Consider the Haber Process:
N2(g) + 3H2(g)
• If we start with a mixture of nitrogen and hydrogen (in any
proportions), the reaction will reach equilibrium with a
constant concentration of nitrogen, hydrogen and
• However, if we start with just ammonia and no nitrogen or
hydrogen, the reaction will proceed and N2 and H2 will be
produced until equilibrium is achieved.
• No matter the starting composition of reactants and
products, the same ratio of concentrations is achieved at
The Equilibrium Constant
A reversible reaction occurs by a one-step mechanism :
The rate of the forward reaction is Ratef = kf[A]2[B] ,
kf = specific rate constants of the forward reaction;
The rate of the reverse reaction is Rater = kr[A2B] ,
kr =specific rate constants of the reverse reactions.
The two rates are equal at equilibrium (Ratef = Rater).
At any specific temperature, both kf and kr are constants,
so kf/kr is also a constant.
This ratio is given a special name and symbol, the
equilibrium constant, Kc or simply K.
Equilibrium constant is the product and ratio of the rate
constants for each step of the mechanism.
• For a general reaction in the gas phase
aA + bB
cC + dD
• The equilibrium constant expression is
where Keq is the equilibrium constant.
• The thermodynamic definition of the equilibrium
constant involves activities rather than concentrations.
• The activity of a component of an ideal mixture is the
ratio of its concentration or partial pressure to a
standard concentration (1 M) or pressure (1 atm).
• Ex : A concentration of 1.20 M becomes 1.20 M/1 M�
= 1.20 ;
• Ex: A pressure of 0.53 atm becomes 0.53 atm/1 atm =
• For now, we can consider the activity of each species to be
a dimensionless quantity whose numerical value can be
determined as follows:
1. For any pure liquid or pure solid, the activity is taken as 1.
2. For components of ideal solutions, the activity of each
component is taken to be the ratio of its molar concentration
to a standard concentration of 1 M, so the units cancel.
3. For gases in an ideal mixture, the activity of each
component is taken to be the ratio of its partial pressure to a
standard pressure of 1 atm, so again the units cancel.
Because of the use of activities, the equilibrium constant
has no units; the values we put into Kc are numerically
equal to molar concentrations, but are dimensionless.
The Magnitude of Equilibrium Constants
• The equilibrium constant, K, is the ratio of products
• Therefore, the larger K, the more products are
present at equilibrium.
• Conversely, the smaller K, the more reactants are
present at equilibrium.
• If K >> 1, then products dominate at equilibrium and
equilibrium lies to the right.
• If K << 1, then reactants dominate at equilibrium and
the equilibrium lies to the left.
Calculating Equilibrium Constants
• Proceed as follows:
– Tabulate initial and equilibrium concentrations (or partial pressures)
– If an initial and equilibrium concentration is given for a species,
calculate the change in concentration.
– Use stoichiometry on the change in concentration line only to
calculate the changes in concentration of all species.
– Deduce the equilibrium concentrations of all species.
• Usually, the initial concentration of products is
zero. (This is not always the case.)
Some nitrogen and hydrogen are placed in an empty 5.00-liter
container at 500°C. When equilibrium is established, 3.01 mol
of N2, 2.10 mol of H2, and 0.565 mol of NH3 are present.
Evaluate Kc for the following reaction at 500°C.
The equilibrium concentrations are obtained by dividing the
number of moles of each reactant and product by the volume,
5.00 liters. Then we substitute these equilibrium concentrations
into the equilibrium constant expression.
The equation for the following reaction and the value of Kc at a
given temperature are given. An equilibrium mixture in a 1.00
liter container contains 0.25 mol of PCl5 and 0.16 mol of PCl3.
What equilibrium concentration of Cl2 must be present?
Because the volume of the container is 1.00 liter, the molar
concentration (mol/L) of each substance is numerically equal to
the number of moles. The equilibrium constant expression and its
numeric value are
In an equilibrium mixture at 500°C, we find PNH3 =
0.147atm, PN2 = 6.00atm, and PH2 = 3.70atm. Evaluate KP at
500°C for the following reaction.
Consider the reversible reaction of sulfur dioxide with oxygen
to form sulfur trioxide at 1500 K.
Suppose 0.400 mole of SO2 and 0.200 mole of O2 are injected
into a closed 1.00 liter container. When equilibrium is
established, we find that 0.056 mole of SO3 has formed and
that 0.344 mole of SO2 and 0.172 mole of O2 remain
unreacted. Calculate the value of the equilibrium constant for
this reaction at 1500 K.
Substituting the numerical values (without units) into the equilibrium
expression gives the value of the equilibrium constant.
* If [SO3]eq is only given, use coefficients of the balanced equation or vice versa.
RELATIONSHIP BETWEEN KP AND Kc
If the ideal gas equation is rearranged, the molar concentration of a
Substituting P/RT for n/V in the Kc expression for the N2–H2–NH3
equilibrium gives the relationship between Kc and KP for this
In general the relationship between Kc and KP is
For reactions in which equal numbers of moles of gases
appear on both sides of the equation, Δn = 0 and
KP = Kc.
From example 1, for the ammonia reaction at 500°C (or
773 K), Kc = 0.286
THE REACTION QUOTIENT
• The reaction quotient has the same form as the equilibrium
constant, but it involves specific values that are not necessarily
• Reaction quotient, Q, as a measure of the progress of the reaction.
• If Q > K then the reverse reaction must occur to reach
equilibrium (i.e., products are consumed, reactants are
formed, the numerator in the equilibrium constant expression
decreases and Q decreases until it equals K).
• If Q < K then the forward reaction must occur to reach
• When all reactants and products are in one phase,
the equilibrium is homogeneous.
• If one or more reactants or products are in a
different phase, the equilibrium is heterogeneous.
CaO(s) + CO2(g)
– experimentally, the amount of CO2 does not seem to depend on the
amounts of CaO and CaCO3. Why?
• The concentration of a solid or pure liquid is its density
divided by molar mass.
• Neither density nor molar mass is a variable, the
concentrations of solids and pure liquids are constant.
• For the decomposition of CaCO3:
[CO 2 ] constant [CO 2 ]
• We ignore the concentrations of pure liquids and pure
solids in equilibrium constant expressions.
• The amount of CO2 formed will not depend greatly on the
amounts of CaO and CaCO3 present.
Le Châtelier’s Principle
• Consider the production of ammonia
N2(g) + 3H2(g)
• As the pressure increases, the amount of ammonia
present at equilibrium increases.
• As the temperature decreases, the amount of
ammonia at equilibrium increases.
• Can this be predicted?
Le Châtelier’s Principle: If a change of conditions (stress) is
applied to a system at equilibrium, the system shifts in the
direction that reduces the stress to move toward a new state
The reaction quotient, Q, helps us predict the
direction of this shift.
Three types of changes can disturb the
equilibrium of a reaction.
1. Changes in concentration
2. Changes in pressure or volume (for reactions that
3. Changes in temperature
Change in Reactant or Product
• Consider the Haber process
N2(g) + 3H2(g)
• If H2 is added while the system is at equilibrium,
the system must respond to counteract the added
H2 (by Le Châtelier).
• The system must consume the H2 and produce
products until a new equilibrium is established.
• So, [H2] and [N2] will decrease and [NH3]
• Adding a reactant or product shifts the equilibrium
away from the increase.
• Removing a reactant or product shifts the
equilibrium towards the decrease.
• To optimize the amount of product at equilibrium,
we need to flood the reaction vessel with reactant
and continuously remove product (Le Châtelier).
• We illustrate the concept with the industrial
preparation of ammonia.
The product gas stream
(containing N2, H2 and
NH3) is passed over a
cooler to a refrigeration
1. N2 and H2 are pumped into a chamber.
the new N2 and
H2 feed gas.
The equilibrium amount of
because the product (NH3) is
continually removed and the
reactants (N2 and H2) are
continually being added.
The catalyst bed
is kept at 460 550 C under
In the refrigeration unit,
ammonia liquefies not
N2 or H2.
gases are passed
through a heating
Consider the following system starting at equilibrium:
Effects of Volume and Pressure
• As volume is decreased pressure increases.
• Le Châtelier’s Principle: if pressure is increased the
system will shift to counteract the increase.
• In a reaction with the same number of product and
reactant moles of gas, pressure has no effect.
For an ideal gas
• The term (n/V) represents concentration, that is, mol/L.
• At constant temperature, n, R, and T are constants.
• Thus, if the volume occupied by a gas decreases, its
partial pressure increases and its concentration (n/V)
• If the volume of a gas increases, both its partial
pressure and its concentration decrease.
• Consider the following gaseous system at equilibrium
• At constant temperature, a decrease in volume (increase in pressure)
increases the concentrations of both A and D.
• In the expression for Q, the concentration of D is squared and the
concentration of A is raised to the first power.
• As a result, the numerator of Q increases more than the denominator
as pressure increases. Thus, Q > Kc, and this equilibrium shifts to the
• Conversely, an increase in volume (decrease in pressure) shifts this
reaction to the right until equilibrium is reestablished, because Q < Kc.
In general, for reactions that involve gaseous reactants or
products, LeChatelier’s Principle allows us to predict the
Effect of Temperature Changes
• The equilibrium constant is temperature dependent.
• For an endothermic reaction, H > 0 and heat can
be considered as a reactant.
• For an exothermic reaction, H < 0 and heat can be
considered as a product.
• Adding heat (i.e. heating the vessel) favors
away from the increase:
– if H > 0, adding heat favors the forward reaction,
– if H < 0, adding heat favors the reverse reaction.
• Removing heat (i.e. cooling the vessel), favors
towards the decrease:
– if H > 0, cooling favors the reverse reaction,
– if H < 0, cooling favors the forward reaction.
Cr(H2O)62+(aq) + 4Cl-(aq)
CoCl42-(aq) + 6H2O(l)
for which H > 0.
– Co(H2O)62+ is pale pink and CoCl42- is blue.
– If a light purple room temperature equilibrium mixture is placed
in a beaker of warm water, the mixture turns deep blue.
– Since H > 0 (endothermic), adding heat favors the forward
reaction, i.e. the formation of blue CoCl42-.
Cr(H2O)62+(aq) + 4Cl-(aq)
CoCl42-(aq) + 6H2O(l)
– If the room temperature equilibrium mixture is placed in a beaker
of ice water, the mixture turns bright pink.
– Since H > 0, removing heat favors the reverse reaction which is
the formation of pink Co(H2O)62+.
The Effect of Catalysis
• A catalyst lowers the activation energy barrier for the
• Therefore, a catalyst will decrease the time taken to reach
• Adding a catalyst to a system changes the rate of the
reaction, but this cannot shift the equilibrium in favor of
either products or reactants. – it affects the activation
energy of both forward and reverse reactions equally
• A catalyst does not effect the composition of the
equilibrium mixture ~ Kc, does not change.
Relationship Between ΔG°rxn and the Equilibrium Constant
The free energy change for any other concentrations or pressures is ΔGrxn
As reaction occurs, the free energy of the mixture and the concentrations change
until at equilibrium ΔGrxn = 0, and the concentrations of reactants and products
satisfy the equilibrium constant. At that point, Q becomes equal to K