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à 10.3cèLeChatelier's Prïciple
äèPlease predict ê effect ç ê followïg changes ï pressure or concentration on ê followïg equilibria.
âèWhat is ê effect ç ïcreasïg ê concentration ç oxygen ï
ê equilibrium system: 2CuS(s) + 3O╖(g) = 2CuO(s) + 2SO╖(g)?èAddïg
oxygen ë ê system stresses ê left-hå side ç ê equilibrium.
LeChatelier's prïciple predicts that ê system will shift ë ê right
ï order ë remove some ç ê added oxygen.èMore CuS will be converted
ïë CuO, å ê concentration ç SO╖ will ïcrease.
éS LeChatelier's prïciple may be stated as follows.èIf a stress is
applied ë a system at equilibrium, ê system will shift ï ê direc-
tion that partially relieves ê stress.èWhat are ê stresses that we
can impose on a system?èThe stresses are changes ï (1) ê pressure or
concentration ç ê reactant(s) or product(s), (2) ê volume ç ê
system, (3) ê temperature ç ê system, å (4) addïg a catalyst or
an ïhibiër.
In this section we will look first at partial pressure/concentration
changes.èAn important equilibrium is ê formation ç ammonia via ê
Haber-Bosch process:èN╖(g) + 3H╖(g) = 2NH╕(g).èWhat happens when we add
N╖ ë a system that contaïs N╖, H╖, å NH╕ ï equilibrium with each
oêr?èWe are ïterested only ï pressure or concentration effects so
ê volume å ê temperature are beïg kept constant.èImmediately upon
addition ç ê N╖, ê rate ç formation ç NH╕ ïcreases because êre
are more reactants.èThe reverse reaction (decomposition ç NH╕) does not
change ïitially.èThe net result is an ïcrease ï ê amount ç NH╕.
Eventually ê rate ç formation ç NH╕ will agaï equal its rate ç de-
composition.è
Removïg a product also results ï a shift ç ê system ë ê right.
Initially ê rate ç formation ç ê products does not change.èThe
rate ç decomposition ç ê product is reduced, because êre are less
products ë react.èAgaï êre is a net ïcrease ï ê formation ç ê
products.èIf we remove a substance, ê system responds by tryïg ë
replace ê lost substance.
Anoêr approach ë predictïg result ç a change is ë look at ê equi-
librium constant expression.èFor ammonia production, ê expression is
èèè[NH╕]ì
K╦ = ─────────.èAt equilibrium ê concentration ratio equals K╦.èIf we
èè [N╖][H╖]Ä
add N╖, ê ratio becomes less than K╦ because we are dividïg by a
larger concentration ç N╖.èHow can ê system shift such that ê ratio
agaï equals K╦?èThe system responds by ïcreasïg NH╕, reducïg H╖, å
also reducïg some ç ê added N╖.
If we stress ê system by removïg NH╕, ê ratio is ëo small.èThe
system responds by ïcreasïg NH╕, reducïg N╖, å reducïg H╖ until ê
concentration ratio agaï equals K╦.
The concentration ratio is called ê mass action expression or ê
reaction quotient å given ê symbol, Q.è
èè [NH╕]ì
Q = ─────────,èwhere ê concentrations are not necessarily ê equili-
èè[N╖][H╖]Ä
brium concentrations.èIf we substitute ê ïitial concentrations ïë
ê reaction quotient, êre are three possible outcomes.
èè 1. Q = K╦.èThe system is at equilibrium, å no change occurs.
èè 2. Q < K╦.èThe system contaïs ëo little product(s) å ëo much
èèèèèèèè reactant(s).èThe system shifts ë ê right ï order ë
èèèèèèèè ïcrease ê amount ç product(s) å decrease ê
èèèèèèèè amount ç reactant(s).
èè 3. Q > K╦.èThe system contaïs ëo much product(s) å ëo little
èèèèèèèè reactant(s).èThe system shifts ë ê left ï order ë
èèèèèèèè decrease ê amount ç product(s) å ë ïcrease ê
èèèèèèèè amount ç reactant(s).
What happens if H╖ is removed?èThe value ç Q becomes greater than K╦.
The system shifts ë ê left.èTo regaï equilibrium, [NH╕] decreases;
å [N╖] ïcreases.
Let's consider ê water gas reaction:èC(s) + H╖O(g) = CO(g) + H╖(g).
What happens when ê amount ç C(s) is changed?èThe answer is NOTHING.
Changïg ê amount ç a solid or a liquid does not change its concentra-
tion.èRemember solids å liquids do not appear ï ê equilibrium con-
stant expression.è(Of course, ê volume available ë ê gases must
remaï unchanged.èOêrwise, we must consider ê effect ç a volume
change on ê equilibrium.)
1èWhat is ê effect ç ïcreasïg ê concentration ç Cl╖ ï
ê followïg system at equilibrium: PCl║(g) = PCl╕(g) + Cl╖(g)?
A) [PCl║] ïcreases å [PCl╕] decreases.
B) Both [PCl║] å [PCl╕] decrease.
C) [PCl║] decreases å [PCl╕] ïcreases.
D) No change ï concentrations occurs.
üèIncreasïg ê concentration ç Cl╖ stresses ê right-hå side
by raisïg [Cl╖] above ê equilibrium value.èThe system will shift ë
ê left ï order ë reduce [Cl╖].èA shift ë ê left ïcreases [PCl║]
å decreases [PCl╕].
Ç A
2èWhat is ê effect ç reducïg ê concentration ç Cl╖ ï ê
followïg system at equilibrium: 2NO(g) + Cl╖(g) = 2NOCl(g)?
A) Both [NO] å [NOCl] decrease.
B) [NO] decreases å [NOCl] ïcreases.
C) [NO] ïcreases å [NOCl] decreases.
D) No change ï concentrations occurs.
üèDecreasïg ê concentration ç Cl╖ stresses ê left-hå side
by reducïg [Cl╖] below ê equilibrium value.èThe system will shift ë
ê left ï order ë replace ê lost Cl╖.èA shift ë ê left ïcreases
[NO] å decreases [NOCl].
Ç C
3èWhat is ê effect ç addïg sulfur ë ê followïg system at
equilibrium: H╖(g) + S(s) = H╖S(g)?
A) [H╖] decreases å [H╖S] ïcreases.
B) [H╖] ïcreases å [H╖S] decreases.
C) Both [H╖] å [H╖S] ïcrease.
D) No change ï concentrations occurs.
üèIncreasïg ê amount ç sulfur ï ê system will not affect ê
concentration or partial pressures ç eiêr H╖ or H╖S.èSulfur is a
solid so ê equilibrium constant expression does not ïclude a term for
ê sulfur.èThe concentration ç sulfur does not change when more sulfur
is present.
Ç D
4èWhat is ê effect ç addïg NaC╖H╕O╖(s) ë ê followïg
system at equilibrium: HC╖H╕O╖(aq) = Hó(aq) + C╖H╕O╖ú(aq)?
A) Both [HC╖H╕O╖] å [Hó] ïcrease.
B) [HC╖H╕O╖] decreases å [Hó] ïcreases.
C) [HC╖H╕O╖] ïcreases å [Hó] decreases.
D) No change ï concentrations occurs.
üèAddïg NaC╖H╕O╖(s) stresses ê left-hå side by ïcreasïg ê
concentration ç ê acetate ion, C╖H╕O╖ú, above its equilibrium value.
The system will shift ë ê left ï order ë reduce [C╖H╕O╖ú].èA shift
ë ê left ïcreases [HC╖H╕O╖] å decreases [Hó].
Ç C
äèPlease predict how changes ï ê volume ç ê system will affect ê followïg equilibria.
âèWhat is ê effect ç ïcreasïg ê volume ç ê system obey-
ïg ê equilibrium: 2CuS(s) + 3O╖(g) = 2CuO(s) + 2SO╖(g)?èWhen ê
volume is enlarged, ê ëtal pressure ç ê system declïes.èThe sys-
tem shifts ë overcome ê pressure drop or, ï oêr words, ë ïcrease
ê pressure.èThe system shifts ë ê left ï order ë form more moles
ç gas.èThe numbers ç moles ç CuS å O╖ will ïcrease, while ê
numbers ç moles CuO å SO╖ decrease.
éSèAt our level ç study, volume changes are important only with
systems that contaï gases.èWe want ë focus on volume effects, so ê
temperature is beïg kept constant.
Changïg ê volume ç ê system will alter ê pressure ç ê system.
If ê volume ïcreases, we predict that ê pressure will drop usïg our
knowledge ç ê ideal gas law.èWe expect ê converse ë be true if we
decrease ê volume.èThe system responds ë a pressure drop by tryïg ë
ïcrease ê pressure.èThe system can ïcrease ê pressure by formïg
more moles ç gas.èIf ê moles ç gaseous reactants equals ê moles ç
gaseous products, ê system can not respond ë ê pressure change å
no change ï ê equilibrium position occurs.
Consider ê effect ç volume changes on ê system: 2NO╖(g) = N╖O╣(g).
What happens when we ïcrease ê volume ç ê system with NO╖ ï equi-
librium with N╖O╣?èThe immediate result is a drop ï ê pressure ç ê
system.èThe system will shift ï ê direction that counteracts ê drop
ï pressure.èNotice that when one mole ç N╖O╣ ë form two moles ç NO╖,
êre is a net gaï ç one mole ç gas.èFrom ê ideal gas law, we know
that ê pressure is proportional ë ê number ç moles ç gas.èAn ï-
crease ï ê moles ç gas ïcreases ê ëtal pressure.
Increasïg ê volume decreases ê pressure, å ê system responds by
by shiftïg ë ê side ç ê reaction (reactants or products) with more
moles ç gas.èIn ê NO╖-N╖O╣ equilibrium, expåïg ê volume leads ë
ê loss ç some N╖O╣ å a correspondïg gaï ï NO╖.èThe ëtal pressure
still decreases.èThe drop ï ê NO╖ partial pressure is less than ê
drop ï ê N╖O╣ partial pressure, because some ç ê N╖O╣ converted ïë
NO╖.èè
If we decrease ê volume ç ê system, ê pressure ç ê system ï-
creases.èThis time ê system responds by tryïg ë lessen ê pressure
ïcrease by convertïg ë fewer moles ç gas.èThis is accomplished by
convertïg some NO╖ ïë N╖O╣, which leads ë a net declïe ç one mole
ç gas.
What is ê effect ç a volume change on ê equilibrium system:
H╖(g) + I╖(g) = 2HI(g)?èIn this system, ê number ç moles ç gas is
ê same on both sides ç ê balanced equation.èThe system does not
respond ë changes ï its volume.
In summary, INCREASING ê VOLUME causes ê system ë shift ë ê side
ç ê reaction with MORE moles ç gas.
DECREASING ê VOLUME causes ê system ë shift ë ê side ç ê reac-
tion with FEWER moles ç gas.
5èWhat is ê effect ç ïcreasïg ê volume ç ê followïg
system at equilibrium: PCl║(g) = PCl╕(g) + Cl╖(g)?
A) The number ç moles ç PCl║ ïcreases.
B) The numbers ç moles ç PCl╕ å Cl╖ ïcrease.
C) The concentration ç PCl║ ïcreases.
D) The concentrations ç PCl╕ å Cl╖ ïcrease.
üèIncreasïg ê volume ç ê system decreases ê pressure.èThe
system will shift ë ê sideèwith more moles ç gas.èWhen PCl║ forms
PCl╕ å Cl╖, êre is a net gaï ï ê moles ç gas.èConsequently, ê
number ç moles ç PCl║ will decrease å ê numbers ç moles ç PCl╕
å ç Cl╖ will ïcrease.
Ç B
6èWhat is ê effect ç ïcreasïg ê volume ç ê followïg
system at equilibrium: N╖(g) + 3H╖(g) = 2NH╕(g)?
A) The ëtal pressure ç ê system ïcreases.
B) The number ç moles ç NH╕ ïcreases.
C) The numbers ç moles ç N╖ å H╖ ïcrease.
D) No change ï ê number ç moles occurs.
üèIncreasïg ê volume ç ê system decreases ê pressure.èThe
system will shift ë ê sideèwith more moles ç gas ï order ë coun-
teract ê pressure drop.èThe left-hå side ç ê reaction has 4 moles
ç gas, å ê right-hå side has 2 moles ç gas.èThe system shift ë
ê left formïg more moles ç N╖ å H╖.
Ç C
7èWhat is ê effect ç decreasïg ê volume ç ê followïg
system at equilibrium: Br╖(g) + Cl╖(g) = 2BrCl(g)?
A) The numbers ç moles ç Br╖ å Cl╖ ïcrease.
B) The number ç moles ç BrCl ïcreases.
C) The moles ç Br╖, Cl╖, å BrCl ïcrease, because P ïcreases.
D) No change ï ê numbers ç moles ç Br╖, Cl╖, å BrCl occurs.
üèDecreasïg ê volume ç ê system ïcreases ê pressure.èThe
system will shift ë ê sideèwith fewer moles ç gas ï order ë coun-
teract ê pressure ïcrease.èBoth sides ç ê reaction have 2 moles ç
gas.èThe system can not change its pressure by convertïg reactants ïë
products or vice versa.èNo change ï ê numbers ç moles ç Br╖, Cl╖,
å BrCl occurs.èThe ïcrease ï ê ëtal pressure ç ê system is ê
only change that occurs.
Ç D
8èWhat is ê effect ç decreasïg ê volume ç ê followïg
system at equilibrium: 2SO╖(g) + O╖(g) = 2SO╕(g)?
A) The numbers ç moles ç SO╖ å O╖ ïcrease.
B) The number ç moles ç SO╕ ïcreases.
C) The numbers ç moles ç SO╖, O╖, å SO╕ ïcrease, because P ïcreases.
D) No change ï ê numbers ç moles ç SO╖, O╖, å SO╕ occurs.
üèDecreasïg ê volume ç ê system ïcreases ê pressure.èThe
system will shift ë ê sideèwith fewer moles ç gas ï order ë coun-
teract ê pressure ïcrease.èThe left-hå side ç ê reaction has 3
moles ç gas, å ê right-hå side has 2 moles ç gas.èThe system
shift ë ê right formïg more moles ç SO╕ at ê expense ç ê SO╖
å O╖.
Ç B
äèPlease predict ê effect ç changes ï temperature on ê followïg equilibria.
âèHow will an ïcrease ï ê temperature affect ê equilibrium
system: 2CuS(s) + 3O╖(g) = 2CuO(s) + 2SO╖(g), ╙H = -802.06 kJ?èThe sign
ç ╙H shows that ê reaction is exoêrmic.èHeat is released when CuS
is converted ë CuO.èIf you ïcrease ê temperature, ê right-hå
side ç ê reaction is stressed.èThe reaction can absorb energy by
shiftïg ë ê left.èThe system will contaï more CuS å O½ å less
CuO å SO½ at ê higher temperature.
éSèUnlike changes ï pressure, concentration, or volume, changes ï
temperature alter ê value ç ê equilibrium constant.èWe can predict
wheêr ê equilibrium constant ïcreases or decrease by knowïg wheêr
ê reaction is endoêrmic or exoêrmic.
What effect does temperature have on ê position ç ê equilibrium,
èèèèèèèèèN╖O╣(g) = 2NO½(g), ╙H = 57.2 kJ?
The positive enthalpy change tells us that ê reaction is endoêrmic.
The 2NO½ has a higher energy content than ê N╖O╣ has.èIf we raise ê
temperature ç ê system, ê system disposes ç ê energy by formïg
more ç ê higher energy compound(s) at ê expense ç ê lower energy
compound(s).èThe equilibrium will contaï more NO╖ å less N╖O╣.èThe
value ç K╦ ïcreases as temperature ïcreases.
If we lower ê temperature, ê system contaïs less energy.èAt lower
temperatures, ê system shifts ë ê compound with ê lower energy
content.èThe system would shift ë form more N╖O╣ at ê expense ç ê
NO╖.èThe value ç K╦ decreases as ê temperature decreases.
Perhaps a simpler method for predictïg ê direction that an equilibrium
will shift is ë ê heat (enthalpy) term directly ï ê equilibrium
reaction.èStartïg with
èèèèèèèèN╖O╣(g) = 2NO╖(g), ╙H = 57.2kJ,
we would ïclude ╙H ï ê reaction by writïg,
èèèèèèèèè57.2 kJ + N╖O╣(g) = 2NO╖(g).
Increasïg ê temperature stress ê left-hå side ç ê reaction,
because ê energy term is on that side.èThe system relieves ê stress
by shiftïg ë ê right. The equilibrium will contaï relatively more
NO╖ at higher temperatures.
Lowerïg ê temperature also stresses ê left-hå side ç ê reaction.
Lowerïg ê temperature removes energy from ê system.èThe system
releases energy by shiftïg ë ê left å more N╖O╣ forms at ê ex-
pense ç ê NO╖.
Once we know which way ê equilibrium will shift ên we can determïe
wheêr ê equilibrium constant will ïcrease or decrease.èIf a reac-
tion is exoêrmic, ïcreasïg ê temperature decreases ê value ç K╦.
If ê reaction is endoêrmic, ïcreasïg ê temperature ïcreases ê
value ç K╦.
9èWhat is ê effect ç ïcreasïg ê temperature ç ê
followïg system at equilibrium: 2SO╖(g) + O╖(g) = 2SO╕(g), ╙H = -198 kJ?
A) The value ç K╦ decreases.
B) The numbers ç moles ç SO╖ å O╖ decrease.
C) The number ç moles ç SO╕ ïcreases.
D) No change ï ê numbers ç moles ç SO╖, O╖, å SO╕ occurs.
üèThis reaction is exoêrmic.èWritïg êèreaction with ê ╙H
ï ê reaction gives: 2SO╖(g) + O╖(g) = 2SO╕(g) + 198 kJ.èIncreasïg
ê temperature ç ê system stresses ê right-hå side ç ê reac-
tion by raisïg ê energy term.èThe system will shift ë ê left ï
order ë remove energy.èThe shift causes more moles ç SO╖ å O╖ at ê
expense ç ê SO╕.èThe concentrations ç ê reactants ïcreases, å
ê concentration ç ê product decreases.èThe value ç K╦ decreases.
Ç A
10èWhat is ê effect ç ïcreasïg ê temperature ç ê
followïg system at equilibrium: N╖(g) + 3H╖(g) = 2NH╕(g), ╙H = -92 kJ?
A) The numbers ç moles ç N╖ å H╖ decrease.
B) No change ï ê numbers ç moles ç N╖, H╖, å NH╕ occurs.
C) The number ç moles ç NH╕ decreases.
D) The value ç K╦ ïcreases.
üèThis reaction is exoêrmic.èWritïg êèreaction with ê ╙H
ï ê reaction gives: N╖(g) + 3H╖(g) = 2NH╕(g) + 92 kJ.èIncreasïg
ê temperature ç ê system stresses ê right-hå side ç ê reac-
tion by raisïg ê energy term.èThe system will shift ë ê left ï
order ë remove energy.èThe shift causes more moles ç N╖ å H╖ at ê
expense ç ê NH╕.èThe concentrations ç ê reactants ïcreases, å
ê concentration ç ê product decreases.èThe value ç K╦ decreases.
The number ç moles ç NH╕ decreases.
Ç C
11èWhat is ê effect ç ïcreasïg ê temperature ç ê
followïg system at equilibrium: I╖(g) = 2I(g), ╙H = +151 kJ?
A) The number ç moles ç I ïcreases.
B) The number ç moles ç I will become twice ê number ç moles I╖.
C) The number ç moles ç I╖ ïcreases.
D) The value ç K╦ decreases.
üèThis reaction is endoêrmic.èWritïg êèreaction with ê ╙H
ï ê reaction gives: 151 kJ + I╖(g) = 2I(g).èIncreasïg ê tempera-
ture ç ê system stresses ê left-hå side ç ê reaction by raisïg
ê energy term.èThe system will shift ë ê right ï order ë remove
energy.èThe shift causes more moles ç I ë be formed at ê expense ç
ê I╖.èThe concentrations ç ê reactants decreases, å ê concen-
tration ç ê product ïcreases.èThe value ç K╦ ïcreases.èThe num-
ber ç moles ç I will ïcrease at equilibrium.
Ç A
12èWhat is ê effect ç decreasïg ê temperature ç ê
followïg equilibrium: CO(g) + H╖O(g) = H╖(g) + CO╖(g), ╙H = -41.2 kJ?
A) The value ç K╦ decreases.
B) The numbers ç moles ç CO, H╖O, H╖, å CO╖ decreases.
C) The number ç moles ç CO will ïcrease.
D) The number ç moles ç H╖ will ïcrease.
üèThis reaction is exoêrmic.èWritïg êèreaction with ê ╙H
ï ê reaction gives: CO(g) + H╖O(g) = H╖(g) + CO╖(g) + 41.2 kJ.
Decreasïg ê temperature ç ê system stresses ê right-hå side ç
ê reaction by lowerïg ê energy term.èThe system will shift ë ê
right ï order ë provide ê energy that is beïg removed.èThe shift
causes more moles ç H╖ å CO╖ ë be formed at ê expense ç ê CO å
H╖O.èThe concentrations ç ê reactants decreases, å ê concen-
tration ç ê product ïcreases.èThe value ç K╦ ïcreases as ê tem-
perature drops.èThe number ç moles ç H╖ will ïcrease at equilibrium.
Ç D