Dario Bressanini1 G spontaneous Extent of Reaction, equilibrium
Reaction Equilibrium for A B (reaction Gibbs energy)
Slide 2
Dario Bressanini2
Slide 3
3 1 B A Even though B has a lower value of the standard molar
Gibbs energy the system can still achieve a lower Gibbs energy by
having some A present in equilibrium with B.
Slide 4
Dario Bressanini4 1 B A Generalizing, we can say that no
chemical reaction of gases goes to completion; however it may be
difficult to detect reactants at equilibrium if the products have a
very much lower Gibbs energy. equilibrium
Slide 5
Dario Bressanini5 Equilibrium Constants at equilibrium, at
equilibrium, K p = K c (RT) n aA + bB cC + dD
Slide 6
Dario Bressanini6 Values for the standard state are included in
the equilibrium constant, but we usually do not write them down
explicitly. Since their numerical value is unity, they really only
serve to make the units work out. In the limit that solutes
approach ideal behavior the activity coefficients approach unity.
In terms of molarity
Slide 7
Dario Bressanini7 Given that the standard Gibbs energy of
reaction at 2000 K for the decomposition of water is 135.2 kJ/mol.
Suppose steam at a pressure of 2.00 bar is passed through a furnace
tube at that temperature. Calculate (a) the equilibrium constant K
at 298 K. (b) degree of dissociation, and (c) mole fraction of O 2
present in the output gas stream. K 1 Reaction favors reactants. In
other words very little water dissociates to hydrogen and oxygen. H
2 O(g) H 2 (g) + 1/2 O 2 (g)
Slide 8
Dario Bressanini8 (b) To simplify notation let p j = p j /p ,
and since the magnitude of p equals unity, it will be dropped from
the equations. H 2 O H 2 O 2 Initial n 0 0 Final (equilibrium) (1 )
n n 1/2 n Mole Fraction p = total gas pressure = 2 bar
Slide 9
Dario Bressanini9 (b) continued Question: Does higher pressure
favor reactants or products?
Slide 10
Dario Bressanini10 (c) Calculate the mole fraction of O 2 (g)
in the output stream.
Slide 11
Dario Bressanini11 The Real K for Gas Reactions From slide two,
K = (a C c a D d ) / (a A a a B b ) From slide two, K = (a C c a D
d ) / (a A a a B b ) For gases the activity may be replaced by the
fugacity. For gases the activity may be replaced by the fugacity.
If the gas is ideal, the fugacity is the same as the pressure.
Otherwise, the fugacity coefficient must be used and K = [( C c D d
)/( A a B b )] [(f C c f D d )/(f A a f B b )] This means that K p
is probably closer the true thermodynamic equilibrium constant that
K c is. This means that K p is probably closer the true
thermodynamic equilibrium constant that K c is. But ONLY b/c the
standard states of gases are usually defined in terms of pressure
rather than concentration.
Slide 12
Dario Bressanini12 Ks for Liquid Solutions The activity of
solute J in solution, in terms of its molality, is given by The
activity of solute J in solution, in terms of its molality, is
given by a J = J b J /b o Here b is the molality. It is divided by
b o, the standard molality (usually exactly one molal) in order to
avoid having any units for the activity. In terms of mole fraction
the activity is given by In terms of mole fraction the activity is
given by a J = J x J This can be used for both gases and liquid
solutions.
Slide 13
Dario Bressanini13 Equilibria with Pure Liquid or Pure Solid
concentrations (or partial pressures) are actually ratios of the
actual concentration to the concentration in the standard state.
concentrations (or partial pressures) are actually ratios of the
actual concentration to the concentration in the standard
state.ACTIVITIES 2HgO(s) 2Hg( ) + O 2 (g)
Slide 14
Dario Bressanini14 Activity BY CONVENTION: for solution
standard state = 1 M; activity = concentration. for solution
standard state = 1 M; activity = concentration. for gas, standard
state = 1 atmos, activity = pressure for gas, standard state = 1
atmos, activity = pressure for solids & liquids: standard state
= pure solid or liquid, whose concentration does not change;
activity = 1 always for solids & liquids: standard state = pure
solid or liquid, whose concentration does not change; activity = 1
always the concentrations of liquids and solids do not appear in
the expressions for K c, K p
Slide 15
Dario Bressanini15 K c = [O 2 ]; K p = p O2 independent of
amount of solid or liquid present Example: evaporation of water
Example: evaporation of water Vapour pressure in sealed vessel in
presence of liquid is always the same (at given temperature)
irrespective of how much liquid water is present cf textbook fig
17.4 Hence for 2HgO(s) 2Hg( ) + O 2 (g) : H 2 O( ) H 2 O(g)
Slide 16
Dario Bressanini16 Le Chateliers Principle A stress applied to
a system at equilibrium will respond in such a way as to relieve
that stress. 1 moles of product compared to 1 mole of reactant. An
increase in pressure at constant temperature favors the formation
of water vapor H 2 O (g). An increase in temperature (thermal
energy) at constant pressure favors the breakup of water to form
hydrogen and oxygen gas (reverse reaction). An increase in
temperature at constant pressure for an endothermic reaction drives
the reaction towards products. In this case the formation of ethene
and hydrogen gas.
Slide 17
Dario Bressanini17 A.Adding or subtracting a product or
reactant if species added is a solid or liquid, there is no effect
on a gaseous equilibrium. if species added or subtracted is a gas
then: a.adding species shifts direction away from the species
added. b.subtracting a species shifts direction towards the species
removed. Add product -> more reactant Remove product > more
product When a reactant is added to a reaction mixture at
equilibrium the reaction tends to form products When a reactant is
removed, more reactant tends to form
Slide 18
Dario Bressanini18 B. Change in volume 1.if volume is
decreased, the reaction proceeds towards the side with least moles
of gas. 2.if volume is increased, the reaction proceeds towards
side with most moles of gas. 3.if in the balanced equation there
are the same number of moles of gas on both sides, a volume change
will not affect the equilibrium. Expansion -> drives reaction in
direction that increases # of gas molecules Compression ->
drives reaction in direction that reduces # of gas molecules
Slide 19
Dario Bressanini19 Changing the volume of the container.
Increase in the volume. Rxn will shift toward the side with the
most moles of gas. Decreasing the volume. Rxn will shift toward the
side with the fewest moles of gas. Changing the volume will make no
difference if there are the same number of moles of gas on each
side.
Slide 20
Dario Bressanini20 C. Change in pressure 1. increase in
pressure shifts equilibrium in direction of decrease in the number
of moles of gas 2. decrease in pressure shifts equilibrium in
direction of increase in number of moles of gas. 3. if in the
balanced equation there are the same number of moles of gas on both
sides, a pressure change will not affect the equilibrium.
Increasing the pressure by adding an inert gas has no effect on the
equilibrium concentration
Slide 21
Dario Bressanini21 Changing the pressure on the container.
Increase in the pressure. Rxn will shift toward the side with the
fewest moles of gas. -Decreasing the pressure. Rxn will shift
toward the side with the most moles of gas. Changing the pressure
will make no difference if there are the same number of moles of
gas on each side.
Slide 22
Dario Bressanini22 H 2 (g) + I 2 (g) 2HI(g). In this example
the number of moles of gas on both sides of the balanced chemical
reaction is the same, so increasing or decreasing the total
pressure will have little or no effect. The equilibrium system
cannot by reaction change the number of moles of gas present. N 2 +
3H 2 2NH 3. The number of moles of gas changes in this reaction,
from 4 to 2 as written. The equilibrium system can by reaction
change the number of moles of gas present, so changing the total
pressure will have a significant effect. Increasing the total
pressure will cause the reaction to proceed to the right,
decreasing total moles of gas, while increasing the total pressure
will cause the reaction to proceed to the left (in reverse),
increasing the total moles of gas.
Slide 23
Dario Bressanini23 N 2 O 4 2NO 2. The number of moles of gas
changes in this reaction, from 1 to 2 as written. The equilibrium
system can by reaction change the number of moles of gas present,
so changing the total pressure will have a significant effect.
Increasing the total pressure will cause the reaction to proceed to
the left (in reverse), decreasing the total moles of gas, while
decreasing the total pressure will cause the reaction to proceed to
the right, increasing the total moles of gas.
Slide 24
Dario Bressanini24 Effect of Pressure on Dissociation The
pressure depen-dence of the degree of dissociation, , at equi-
librium for the reaction A(g) 2B(g) for different values of the
equilibrium constant K. The pressure depen-dence of the degree of
dissociation, , at equi- librium for the reaction A(g) 2B(g) for
different values of the equilibrium constant K. The value = 0
corresponds to pure A = 1 corresponds to pure B. 1 =. (1 + 4p/K) =.
(1 + 4p/K)
Slide 25
Dario Bressanini25 Response of Equilibria to Temperature ln K
1/T Reaction enthalpies can be determined based on knowledge of
equilibrium concentrations and temperature ln K = H o /R (1/T)
Slide 26
Dario Bressanini26 Determination of K at different
temperatures. Assuming that the reaction enthalpy does not vary
much over the temperature range of interest we get the following
result.
Slide 27
Dario Bressanini27 A container holds a mixture of the isomeric
compounds cis-2-butene and trans-2-butene at 400 K. What is the
ratio of trans to cis isomer? cis trans r G can be found from the
thermodynamic tables in your book. r G at 298 K
Slide 28
Dario Bressanini28 r H can be found from the thermodynamic
tables in your book. Now calculate K(T 2 ).
Slide 29
Dario Bressanini29 Optimising Reaction Conditions CONCENTRATION
CONCENTRATION remove the product or increase the reactants, ie make
Q c smaller and push the reaction toward products eg liquefy the
ammonia and recycle the N 2 and H 2 PRESSURE increasing the total
pressure (for this reaction!) makes Q p smaller since there are 4
moles of gaseous reactants to 2 moles gaseous product; reaction is
shifted toward products CATALYSTS No effect on value of K c or
position of equilibrium but increases rate at which equilibrium is
attained. N 2 (g) + 3H 2 (g) 2NH 3 (g) Run at high T, high pressure
with V2O5 catalyst Run at high T, high pressure with V2O5 catalyst
Gives compromise between rate and yield (ie, how fast reaction goes
to equilibrium vs optimal concentrations at equilibrium Gives
compromise between rate and yield (ie, how fast reaction goes to
equilibrium vs optimal concentrations at equilibrium
Slide 30
Dario Bressanini30 The Haber Process N2 + 3H2 2NH3 forwards
reaction is exothermic Conditions for best yield of ammonia: a)
Temperature - high temperature would favour the endothermic
process, so if it's too high, this could decompose too much of the
ammonia. If it's too low, the production would be too slow, or not
enough ammonia would be produced - a compromise must be reached. b)
Pressure - increased pressure will produce more ammonia, but it is
expensive in machinery and energy and it can also be dangerous. At
1 atmosphere pressure the yield is very low, and so high pressure
is necessary. c) Concentration - if the ammonia could be
continuously removed, the equilibrium would shift to the right hand
side. If extra nitrogen or hydrogen were added, the equilibrium
would again shift to the right hand side. d) Catalyst - the
addition of a catalyst doesn't increase the yield in any way,
however, it does make the overall reaction proceed much quicker.
Solution: A temperature of 500C, a pressure of 200 atmospheres, and
an iron catalyst are used. The ammonium is removed by liquifying
(condensing) and the unreacted nitrogen and hydrogen is
continuously recycled. A yield of 15% is achieved, but it is quick
and continuous and there is no wastage and so it costs relatively
little. Equilibrium is never achieved.