Equilibrium 1
Eqllllb"l..
~
Equilibrium: the forward and reverse reactions occur at the same rate
(represented by a double arrow between reactants and products)
-4 .,..
e>~
AB
~
ArB';;: A8
1. General Information
A. Can ONLY occur in a closed system
B. Equilibrium is dynamic: there is constant interaction between reactants and
products, but the quantities of the reactants and products are usually not
equal- only the rate of change
Example: Rate ofevaporation and condensation ofwater in a closed jar is at
equilibrium BUT the amount of water occurring as liquid and vapor
are NOT equal
~
..
.8..-
L.) ("') ­
00"
')0 C)
2. Physical Equilibrium
A. Phase Equilibrium
1. Solid -liquid equilibrium: The rate of melting = the rate of freezing
H-JO(s)
~ 11)0(1)
2. Liquid - gas equilibrium: The rate of evaporation = the rate of
condensation
~
~fJO(t)
<= H ~O(3)
Equilibrium 2
B. Solution Equilibrium (NOTE: MUST BE SATURATED SOLUTIONS!!!)
1. Gases in Liquids: The rate of gas dissolving in a liquid = the rate of gas
~
escaping the liquid.
2. Solids in Liquids: the rate of dissolving = the rate ofcrystallization
, ox) sed" :
~ (OJ(Q1;)
CO)(9)
J
t..,{)
*
S(rLrot-t~
tNeJu ;
• ..j,
AI"c.itS) ~ N:/
t- I
e:-
DI'5solveJ
t:%«2« ()1<- u.., J1<;
3. Chemical Equilibrium
ta­
\<> I
A. The point at which the concentrations of all reactants and products cease to
change with time.
B. The rate at which reactants are making products equals the rate at which
products are making reactants.
C. When reactants are first mixed, only the forward reaction occurs. Over time as
products are formed, the rate ofthe forward reaction slows as reactants are
used up (concentration decreases) and the reverse reaction speeds up because """
the concentration ofproducts increases. This occurs until the rates of the two
reactions are equal.
~iB~AB
Equilibrium Reverse Reaction Forward Reaction
~~
Time
Time
~
5V1lh4 ~
.:f¥J,-I,~ no
48
rcti:.f lMl/\to KS q <;
At-B)
n'"ie. "- I
j
,,~
_J
..
I
J S
'<,.',",\,..i
<.
")~.
'-'(
.,.;,
rt q\.. ~~.tlf4 v~d
()P
+(oj iJC ~. <;
I
t'Of Yh
..
Time
fi;yW4 r d f
rC liC;, ~)e
rl t..,..U. . )( Pee Dyr4Z
€raJ
~
Equilibrium 3
4. Law ofMass Action
\.,
A. Founded by Maximilian Goldberg and Peter Waage in 1864
B. Shows the relationship between the concentration of the reactants and the
concentration of the products for a reaction at equilibrium.
C. aka: Equilibrium Expression
41;. 0 3 + 6Noc.e
1. Given the hypothetical equation: aA + bB
~
~ JA /(J3
t
3Na.O
cC + dD
a. Lower case letters represent the coefficient (# of moles)
b. Upper case letters represent the compound
2.
Kc =K
.q
= [Ct[Dt
**Products over reactants
~ = fA 1{J~l"-fJJ()ffJJ'3 ~
[At[Bt
[J)~~J [N~(JJo
a. The square brackets indicate the equilibrium concentration
b. The exponents are the coefficient$
c. Pure solids and liquids are not included in the equilibrium
expression because their concentrations are constant
\..,. 3. Kc = Keq = equilibrium constant
a.-tm ttnit& vnfrs &zt~4
(Yv?
/1It. ,t(ul,~
b. independent of starting reactant concentrations
4. Interpreting the magnitude of Keq
a. When Keq > 1 the reaction is product favored
b. When Keq < 1 the reaction is reactant favored
,
PY'C' i!
th-,
c.c
r
u,,;(f /'1
5. Calcolatmg Keq
A. If all equilibrium concentrations are known simply plug numbers into the Keq
expression and solve
B. If all concentrations are not known - use ICE box method
I = initial concentrations
C = change in concentrations
E = concentrations at equilibrium
\.,
1. The change in concentration of reactants is negative
2. The change in concentration of the products is positive
3. The change in concentration corresponds to the mole ratios (coeffs)
.
I
wCr} "<:t~[()
fI?
•
Narne
EQUILIBRIUM CONSTANT (K)
Iav
A1Sllr<-Y
Write the expression for the equilibrium constant K for the reactions below.
1.
N 2(g) + 3H 2 (g)
~
2NH 3(g)
Ke : [ NIi ~ ] fJ.
o
2.
t6
MM-MPY·
[iJJ ] ' [ H,,]3
~
2KCIOis)
.~
2KCI(s) + 30 2(g)
= f5aJtO)jJ ::: [O;]s
~
3.
H20(I)
Jl
Ke; = [WJf2j-r
~
"SolidS
d~~
/11'1
LH'][OH']
'cc
'ff'~~
(tJUA 2CO(g) + 02(g) ~ 2C0 2(g)
n
K~
5.
P4f'7(
H+(aq) + OH-(aq)
~
I
4.
MMMt:Jdcf
-=:
Li 2C0 3(S)
keD
0.
~
[COJ ] '
[COJ~ [0)J
/'-1t"'\ ~-\
1
2Li+(aq) + C0 3-2(aq)
CU' J :>[cO~-'J
[L~
.
I
= [LJ <] J [CO~ -']
..
-
l""l"II~..L..L
i
f1 /1.-1/1
501lds
d~/L~
___ 1_'&'_.
f--1.
70
4
CCllndn Irtinnni Fnir. Inc
Name _ _ _ _ _ _ __
CALCULATIONS USING
lIiE EQUILIBRIUM CONSTANT
Using the equilibrium constant expressions you determined on page 79, calculate the
value of K when:
[NH3J = 0.0100 M, [N 2] = 0.0200 M, [H 2]
1.
= 0.0200 M
kit : - fOIJ~
::::
~
(; [o~r[9JJ'
b~}S :?'t­
2.
[°2 1 = 0.0500 M
Kt6 ;;. J~(OS]:
CK~
B.
Pro/u is· {;,l/fJY(.J
:::: /;) C, )< 10 -'I f'{~
~." itda<-t:> ftr
VdY(
J
[H+) = 1 X 10,8 M, [OH'] = 1 X 1.0.6 M
K% -=. 0 10 'R)DXlJtJ ;:. /" 'D-'~ W
qa
l
/~/<; (qIJort:'d
~
~/
4.
[CO] = 2.0 M, [02] = 1.5 M, [C02] = 3.0 M
kt~ - [ -:; J')' .
()
5.
[Li+]
r . .;!L...... "
L-~
J
I
~J
_I
-:: '/5 M
i/o!u/ c;
J:rVdY(c/
= 0.2 M, [C03'2] = 0.1 M
YI''b ­
[, ~ ] l[!J -=, (JO L1 M1
;;'
,­
~ tf ( 1tL.,,,l,,~ /'1 (
f
(
d
~
I
0('\
,;;::.,.1 __ ",,_•• _.Lt _ _ _ I ,..
_t~
I ~ ._
;:c[. Eo):
C. Practice 1
l, ;71 ,
?11>~{e-f?ru
Equilibrium 4
aJ
~i\ ,,0.1
-7
///
A mixture of 5.00xlO-3 mole ofH2 and 1.000xlO-2 moles of 12 is placed in a
5.000L container at 448°C and allowed to come to equilibrium. Analysis
of the equilibrium mixture shows that the concentration of HI is 1.87xlO·3 M.
~'l.i/;. eyruw>
Calculate Keq at 448°C for the reaction...
'. I;' ~
',' /
/
U~
IS + ­
JIVe"y,
-fu
mfo +0
_ r:~.,
/1:v
(>
,'"
<
Molan Ius. tv sfart fJ III':} "'"
tx)Xes:
r-1 d
l.,
~
.1:
J
dhO ­
r-t
"/ 'p(
'J
,
F- d!.
C/r'J
,QJ25M
51.­
,0
..
~-
~ .001 N
(f\I(}.P_. ::-
5L
~
tl.Jha..f yfJlA knew
DOd H
t'cboz·d
)y"711) v·--t:"r(,. ,)){:::;
Vvt');.( (
- 1tv --ll. 0/tL. ctll {jC{~fs UfD f r Dd u( i S r: t' £<~ { t:;
- f(~J!{tCVrLta ({~!J~ s cA1M7F fJ!j tkcyeq~//J C/.;.~~
·~/d
_
f" c' dvet s eJ.M/tLj S i" t.u t1 &t
Oetcll11/Itt. hOlV-ectLh rroduc.J cv~cd r.((7(~ftt-J;;I c.J,~1<lcfd; ,
l.?ilI11bY/vj/V} ::: 1f)/f7ed
C.11~~~-e .7flL ('({flo 4:tr,c (~1QitS
t~ (;q~-'cd (rYl #\.l l~I(ltitJ ~'Y.~ fu 6a./~ct J yt.: Ih~¥l
(0) ( v loJJ.
e~ util byI U ¥l1 v~LlA. us
f_
1
lrti~
­
~ '" ­ /A1Ir, +e, tN. ~ r~ e. '1(U5o.h"'1,
Ke
'b
~
[ HJ J~
pI II)
_ CJ)() r~7) ;)
[14,1 rrJ - (GS,/v-')O,{h,IO-'):;
'"
,;1/",6(, So
[50. 5]
(
~ solve.
.-'1
Equilibrium 5
D. Practice 2
Sulfur trioxide decomposes at high temperature in a sealed container
~
according to the following reaction: 2803 (g)!:; 2802 (g) + ~ (g)
Initially the vessel is charged at lOOOK with 803 (g) with a concentration of
1'.1 /1 6.09xlO-3 M.
\~\ At equilibrium the S03 (g) concentration is 2.44xlO-3 M. r-:
\~
,: 6t'/~i""'v1
Calculate the value ofKc at lOOOK.
;/503('3) ~ ~5();)(t) ~,C?dCs.}_._
--( \I<;I--~-;-.'oa-;O~ 3M'-";~;-'- O---'--~'-~![ Jl'c. 0 M
~
b,U1
",) "-i
;D~
~,b<'1-'IJi)-
,}
-; ,::nt:, .,.1)
K~ (
'x ~
I
__
-;> 3.b<;-'" 10'5 M+'i.bs~,D-~~l§~ID_ ~ . l l <
C ,
J
rJ,-~q~IO-~ M "3.1,'5 ~ 10" MIJ.S~'IO' f'I
t
'~~ [55-:1 ) [0.1
, [C."
n '1;;'
..-J''''''J' .
7>
Kc
7,'1 ' /
cA1rf"1
y:'
''!I'"' ,
._ C:" .-::.2j'(I'6~.tiQ~
'-----;\0<,
1 \.f Y 'i IO'S) ~
I
\,
f'-"''''''--'''' ,,"
~ {uO~
"'"'
rl
"" ~ A;", "~w...
I'J~ "'-~
~'dl '~_,c,,I', "~"~
~,' f~bl&I}1f'!2
.'.,
_ ~ _,~ •. r~~
P'
,
;.; ',,,,ll " :.nilt" ~ t ;:~l
~
lYY
i~!i'~ ;:'_i{'L~;,\
',:
'jiJ- 'I ...
·
'
..
-
_""
}~Q Yr,~.y.,_b(' ',' v~
'~
j/
S ~~I
,
t
1. Hydrogen chloride reacts with oxygen to yield chlorine and water. Equilibrium can be I
established with this reaction. An experiment was performed in a closed vessel starting with a '-' mixture of 0.50 MHel and 0.050 M O2. The amount of chlorine was monitored until no
change was observed, at 0.048 Mchlorine. What is Keq for this reaction?
~hre
~J.Cl;) + J.J-/ 0
~l"-"-'<"'--'--""
~" ..... /. ~ ,-.,}.~,5 1 ,DS f V
0
0d
-f
--,096-r,--ID)~--J-'-'-,~-Ul~- -'I' +-·C~\- \',
,.___._.. '_.-_.H _".
...
~'
.
~
i,
___.
. . . _,,\
'f.
~
02.4
kif :; [.o'fa) Lcq~L:>
J. O;U· t 'O\i'\, I. .O'{g j
:_~o_~
: : .oorz
[.'full) '1[. t;j;;JCJ'
2. Using the following reaction: H2 + h ~ 2 HI calculate all three equilibrium concentrations
fHi:.~!o. ~ 0.200",M and ~ = 64.0
~r;.If+i ~ when
\.,
. '/
"",
,r
),
~~'~,~,?,'",J-"
-~j t, ~~,J..Y, ,~-x ,1-'1
;,;," y-
,,-.
-:
'X
" _.,_ _ . "",, __- ' 6'1:; .:..,CJx, ­
b " J-x
)(f .2'~)
[<"~! I 1"0: .) - ·1b ;:. ,0'1 H
'. Ke:::
&= -J-x
• ~-)<
,. b - ilC =Ax
=)
/'.6 ::: It?)<
I
=-
II;
C'Ta )'Z =,;; *, 16
[H1]f't
=:
=- .0411
J{lh) ~ ,3;1'1
'X
3. Given this equation: H2 + Br2 ~ 2 HBr calculate all three equilibrium concentrations when
0.500 mole each of H2 and Br2 are mixed in a 2.00 Lcontainer and Keq = 36.0
,'S
-'lC All
'
-. - ~
rtl)' I
d ­
--- -
I · -'''«f ..c--,-----,--··-~,l..-'·-~'· .. -'-~"
; -; ;- "-;~ .:: - -1f
A) .... 'J..
, (").') -'AXJ.C;-",)
,J :=C/
b-
f& 7'; -=- ·06)<:; M
[8f)J ~ ·V6dSM
I J )<.
[HI$(J'~ J(./~7S):' ~ 37S
--'->-----:-".--..-..
l,/(cr % '" J}~) ~
#.
l.#- i}.;x
~~----·-1~-···~··~·~)<; - 'j.
[ Jf).1" =,J ")
-.-'~
,
h- ~>
.1) - y.
I ,S -,by." ") .. J <; ::. '6 ~ • I~ 75 :::. )(
H
I­
I
4. Calculate the equilibrium constant I\:eq at 25°C for the reaction: 2NOCI(g) :t:; .2NO(g) + Cl2(g)
using the following information. In one experiment 2.00 mol of NOCI is placed in a 1.00L
flask, and the concentration of NO after equilibrium is achieved is 0.66 mol/L.
c;(NOLR ~ o?NO i (.Q ~
. ~.~-Z~q
'-_~.~:
,
Ki!'1£
()
(
~
,~, ~~
I ,'?!j
)~
-= ,,;';33)'{ 6h _ -:: , DgO I
.
I,~"'l
5. Amixture of 0.75 mol of N2 and 1.20 mol of H2 are placed in a 3.0 liter container. When the
reaction: N2 (g) + 3H2 (g) !:+ 2NH3 (g) reaches equilibrium, [H?J = 0.100 M. What are the values
of [N:) and rNHsJ at equilibrium?
IV:,
:!.5N(} f 3
H -;;;;;:i!::t__N
H~
~
.; ~ . J s: 1"\
_"_"___
....:c.....
~_..,__ ~,
HI
~
¥:::, 't
M
I
-
·
'1
;}.~.
0
I
/!-..~ -: ~. ~___.::. ~_,,___
1.f5· 1
i- . ".~---'"'----
eN ,J~b
f:f
b - :> (; 7
.'d­ .... ,.. . _­
-~~. ~ ~ -~-
~
- " ~- M
[IVH~j:.
,JM
6. Amixture of 2.5 moles H20 and 100 g of Care placed in a 50-L container and allowed to
come to equilibrium subject to the following reaction: C(s) + H20(g) ~ CO(g) + H2(g)
The equilibrium concentration of H1:]dIOgen is found to be fH?J = 0.040 M. What is the
equilibrium concentration of water, [H 20]?
HJo;
J5- -: to S 1"\
SO
( .. !:l!So -: ,17M
C(~~ + /1)0(9) ;?
,...-"
.
_../ 7
-
. <>.
~~
.",,""
,­
'.-'
(0 (:V
.05
-,04
.,.. 0'1
, 01
·O~
-~-
1--14- (<j)
7
-.~
0
O
~~ ,.-~--J
r
I
-­,
.0'1 :
._--1
l
I
O'-i
"-'"
[Hj)Je~ - ,01 ~ ~
~
I
Equilibrium 6
6. Reaction Quotient (Q)
A. The result of substituting concentration values into the equilibrium expression
l,
B. If Q equals Keq when the reaction is at equilibrium
«..
C. If Q is greater than Keq, the reaction will move from right to left until equilibrium
~
I
is reacted
D. If Q is less than Keq, the reaction will move from left to right until equilibrium ,is
reached
E. Practice:
1. At 100°C the reaction COC12 (g) !; CO(g) + Cl2 (g) has an equilibrium constant
of 2 .19x 10- 10 • Are the follOwing mixtures at equilibrilPll? If not, indicate the
direction that the reaction will proceed to reach equilibrium.
~fO::
[(DJ[Ci~J
[COU)]
A. [COCI 21= 5.00xlO- 2 M
[CO] = 3.31x10-6 M J I q ;X ((j , 10
I
[CI2] = 3.31xl0-6 M l.,
B. [COCI2] = 3.50xlO- 3 M
5
[CO]=1.11x10- M
[CI 2]
/.03x/O-'i
> Keg,
~
•
CJ
=3.25xl0-6 M
C. [COCI21= 1.45 M
[CO] = 1.56xlO-6 M
/, t/6'}}D- 1) <. k~
~
,"
[CI2] = 1.56xlO-6 M
2. At 448°C, Keq = 51 for the reaction: H2 (g) + 12 (g) !; 2HI(g)
2
Predict how the reaction will proceed if 2.0xlO- 2 moles of HI, 1.Oxl0- moles
ofH 2 and 3.0xlO-2 moles 12 are in a 2.0L container.
HJ- : ~::
HJ
l.,
r)
,!21-J.
I~
d-
=
,01
,005 M
Q=
[HI]
,01')
.:: /,33
­
[f-td] [ I d1- (oosyo/S)
;l
Ol5M
-
-
f1
I
Q
L.
Ke
,
b
I
•
--7
"
Chemistry
~
WORKSHEET - 16-6
Calculating Reaction Quotients
';":':":';":":":"::':":":":':':":";':':":':':'[::':':':':':':':':"(::':':':':':':':'~:-~~:':'~:::':'::::':':':':':':':';':';'::':':':':'::':':'::"~:;'~:-:':'::':-!':':~':':':':':':':':':':':';~:':~':':';'; ':':':';':':':':':':~':':':':':'{:':':':':':'::'~':':'~:':':':':':':':':':':~':1';';';';::':';':'~:':';::"(;::':':'':-::'::::::::':':':':':':';'~:':':'::':':':':':':";"!"::";":!':":";";':":':
Answer the following questions about equilibrium.
1. At 740°C, Ke q = 0.0060 for the
decomposition of calcium carbonate
(CaC03). Find Q and predict how the
reaction will proceed if [C02] = .0004M.
Ca9J1(s) H C.aO(s) + C02 (g)
G=.~ ... <y~ - }
~ The equilibrium constant for the
following reaction at 527°C is 5.10. If
[CO] = 0.15 M, [H20] = 0.25 M, [H2] =
0.42 M, and [C02l 0.37 M, calculate Q
and determine how the reaction will
proceed.
.,"(tf)~ IJ 111
,. 1
.1<;(.').5}
5. At 500°C, the equilibrium constant for
the following reaction is 0.080. Given
that [NH3] = 0.0596 M, [N2] 0.600 M,
and [H2] = Q.420 M, fmd Qand predict
how the reaction will proceed.
( 0 \;1.y..
N2
'-'
F~s) + 3H2 (g) H 2Fe(s)+ 3H20 (g)
Q)kllf
0<:--
,p'
,'-Ie;..
::
.S',
SbCls (g) H SbCh (g) + Ch (g)
(g) + O 2 (g) H 2NO (g)
loooec, Keq =
I
~~ (H) _ tJ{OIf
;}3 :1
2HF (g) H H2 (g) + F2 (g)
8. At 1227 °C, Keq for the following
reaction is 0.15. If [S02] =0.344 M, [02]
= 0.172 M, and [S03] = 0.056 M, find Q
and determine how the reaction will
proceed.
~
:-ii(.7S)
~
Chemistry
WSl6-6ReactionQuoffent
Q <: Ket
~
- ''5
2802 (g) + O2 (g) H 2803 (g) .N 1 "J.( /7)) .
Q--.;k%
::, 00/5
,DS'l
, oe,f,,­
-?
1.0 X 10-13 for the
following reaction. If [HF] = 23.0 M,
[H2] = 0.540 M, and [F2] = 0.38 M,
determine the value of Q and predict
how the reaction will proceed.
7. At
, (I"'~H"
-"",",,-: , () 1/
Q. '> Kl'b ~
4. The equilibrium constant for the
following reaction at 2130°C is 0.0025.
If [N2] = 0.81 M, [02] = 0.75 M, and
[NO] = 0.030 M, fmd Q and determine
the direction in which the reaction will
proceed.
N2
Ket,
Q<kl!t
0.064 for the reaction of
. rust with hydrogen gas. Given the [H2] =
0.45 M and [H20] = 0.37 M, fmd Q and
predict how the reaction will proceed.
,08
6. For the decomposition of antimony
pentachloride (SbCl s) Keq = 0.0251.
What is the value of Q if [SbCls] = 0.095
M, [SbCh] = 0.020 M, and [Ch] = 0.050
M! How will this reaction proceed?
'/)
~
3:: At 340°C, Keq =
.:
2
Q
CO (g) + H 20 (g) H H2 (g) + CO2 (g)
Q<k',,'t,
(g) + 3H (g) H 2NH3 (g) ." (./of;)) ~
Equilibrium 7
7. Solubility Product Constant (Ksp)
A. For saturated solutions equilibrium is established between the ions
in solution and the excess solid.
CJ
B. The solubility constant expresses the degree to which a solute
dissolves in water.
1. The lower the value of Ksp, the lower the dissolved amount
2. The higher the value of Ksp, the higher the dissolved amount
C. Given the hypothetical equation: AaBb (5) !::+ aA\q) + bB-(aq)
1. Lower case letters represent the number of moles
2. Upper case letters represent the compound
D. Ksp - [A+]"[B_]b
**Products over reactants
1. The square brackets indicate the concentration (molarity, M)
2. The exponents are the coefficients
3. Pure solids and liquids are not included in the equilibrium
expression becaus.e their concentrations are constant
E. Interpreting the magnitude of Ksp
u
1. When the ion product < Ksp NO PPT will form
2. When the ion product = Ksp NO PPTwill form
3. When the ion product> Ksp PPT WILL form
u
Odd.
WORKSHEET Solubili1y Equilibria
\.,
16~9 C,1/~
t'1)
j'J
Chemistry
I7Pif!..
':':':':':';'::':':':':':':':':':':':':':':':':':':~':':':':':':';':':':':':':':':':':':':':':'~:':';':':':':':';':~':';'!':';':':':':':':':':':':'~:'~:':':';':':':':':':';':':':':':';':~;~~':(::';':';';':':':':':':';':':':':~':':':':':':':';';':':':':' ;'?:':':':':':';":':':':';':':':':':':':';':';':';';':':':~':':':':':':':'~:':'::':':':':':':';':':';':':':::':':':';'::':':':':;';:-::':':':'~:'~,?:'~:':':':':;':':':
Answer the following questions about equilibrium.
~
(
19'~ product
Write the expression for the solubility
constant for MgCh.[!1J 'Jflat
v
f
y!
8
10-7 M. What is the value of Ksp for
AgBr? /7,1) 7/'10 - "'r~:: 50)( 10 ~ I~ Write the solubility equilibrium for FeS04. [Ft~IJC50; -;J
lILA sample of MgC03 (s) is added to pure water and allowed to come to ~:) Write the expression for the solubility
equilibrium at 2SoC. The concentration ,
product constant for Zn(OH)2. [z"oJ[fOfd}
of Mg2+ is 6.3 x 10-3 M at equilibrium. rJ /4. Write the solubility equilibrium for if
What is the value of Ksp for MgC03? , CU(OH)2. LCq°J [OItJ~ \ '
~)< f6- sJO
\, 12.At 2SoC, the concentration of S~+ ions
,;'/
1,
j", Write the expression for the solubility
in a saturated solution of Sr(OH)2 is 4.31
2
,
product constant for Ah(S04)3. CAI :J[50tJ3
X 10- M. What is the value of Ksp for
r ) s, :p.;
~, §; Write the solubility equilibrium for
Sr(OH)2? Kft'-::[:5f J[oHl:i;)
Ca3(P04)2. [Cq '~:rL.p()'1·:J';>
_,~
k5f ~[O'l~IJ[)iU4-:'f) :;; 5,dX 10 I
I L 13.What are the equilibrium concentrations
7. A sample of srCo3 (s) is added to pure v
of the dissolved ions in a saturated ' ".~
water and allowed to come to[5r~ ={(chj
solution ofFe(OH)2 at 2SoC? (Ksp = 1.8
Ce.~~at 2S~C. The co~~en~tion
x 10-15) ,K~f :;. [r(] [OHJ;I ':.. j«(,})..)~_~ 1.,5~ I~-;o
+ls4.0x 105 MateqUlhbnum.
7 bl,xl[) -"
't);~ '?}\/~ I::. What is the value ofKsp for srCo3?
\\ 14.What are the equi1ibrf~~Srtcen'fi.~fi~n'S (,A It.' -';. k y =[5r"'J[((/3 ''j :;. ('!r//)-sl'ld()'S)
of the dissolved ions in a saturated
2
\_\ 8. At 18°C,t~e conce~tration of Pb + ions
solution of ~g2S0~a~~?oC? (Ksp ~ 1.~ x
f.(~ rM
5
in a~jea::solut1on of lead oxalate
10- ) ~'
i:./ :l.i a /f L
(PbC20 4) is S.23 x 10-6 M. What is the
. Od-~
~ ,'f ~~ ~. ~~ >K<:p
value of Ksp for!bC 204?
(~lS.What are the equilibrium concentrations
k~f:::[rJPJ@:.jO~3 ::. LS,) 3)(JO''..[C;,)?40 -'j!' ). J'I~O I ofPb2+andS2- in a saturated solutionof ,at",J ,/1
;. ("
. ~~) A sample ofCu3P04 (s) is added to pure v
PbS at 25°C? (Ksp = 7.0 x Io-2~}6
,,.'.A
'-....1
-water and allowed to come to
\1'5' ,"'/1 It> -I~
1'/1 (o/"'J
equilibrium at 2SoC. The concentration ,\ 16.What are the equilibrium concentrations
I . Lf i "I' b
5
of Cu+ is 1.OS x 10- M at equilibrium.
of Ag+ and S~ in a saturated solution
of silver thiocyanate (AgSCN) at 2S0C?
What is the value of Ksp for CU3P04?
~7P -=. [ev 1~[vo ..,-1 :: [,·os" 'b -~] ~[I o'S :<Ilr'l-; ] :: Lf .05",0-.71 (Ksp = 1.16 x 10-12) 1./>, 10 - ,
IO.At 2SoC, the concentration of Ag+ ionS~)
,'" . ~
ina saturated solution of AgBr is 7.07 x
Yb S'X I()- -. I." !l t //:/ f;~1 "1k! ,,;
1
l­
L'
l" WSl6-9So/ubilityEquilibria
Chemistry
'J
<::­
I, ) xl/)-S
J
J .
_ I}x]
)r5~
J
>
<:,
).
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u}(
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(x ) • ,;}
.01)
7
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r
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8 . Le Chatelier's Principle - When a stress is applied to a system, the reaction will try to shift in a direction that
will relieve the stress
Ex:
02(g) +
1hli1k
2H2(g)
~
db
ftJl.J///Jlivm Ilk~ a
Set'5~
ft·vd
2H20(l) + 572 kJ
ZI
A. Temperature
l' T -=:. 1ltid eIL{Y'jY .t T :;: J. AlJ ~"j'f vy.) l'
sfr'ecne5 ~rI8h+ ~ /, rettchi:J..", sh,+1-S ~
• ( Irllu--.p t.L.J ~)
-Hv.. sfr~ S!;
-r
to
Otl.er(l/YJrl
B. ConcentratIon
When the concentration of a reactant or a product is changed, the system will shift in order to restore the
original concentration as closely as possible
-Increasing the concentration of one substance, causes the reaction to shift in the opposite direction.
-Decreasing the concentration of one substance, causes the reaction to shift in the direction where the
concentration decreased
"-1-)
~ [O~]
J, ['1,01
"!lty-t S5(' 'S
':)fye~~
Ctp
~
r-tl1ch()'h
5)"fIs
-->
njh+ ~ :, rt~chO">1 si1,Hs ~
C. Pressure - Affects gases only!
-Increasing pressure pushes molecules closer together - increases concentration.
'-?) l' P ~+re$'.){ S ltp"'?lck 6.t C 4 iI~
fr/()f1l
YI'/b i.e s
&/6 ;) ct s "L .
#
~
-Ifboth sides of the reaction contain gases, the side with more moles will be affected more.
~~rt.
~f\1~ Ex) 2803 (g) ~ 2S~ (g) + 02 (g) J P Str-esz;e '5 n~hf ~ r: ---7
- 1+ ~t.. / 5 ~ e'rvtJ. 4) /?flit s tr/;
*'
fre95VrL 1<:, nt1 Jf $frts~
qc; 5
(h!
60 fll '!ilks
4
d,aArtF-
~
J
D. Catalyst - Increases the forward and reverse rates equally, so there will be no shift in the system!
Here's a biological example ofan application ofLe Chatelier's principle:
Hemoglobin (Hb) reacts with oxygen to form Hb02, a substance that transfers oxygen to the tissues in the
body. Carbon monoxide (CO) also reacts with Hb02 by the process below.
tC'1LU J to lite.
~
Le Chatelier's Principle Demonstration
Name:
An5Wf.)'" kt /V '--'
Date: -----------------
CoCh . 6H20 (aq) + 4cr (aq) + 50 kJ
~
CoCh (aq) + 6H2 0 (1)
(pink)
(blue)
Predictions:
1. IfHCI is added to the system, what prediction do you make for the color of the solution? Explain your
prediction using Le Chatelier's principle.
Blu~
~
2. If water is added to the system, what prediction do you make for the color of the solution? Explain your
prediction using Le Chatelier's principle.
fink
~
3. If heat is added to the system, what prediction do you make for the color of the solution? Explain your
prediction using Le Chatelier's principle.
Blue
---7
~
4. If this system is put into ice-water and cooled, what prediction do you make for the color of the system?
Explain your prediction using Le Chatelier's principle.
Pll1k
<­
Actual Results:
1. What color is the solution at room temperature?
2. What color is the solution when HCI is added? Compare it to the room temperature solution.
3. What color is the solution placed in hot water? Compare it to the mom temperature solution.
l,
4. What color is the solution placed in ice-water? Compare it to the room temperature solution.
L
1. Given the solution at equilibrium:
La. Ammonia is produced commercially by the Haber
reaction:
Pbl 2(s)
f-'j>
Pb2+(aq) + 2naq)
N2(g) + 3 H2(g)
The addition of which nitrate salt will cause a
decrease in the concentration of naq)?
(1) Pb(NO"2
(3) LiNOs
(2) Ca(NOS)2
(4) KNOs
_1_2. Given the reversible reaction
A(g) + B(g)
f-'j>
C{g) at equilibrium.
~7. Given the system at equilibrium:
2POCla(g) + energy
2HI(g)
~8. Given the reaction at equilibrium:
N2(g) + 02(g) + energy f-'j> 2 NO(g) f-'j>
A2Bs(g) + heat
Which change will not affect the equilibrium
concentrations of A(g), B(g) , and A2Bs(g)?
(1)
(2)
(3)
(4)
adding more A(g)
adding a catalyst
increasing the temperature
increasing the pressure
. , 5. Given the reaction at eqUilibrium:
4 HCI(g) + 02(g)
f-'j>
2 CI2 (g) + 2 H20(g)
If the pressure on the system is increased, the
concentration of CI 2(g) will
(1) decrease
(3) remain the same
(2) increase
'1
Which change will result in a decrease in the amount of NO(g) formed? (1) decreasing the pressure
(2) decreasing the concentration of N2(g)
(3) increasing the concentration of 02(g)
(4) increasing the temperature
__4. Given the reaction at equilibrium:
2 A(g) + 3 B(g)
2PC1ig) + 02(g)
Which changes occur when 02(g) is added to this
system?
(1) The equilibrium shifts to the right and the concentration of PCls(g) increases. (2) The equilibrium shifts to the right and the concentration of PCls(g) decreases. (3) The eqUilibrium shifts to the left and the concentration of PCls(g) increases. (4) The equilibrium shifts to the left and the concentration of PCI3 (g) decreases. ;). 3. Given the equation representing a reaction at eqUilibrium: Which change favors the reverse reaction?
(1) decreasing the concentration of HI(g)
(2) decreasing the temperature
(3) increasing the concentration of 12(g)
d.(4) increasing the pressure
."
(1) an increase in pressure
(2) a decrease in pressure
(3) removal of N2(g)
(4) removal of H2(g)
(1) the rate of the forward reaction
(2) the value of the equilibrium constant
(3) the activation energy
(4) the concentration of B
f-'j>
2 NHs(g) + heat The formation of ammonia is favored by
If the concentration of A is increased at constant
temperature and pressure, which will also increase?
H2(g) + 12(g) + heat
f-'j>
'I
9. Given the system at equilibrium:
N20 4 (g) + 58.1 kJ
f-'j>
2 N02{g)
What will be the result of an increase in temperature
at constant pressure?
(1) The equilibrium will shift to the left, and the concentration of N02(g) will decrease. (2) The equilibrium will shift to the left, and the concentration of N02{g) will increase. (3) The eqUilibrium will shift to the right, and the concentration of N02(g) will decrease. (4) The eqUilibrium will shift to the right, and the concentration of N02(g) will increase. ~
Le Chatelier's Principle
0~ame:
_ _ _ _ _ _ _ _ _ _ __
Date: - - - - ­
- When a stress is applied to a system at equilibrium, the reaction will try to
shift in a direction that will relieve the stress.
Ex)
la) t Concentration of
reactants
Ib) t Concentration of
products
'"
2a) 1 Concentration of
reactants
~
2b) 1 Concentration of
products
"'---7
3b) 1 Temperature
of system
5) Add a catalyst
.....
-:;:-
'
...
~
-
4a) t Pressure of
system
4b) 1 Pressure of
system
Effect
-
.~
: 3a) t Temperature
of system
L
2NH3 (9) + Heat
Shift in Equilibrium Stress
~
~
N2 (9) + 3H 2 (9)
...
~
.. .....
~
~---' I
-
tp~ t
(tlJ.c/
l'
t~~~+ l' ({'cd t
Itoct
J
t fr()J
I
t
froJ
l'
("(€td
I
I
fQst~ F~' re. LeChatelier Practice
I. Consider the following equilibrium system in a closed container:
Ni(s) + 4 CO(g)
~
t1H
Ni(CO)4(g)
-161k1
~
In which direction will the equilibrium shift in response to each change, and what will be the effect on the
indicated quantity?
Change
Direction
Effect on
Effect
of Shift
Quantity
(increase, decrease,
or no change)
( --..; -t-; or no chan!i!e)
(a) add Ni(s)
[Ni(CO)4(g)]
->
(b) raise temperature
[CO(g)]
~
(c) add CO(g)
~
[ Ni(s)]
(d) remove Ni(CO)4(g)
~
[CO(g)]
(e) decrease in pressure
<E-­
(f) lower temperature
[Ni(CO)4(g)]
[CO(g)]
~
(g) remove CO(g)
~
-
[Ni(CO)4(g)]
l'
l'
-l­
,t,
t
J,
f'
~
2. The manufacture of ethanol is done as follows:
CH2=CH2(g) + H20(g)~ CH3CH20H(g) + 46 k1
(ethene) (ethanol)
~
a. To get the maximum percentage conversion of the ethene into ethanol is a high or®emperature best? Explain why. :7
b. Conversion of ethene into ethanol is favored by high pressures. Explain why. ----3J MD(('
#1"~S.
:JGS
fh"
fe/f
c. \Vould adding extra steam favor the creation o f e or less ethanol? \Vhy?
~
~