Equilibrio

Equilibrio Químico

17-1

Kinetics applies to the speed of a reaction, the concentration of
product that appears (or of reactant that disappears) per unit time.
Equilibrium applies to the extent of a reaction, the concentration
of product that has appeared after an unlimited time, or once no
further change occurs.
At equilibrium:

rateforward = ratereverse

A system at equilibriumis dynamic on the molecular level;
no further net change is observed because changes in one
direction are balanced by changes in the other.

17-2

Reaching equilibrium on the macroscopic and molecular levels.
N2O4(g)

17-3

2NO2(g)

Equilibrium ≠ Equal
•  the rates of the forward and reverse reactions are
equal at equilibrium
•  but that does not mean the concentrations ofreactants and products are equal
•  some reactions reach equilibrium only after almost
all the reactant molecules are consumed- we say
the position of equilibrium favors the products.
•  some reactions reach equilibrium only after small
percentage of the reactant molecules are consumedwe say the position of equilibrium favors the
reactants.
17-4

If rateforward = ratereverse thenkforward[reactants]m = kreverse[products]n

kforward
kreverse

=

[products]n
[reactants]m

=K

the equilibrium constant

This is also known as the LAW OF MASS ACTION.
The values of m and n are those of the coefficients in the balanced
chemical equation. Note that this is equilibrium, not kinetics. The rates of
the forward and reverse reactions are equal, NOT the concentrations of
reactantsand products.

17-5

The range of equilibrium constants

K = [Productos]m
[Reactivos]n

small K

large K

intermediate K

17-6

Calculate K for each experiment.
Compare equilibrium concentrations.
Initial and Equilibrium Concentration Ratios for the N2O4-NO2 System at 2000C(473 K)

Initial
Experiment [N2O4] [NO2]

Equilibrium
[N2O4]eq [NO2]eq

[NO2]eq2
[N2O4]eq

10.0000

3.57x10-3

0.193

10.4

2

0.0000

0.1000

9.24x10-4

9.82x10-2

10.4

3

0.0500

0.0500

2.04x10-3

0.146

10.4

4

17-7

0.1000

0.0750

0.0250

2.75x10-3

0.170

10.5

=K

Sample Problem 17.1

Writing the Reaction Quotient from the Balanced
Equation

PROBLEM: Write the reaction quotient, Qc, for each of the following reactions:(a) The decomposition of dinitrogen pentoxide, N2O5(g) ⇌ NO2(g) + O2(g)
(b) The combustion of propane gas, C3H8(g) + O2(g) ⇌ CO2(g) + H2O(g)
PLAN: Be sure to balance the equations before writing the Qc expression.
SOLUTION:

[NO2]4[O2]

(a) 2 N2O5(g) ⇌ 4 NO2(g) + O2(g)

Qc =

[N2O5]2
[CO2]3[H2O]

(b)

17-8

C3H8(g) +

5O2(g)

⇌

3CO2(g) +

4H2O(g)

Qc =

[C3H8][O2]54

Q - The Reaction Quotient
At any time, t, the system can be sampled to determine the
amounts of reactants and products present.
Q is calculated in the same manner as K
c

[D]d

[A]

Q=

[C]

a

[B]b

We use the molar concentrations of the substances in the
reaction. This is symbolised by using square brackets - [ ].
For a general reaction
aA + bB ⇌ cC + dD where a, b,c,
and d are the numerical coefficients in the balanced equation, Q
(and K) can be calculated as
17-9

Calculating Variations on Q and K

aA + bB ⇌ cC + dD

[C] c [D]d
Qc =
[A]a [B]b

cC + dD ⇌ aA + bB

Q’ =

1
Qc

n

aA + bB ⇌ cC + dD

Qc’ = (Qc)n

For a sequence of equilibria, Qoverall = Q1 x Q2 x Q3 x …

17-10

Sample Problem 17.2

Writing the Reaction Quotientfor an Overall Reaction

PROBLEM: Understanding reactions involving N2 and O2, the most abundant
gases in air, is essential for solving problems dealing with
atmospheric pollution. Here is a reaction sequence between N2 and
O2 to form nitrogen dioxide, a toxic pollutant that contributes to
photochemical smog.
(1) N2(g) + O2(g)

⇌ 2NO(g)

(2) 2NO(g) + O2(g) ⇌ 2NO2(g)

Kc1 = 4.3 x...