Doc Brown's Chemistry Advanced A Level Notes 
TheoreticalPhysical Advanced Level Chemistry Equilibria Chemical
Equilibrium Revision Notes PART 2
2c. K_{p} chemical equilibrium expression, K_{p}
equilibrium constant and K_{p} equilibrium calculations
email doc
brown  comments  query?
Index of ALL my chemical equilibrium
context revision notes Index
ALL my advanced A
level theoretical
chemistry revision study notes
Use your
mobile phone or ipad etc. in 'landscape' style
This is a BIG
website, you need to take time to explore it [SEARCH
BOX]
How do we write out partial pressure chemical
equilibrium expressions. What is the equilibrium constant Kp? How do you
do partial pressure equilibrium calculations? All is explained with
examples including the units for or partial pressures (e.g. atm, Pa or
kPa) and how to solve equilibrium problems using partial pressure units.
The equilibrium constant K_{p} is deduced from the balanced
chemical equation for a reversible reaction, NOT experimental data as is
the case for rate expressions in kinetics.
2.c
Partial pressure equilibrium law
expressions

K_{p} partial
equilibrium expression INTRODUCTION

For any gaseous reaction: aA_{(g)}
+ bB_{(g)} + cC_{(g)}
etc.
tT_{(g)} + uU_{(g)} + wW_{(g)}
etc.

K_{p} =

p_{T}^{t}
p_{U}^{u} p_{V}^{v} etc. 
(for units see later) 
p_{A}^{a}
p_{B}^{b} p_{C}^{c} etc. 

p_{x}
indicates the partial pressure of x, usually in atm
(atmospheres) or Pa (pascals).

Do NOT put square brackets in K_{p}
expressions!

AND again note that
K_{p} (like K_{c}) is only constant for a
specific constant temperature at which the partial pressures of
the component gases might vary from one equilibrium situation to
another at the same temperature.

The 'rule' for the trend in K_{p}
value change is provided by Le Chatelier's Principle.

(i) If the forward reaction
is endothermic, then K_{p} increases with
increase in temperature (so K_{p} decreases if
temperature decreased).

(ii) If the forward reaction
is exothermic, then K_{p} decreases with
increase in temperature. (so K_{p} increases if
temperature decreased).

The partial pressure of a gas is
defined as the pressure a gaseous component in a mixture would
exert, if it alone occupied the space/volume in question.

e.g. in air, 21% by volume is
oxygen and 79% is nitrogen etc.

Therefore the fraction of
molecules which are oxygen and nitrogen is 0.21 and 0.79
respectively.

Since air has a total pressure of
1 atm. or 101325 Pa, the partial pressures of oxygen and nitrogen in
air are:

p_{O2}
= 0.21 x 1 = 0.21 atm or p_{O2} = 0.21 x
101325 = 21278 Pa,

p_{N2}
= 0.21 x 1 = 0.79 atm or p_{N2} = 0.21 x
101325 = 80047 Pa,

in other words, the partial
pressure of a gas in a mixture p_{gas}
= its % x p_{tot} / 100

In a gaseous mixture the total
pressure equals the sum of all the partial pressures:

Some other useful mathematical
ideas and expressions for gas mixtures:

Equilibrium example
2.1b.1 The formation of nitrogen(II) oxide (e.g. in
car engines)

Equilibrium example
2.1b.2 The synthesis of sulfur(VI) oxide in the manufacture of
sulfuric acid

Equilibrium example
2.1b.3 The synthesis of ammonia (Haber Process)

Equilibrium example
2.1b.4 The manufacture of hydrogen (e.g. for ammonia
synthesis)

Equilibrium example
2.1b.5 The synthesismanufacture of methanol (gas
phase)
 CO_{(g)} + 2H_{2(g)}
CH_{3}OH_{(g)}

K_{p} =

p_{CH3OH}

(units atm^{2} or Pa^{2}) 
p_{CO}
p_{H2}^{2} 
 imagine p = partial pressure units
 units = p/(p x p^{2}) = p/p^{3}
= p^{2}
TOP OF PAGE
2.2b
K_{p}, partial pressure
calculations and applying Le Chatelier's Principle

Example Q
2.2b.1
Nitrogen(IV) oxidedinitrogen tetroxide equilibrium
(nitrogen dioxide <=> dimer)

For the equilibrium between
dinitrogen tetroxide and nitrogen dioxide:

the value of the equilibrium
constant K_{p} = 0.664 atm at 45^{o}C.

(a) If the partial pressure
of dinitrogen tetroxide was 0.449 atm, what would be the equilibrium
partial pressure of nitrogen dioxide and the total pressure of the
gases? and what % of the dinitrogen tetroxide is dissociated?

K_{p} =

p_{NO2}^{2}


p_{N204} 

Rearranging the

p_{NO2}^{2}
= K_{p} x p_{N204}, so, p_{NO2} = √(K_{p}
x p_{N204} ) = √(0.664 X 0.449) = 0.546 atm

p_{tot}
= p_{NO2} + p_{N2O4} = 0.546 + 0.449 = 0.995 atm

The pressure of NO_{2}
is 0.546, which is equivalent to 0.273 atm of N_{2}O_{4}
before dissociation.

Therefore the
% N_{2}O_{4}
dissociated = 0.273 x 100 / (0.449 + 0.273) = 37.8%

(b) In another experiment at
77^{o}C and a total pressure of 1.000 atm, the partial
pressure of dinitrogen dioxide was found to be 0.175 atm. Calculate
the value of the equilibrium constant at 77^{o}C and the %
dissociation of the dinitrogen tetroxide.

p_{tot} = p_{NO2} +
p_{N2O4}, so p_{NO2} = p_{tot}
p_{N2O4} = 1.000 0.175 = 0.825 atm

From the equilibrium expression
above:

K_{p}
= 0.825^{2} / 0.175 = 3.89 atm

The pressure of NO_{2}
is 0.825, which is equivalent to 0.4125 atm of N_{2}O_{4}
before dissociation.

Therefore the % N_{2}O_{4}
dissociated = 0.4125 x 100 / (0.175 + 0.4125) = 70.2%

(c) At a total pressure of
102000 Pa, dinitrogen tetroxide is 40% dissociated at 50^{o}C.

(d) Explain the effect, if
any, of increasing the total pressure of the system on the position
of the N_{2}O_{4}/NO_{2}
equilibrium.

Increasing pressure will favour the
LHS, i.e. more N_{2}O_{4} and less NO_{2},
because the system will move to the side with the least number of
gaseous molecules to try to minimise the increase in pressure (1 mol
gas <== 2 mol gas).

(e) From information from
parts (a)(c) is the dissociation exothermic or endothermic? and
give your reasoning.

The dissociation is endothermic
because on raising the temperature from 45^{o}C to 77^{o}C
the value of K_{p} has increased.

i.e. p_{NO2}
has increased in value and p_{N2O4} decreased in value from
more dissociation.

An equilibrium system moves in the
endothermic direction on raising the temperature to absorb the
'added' heat to try to minimise the temperature rise.

Example Q
2.2b.2
Ammonia synthesis

Ammonia is synthesised by combing
nitrogen and hydrogen, but the reaction is reversible.

In an experiment starting with a
1:3 ratio N_{2}:H_{2} mixture, at 400^{o}C,
the final total equilibrium pressure was 200 atm. (See graph of
yields in
Haber synthesis
section)

The final equilibrium mixture was
found to contain 36% by volume of ammonia.

Assume the gases behave ideally.

(a) Write out the
equilibrium expression for K_{p} and quote some typical
units.

K_{p} =

p_{NH3}^{2}

(units atm^{2} or Pa^{2}) 
p_{N2}
p_{H2}^{3} 

(b) What was the % by volume
of nitrogen in the original mixture and how will it change?

from the 1:3 ratio, % ammonia =
100 x 1/4 = 25% N_{2}

Since some nitrogen will combine
with the hydrogen, its percentage will decrease.

(c) Predict and explain any
difference between the initial and final pressures of the system.

Since the equilibrium has moved
to the right i.e. ammonia formed, 4 molecules of gaseous reactants
gives 2 molecules of product.

Since the total equilibrium
number of molecules is reduced, the initial total pressure must
have been greater than 200 atm at the start.

(d)
Calculate the partial pressures of nitrogen, hydrogen and ammonia and
the % nitrogen and hydrogen in the final equilibrium mixture.

36% of ammonia means its mole
fraction is 0.36, so p_{NH3} = 0.36 x 200 = 72
atm

The other 64% of gases must be
split on a 1:3 ratio between nitrogen and hydrogen.

So mole fraction of nitrogen x p_{tot}
= p_{N2} = ^{64}/_{100}
x ^{1}/_{4} x 200 = 32 atm,

and mole fraction of hydrogen x p_{tot}
= p_{H2} = ^{64}/_{100}
x ^{3}/_{4} x 200 = 96 atm

check: p_{tot} = p_{N2}
+ p_{H2} + p_{NH3} = 32 + 96 + 72 = 200 atm !

(e)
Calculate the value of the equilibrium constant at 400^{o}C
and give its units.

K_{p} =

p_{NH3}^{2}


p_{N2}
p_{H2}^{3} 

K_{p} =

72^{2}

= 1.83 x 10^{4}
atm^{2}

32 x 96^{3} 
 (f) What will be the effect on ammonia
yields and the value of K_{p} by (i)
raising the pressure and (ii) raising the temperature. Give
reasons for your answers.
 (i) The reaction to form ammonia is
exothermic, so higher temperature will favour its endothermic
decomposition back to nitrogen and hydrogen. So the yield of
ammonia and the value of K_{p} will be reduced as the
system will absorb heat energy to attempt to minimise temperature
rise.
 (ii) Higher pressure will favour a
higher yield of ammonia because 4 mol gaseous reactants ==> 2
mol gaseous products, so if the system is subjected to higher
pressure it reduces the number of gas molecules to attempt to
minimise the enforced increase in pressure. Assuming the gases
behave ideally, the value of the equilibrium constant K_{p}
is unaffected by pressure changes, only temperature affects the
value of K_{p} in an ideal gas mixture.

Example Q
2.2b.3 Phosphorus(V) chloride
(phosphorus
pentachloride)
dissociation

At high temperatures vapourised
phosphorus(V) chloride dissociates into gaseous phosphorus(III)
chloride (phosphorus trichloride) and chlorine.

PCl_{5(g)}
PCl_{3(g)} + Cl_{2(g)}
(a) Write the equilibrium
expression for this reaction in terms of K_{p}
and partial pressures.
K_{p} =

P_{PCl3}
x P_{Cl2} 
(units atm, Pa, kPa etc.) 
P_{PCl5} 

(b) At a particular
temperature 40% of the PCl_{5} dissociates into PCl_{3}
and Cl_{2}. If the total equilibrium pressure is 210 kPa
calculate the value of K_{p} at this temperature quoting
appropriate units. The solution is set out below in a series of
logical steps, either thinking from the point of view of
starting with n/1 moles of PCl_{5} or
starting with 100/100% of PCl_{5}
molecules.

PCl_{5(g} 

PCl_{3(g)} 
Cl_{2(g)} 
comments, z
refers to any component in the mixture 
n(1x)
(1x)
can think of as (100
x%) 

nx
x
can think of as
x% dissociated 
nx
x
can think of as
x% dissociated 
n = initial moles of undissociated
PCl_{5} for a general solution, but consider 1 mole
of PCl_{5} (n = 1) of which fraction x
dissociates giving a total of (1 + x) moles of gas
from 1 mol of PCl_{5}
This is the logic
thinking in terms of 100/100% undissociated P_{PCl5}
molecules and x% is the percentage of PCl_{5}
molecules dissociated. 
(1x)/(1 + x) 

x/(1+x) 
x/(1+x) 
calculation of mole fractions
mole fraction z = moles z/total moles 
0.6/1.4 = 0.4286
can think of as 60/140 =
0.4286 

0.4/1.4 = 0.2857
can think of as 40/140 =
0.2857 
0.4/1.4 = 0.2857
can think of as 40/140 =
0.2857 
since x = 0.4 (= 40% dissociated)
It works out the same in
terms of the original 100/100% undissociated PCl_{5} 
P_{PCl5} = 0.4286 x 210 = 90.0 

P_{PCl3} = 0.2857 x 210 = 60.0 
P_{Cl2} = 0.2857 x 210 = 60 
partial pressure of component z
P_{z} = mole fraction z x P_{tot}
note 90 + 60 + 60 = 210 check! 

K_{p} =

60.0 x 60.0 
= 40.0 kPa 
90.0 
advanced level
chemistry how to write equilibrium expressions and do
equilibrium calculations for AQA AS chemistry, how to write
equilibrium expressions and do equilibrium calculations for
Edexcel A level AS chemistry, how to write equilibrium
expressions and do equilibrium calculations for A level OCR
AS chemistry A, how to write equilibrium expressions and do
equilibrium calculations for OCR Salters AS chemistry B, how
to write equilibrium expressions and do equilibrium
calculations for AQA A level chemistry, how to write
equilibrium expressions and do equilibrium calculations for
A level Edexcel A level chemistry, how to write equilibrium
expressions and do equilibrium calculations for OCR A level
chemistry A, how to write equilibrium expressions and do
equilibrium calculations for A level OCR Salters A level
chemistry B how to write equilibrium expressions and do
equilibrium calculations for US Honours grade 11 grade 12
how to write equilibrium expressions and do equilibrium
calculations for preuniversity chemistry courses
preuniversity A level revision notes for how to write
equilibrium expressions and do equilibrium calculations
A level guide notes on how to write equilibrium expressions
and do equilibrium calculations for schools colleges
academies science course tutors images pictures diagrams for
how to write equilibrium expressions and do equilibrium
calculations A level chemistry revision notes on how to
write equilibrium expressions and do equilibrium
calculations for revising module topics notes to help on
understanding of how to write equilibrium expressions and do
equilibrium calculations university courses in science
careers in science jobs in the industry laboratory assistant
apprenticeships technical internships USA US grade 11 grade
11 AQA A level chemistry notes on how to write equilibrium
expressions and do equilibrium calculations Edexcel A level
chemistry notes on how to write equilibrium expressions and
do equilibrium calculations for OCR A level chemistry notes
WJEC A level chemistry notes on how to write equilibrium
expressions and do equilibrium calculations CCEA/CEA A level
chemistry notes on how to write equilibrium expressions and
do equilibrium calculations for university entrance
examinations with advanced level chemistry practice
questions for solving equilibrium problems, how do you write
Kc equilibrium expressions? how do you write Kp equilibrium
expressions? what are the units for Kc equilibrium
expressions? what are the units for Kp equilibrium
expressions? how to do Kc equilibrium calculations using
molarity concentrations, how to do Kp equilibrium
calculations using partial pressures, how to use partial
pressure and molarities in equilibrium expressions,
explaining the carboxylic acid alcohol ester water
equilibrium experimental procedure, explaining methods of
solving equilibrium problems, solving hydrogen iodide
hydrogen iodine equilibrium problems using the equilibrium
expression, how to solve PCl5 phosphorus(V) chloride PCl3
phosphorus(III) chloride Cl2 chlorine problems calculations
from the equilibrium expression, solving the ammonia
hydrogen nitrogen equilibrium expression, how to do
calculations from the formation of ammonia equilibrium
expression, solving the phosphorus fluorine phosporus(5)
fluoride equilibrium expression units calculations, how to
write equilibrium expressions for complex ion formation, how
to work out units for equilibrium expressions, solving the
nitrogen oxygen nitrogen(II) oxide equilibrium expression
calculations, solving the equilibrium expression for the
contact process, solving the equilibrium expression for the
synthesis of methanol CH3OH from hydrogen H2 and carbon
monoxide CO
explain the effect of temperature on equilibrium constants,
how do you write Kc equilibrium expression? how do you solve
numerical equilibrium problems? how do you write out Kp
equilibrium expression in partial pressures? how do you
solve equilibrium problems using partial pressures? how do
you calculate the Kc equilibrium constant from
concentrations in mol/dm3? how do you calculate Kp
equilibrium constants using partial pressures atm Pa?
WHAT NEXT?
Advanced Equilibrium Chemistry Notes Part 1. Equilibrium,
Le Chatelier's Principlerules
* Part 2. K_{c} and K_{p} equilibrium expressions and
calculations * Part 3.
Equilibria and industrial processes * Part 4
Partition between two
phases, solubility product K_{sp}, common ion effect,
ionexchange systems *
Part 5. pH, weakstrong acidbase theory and
calculations * Part 6. Salt hydrolysis,
acidbase titrationsindicators, pH curves and buffers * Part 7.
Redox equilibria, halfcell electrode potentials,
electrolysis and electrochemical series
*
Part 8.
Phase equilibriavapour
pressure, boiling point and intermolecular forces watch out for subindexes
to multiple sections or pages
TOP OF PAGE
















Website content © Dr
Phil Brown 2000+. All copyrights reserved on revision notes, images,
quizzes, worksheets etc. Copying of website material is NOT
permitted. Exam revision summaries & references to science course
specifications are unofficial. 
Chemical
Equilibrium Notes Index
