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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
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.
Chemical Equilibrium Notes Index
Part 2a. How to
write equilibrium
expressions and the effect of temperature on the equilibrium
constant K
Part 2b. Writing K_{c}
equilibrium expressions, units and exemplar K_{c}
concentration calculations
including esterification
Part 2c.
Writing K_{p}
expressions, units, and exemplar K_{p}
partial pressure equilibrium calculations
Part 3.
Applying Le Chatelier's Principle
and equilibrium expressions to Industrial
Processes
The equilibrium constant K_{p} is deduced from the balanced
chemical equation for
a reversible reaction, NOT experimental data as 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 forwards 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}
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.

N_{2(g)} + 3H_{2(g)
}
2NH_{3(g)} (ΔH = 92 kJ mol^{1})

In an experiment
starting with a 1:3 ratio H_{2}:N_{2} mixture,
at 400^{o}C and total pressure of 200 atm, the
equilibrium mixture was found to contain 36.3% by volume of ammonia.

(a) Write
out the equilibrium expression for K_{p}.

K_{p} =

p_{NH3}^{2}

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

(b)
Calculate the partial pressures of nitrogen, hydrogen and ammonia.

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 !

(c)
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} 
 (d) 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 


Example Q
Part 2a. How to
write equilibrium
expressions and the effect of temperature on the equilibrium
constant K
Part 2b. Writing K_{c}
equilibrium expressions, units and exemplar K_{c}
concentration calculations
including esterification
Part 2c.
Writing K_{p}
expressions, units, and exemplar K_{p}
partial pressure equilibrium calculations
Part 3.
Applying Le Chatelier's Principle
and equilibrium expressions to Industrial
Processes
Chemical Equilibrium Notes Index
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Chemical
Equilibrium Notes Index 