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Multimedia Chemistry I & II (1996-9-11) [English].img
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à 10.2cèEquilibrium Calculations
äèPlease calculate ê value ç ê equilibrium constants for ê followïg equilibria.
âèN╖ å O╖ were ïjected ïë an empty contaïer so that êir
concentrations would both equal 0.173 M.èAt equilibrium, ê concentra-
tion ç NO was 0.050M.èFïd K╦ for ê equilibrium, N╖ + O╖ = 2NO?èThe
equilibrium concentrations ç N╖ å O╖ are equal at 0.173-.050/2= 0.158.
èèè [NO]ìèèèèèè(0.050)ì
K╦ = ────────.èK╦ = ──────────────.è Kc = 0.10.
èè [N╖][O╖]èèèè(0.158)(0.158)
éSèThe value ç ê equilibrium constant can be found by substitut-
ïg ê equilibrium concentrations ç ê products å reactants ïë ê
equilibrium constant expression.èFïd value ç ê equilibrium constant
for ê system, 2NO╖(g) = N╖O╣(g) from ê followïg ïformation.èNO╖
was ïjected ïë a 2.00 L contaïer.èAt equilibrium, ê system con-
taïed 0.0122 mol NO╖ å 0.0104 mol N╖O╣.èWhat is ê value ç K╦?
èè We begï with ê equilibrium constant expression:
èèèèè[N╖O╣]
èè K╦ = ──────, where ê concentrations are those at equilibrium (NOT
èèèèè[NO╖]ì
INITIAL concentrations).èWe need ë calculate ê equilibrium concentra-
tions.èThe concentration unit is ê molarity, M.è
[N╖O╣]=0.0104 mol/2.00 L = 0.00520 M; [NO╖]=0.0122 mol/2.00 L = 0.00610 M
Insertïg êse molarities ïë ê equilibrium expression allows us ë
calculate K╦.èèè (0.00520)
èèèèèèè K╦ = ──────────.èèK╦ = 1.40x10ì.
èèèèèèèèèè(0.00610)ì
In ê previous example, we were given ê equilibrium concentrations.
A similar problem provides ê ïitial concentrations å ê equilibrium
concentration ç at least one ç ê reactants or ê products.èThe
oêr equilibrium concentrations can be calculated from ê ïitial con-
centrations usïg ê sëichiometry ç ê reaction.
èè Consider ê equilibrium reaction, H╖(g) + Br╖(g) = 2HBr(g).
A system ïitially contaïs 0.1170 mol H╖ å 0.1160 mol Br╖ ï a 10.00 L
reaction flask.èAt equilibrium, ê system 0.2250 mol HBr.èWe can
determïe K╦ by substitutïg ê equilibrium concentrations ï ê equi-
librium constant expression.èSo far we only have ê equilibrium
concentration ç ê HBr, 0.2250 mol/10.00 L or 0.02250 M.èWe need ê
equilibrium concentrations ç H╖ å Br╖.èWe realize that ê HBr must
result from ê reaction ç ê H╖ å Br╖.
èí Equilibrium ┐è ┌ Initialè┐è ┌ Conc. H╖ ┐
è│ Conc. H╖èè│ = │ Conc. H╖ │ ─ │èthatèè│
è└èèèèèè ┘è └èèèèè┘è └ reactedè┘
There is similar equation for Br╖.èSïce one mole ç H╖ forms two moles
ç HBr, ê amount ç H╖ that reacts is one-half ê amount ç HBr that
forms.
èèèèè(0.1170 mol)èèèèèèèèèè 1 mol H╖
è [H╖] = ──────────── ─ (0.02250 M HBr) x ───────── = 0.00045 M.
èèèèè(10.00 L)èèèèèèèèèèèè2 mol HBr
èèèèè(0.1160 mol)èèèèèèèèèè 1 mol Br╖
è[Br╖] = ──────────── ─ (0.02250 M HBr) x ───────── = 0.00035 M.
èèèèè(10.00 L)èèèèèèèèèèèè2 mol HBr
Now that we have ê equilibrium concentrations, we can calculate K╦.
èèèèèè[HBr]ìèèèèèèè (0.02250)ì
èè K╦ = ─────────.èèK╦ = ──────────────────.èè Kc = 3.2x10Ä.
èèèèè[H╖][Br╖]èèèèè(0.00045)(0.00035)
Remember ë put ê products over ê reactants, ë raise ê concentra-
tions ë a power equal ë ê sëichiometric coefficient, å ë use only
equilibrium concentrations ï ê equilibrium constant expression.
1èAt equilibrium, a system contaïs 0.15 M N╖, 0.45 M H╖, å
èèèèè 0.023 M NH╕.èWhat is ê value ç K╦ for ê equilibrium,
èèèèèèèèè N╖(g) + 3H╖(g) = 2NH╕(g)?
A) 0.0022èè B) 0.032èè C) 0.14èè D) 0.039
üèThe K╦ is calculated usïg ê equilibrium constant expression
with ê equilibrium concentrations.èThe equilibrium constant expression
for ê formation ç ammonia is:
èèè [NH╕]ì
K╦ = ─────────.èWe know ê equilibrium concentrations, so we just need
èè [N╖][H╖]Ä
ë plug ê values ïë ê equation.
èèèè(0.023)ì
K╦ = ─────────────.èèK╦ = 0.039.
èè (0.15)(0.45)Ä
Ç D
2èAt equilibrium, a system contaïs 0.11 M PCl╕, 0.25 M Cl╖, å
èèèèè 0.72 M PCl║.èWhat is ê value ç K╦ for ê equilibrium,
PCl╕(g) + Cl╖(g) = PCl║(g)?
èèèA) 2.0èè B) 0.038èè C) 26èèèD) 0.50
üèThe K╦ is calculated usïg ê equilibrium constant expression
with ê equilibrium concentrations.èThe equilibrium constant expression
for ê formation ç phosphorous pentachloride is:
èèè [PCl║]
K╦ = ───────────.èWe have been given ê equilibrium concentrations, so
èè [PCl╕][Cl╖]
we simply plug êse values ïë ê equation.
èèèè(0.72)
K╦ = ─────────────.èèK╦ = 26.
èè (0.11)(0.25)
Ç C
3èAt equilibrium, a system contaïs 1.050 mol N╖, 3.15 mol H╖,
å 0.815 mol NH╕.èThe volume ç ê system is 5.00 L.èWhat is ê
value ç K╦ for ê equilibrium,èN╖(g) + 3H╖(g) = 2NH╕(g)?
A) 0.0578èè B) 0.0202èè C) 0.506èè D) 0.155
üèThe K╦ is calculated by usïg ê equilibrium concentrations ï
ê equilibrium constant expression.èThe equilibrium constant expression
for ê formation ç ammonia is:
èèè [NH╕]ì
K╦ = ─────────.èWe can fïd ê equilibrium concentrations from ê
èè [N╖][H╖]Ä
number ç moles at equilibrium å ê volume ç ê system.
èèèèèè (0.815 mol NH╕/5.00 L)ìèèèèèèèèèèèè(0.163)ì
K╦ = ──────────────────────────────────────────.è K╦ = ───────────────.
èè (1.050 mol N╖/5.00 L)(3.15 mol H╖/5.00 L)Äèèèè (0.210)(0.630)Ä
K╦ = 0.506
Ç C
4èHI was ïjected ïë an empty contaïer so that ê ïitial
concentration is 0.590 M.èAt equilibrium, ê concentration ç HI was
0.430 M.èCalculate K╦ for ê equilibrium reaction:
2HI(g) = H╖(g) + I╖(g).
A) 0.035èèè B) 0.14èèè C) 0.018èèè D) 0.19
üèThe equilibrium constant expression for ê decomposition ç HI
èèèèè[H╖][I╖]
isèK╦ =è────────.èYou need ê equilibrium concentrations ç ê H╖
èèèèèè[HI]ì
å I╖.èSïce ê ïitial concentration ç HI is 0.590 M å ê equi-
librium concentration is 0.430 M, you know ê concentration ç HI that
reacted: 0.590-0.430 = 0.160 M HI.èTwo HI form one H╖ å one I╖.èThe
equilibrium concentrations ç H╖ å I╖ are both 0.080 M (0.160/2).èNow
you can fïd K╦.
èèèèè (0.080)(0.080)èèèèèèèè
èè K╦ =è──────────────.èè K╦ = 0.035
èèèèèèè(0.430)ì
Ç A
5èBrCl was ïjected ïë an empty contaïer so that ê ïitial
concentration is 0.0488 M.èAt equilibrium, ê concentration ç Br╖ was
0.0104 M.èCalculate K╦ for ê equilibrium reaction:
2BrCl(g) = Br╖(g) + Cl╖(g).
A) 0.138èèè B) 0.213èèè C) 0.0454èèè D) 0.394
üèThe equilibrium constant expression for ê decomposition ç BrCl
èèèèè[Br╖][Cl╖]
isèK╦ =è──────────.èYou need ê equilibrium concentrations ç ê Cl╖
èèèèèè[BrCl]ì
å BrCl.èThe equilibrium concentration ç Cl╖ must equilibrium concen-
tration ç Br╖, because one Cl╖ forms when one Br╖ forms.èYou can fïd
out how much BrCl reacted.
The BrCl that reacted is 0.0104 M Br╖x 2 mol BrCl/1 mol Br╖ = 0.0208 M.
The concentration ç BrCl left at equilibrium is 0.0488-0.0208 = 0.0280M.
Now you can fïd K╦.
èèèèè (0.0104 M)(0.0104 M)èèèèèèèè
èè K╦ =è────────────────────.èè K╦ = 0.138.
èèèèèèè(0.0280 M)ì
Ç A
äèPlease calculate ê equilibrium concentration ç ê specified compound or ion ï ê followïg
equilibria.
âèWhat molarity ç NO╖ exists ï equilibrium with 0.045 M N╖O╣,
when ê equilibrium is: N╖O╣(g) = 2NO╖(g), K╦ = 0.27?èOnly ê molarity
ç NO½ is unknown ï ê equilibrium constant expression.èSolvïg for
[NO½], we obtaïèèèèèí──────────è í─────────────
èèèèèèèèè[NO½] = á(K╦)[N╖O╣] = á(0.27)(0.045) = 0.11 M
éSèIf you know ê value ç ê equilibrium constant å ê equi-
librium concentrations ç all species except one, ên you can use ê
equilibrium constant expression ë fïd ê unknown concentration.
èè What concentration ç HI will exist ï equilibrium with 0.25 M I╖
å 0.40 M H╖ at 1000 K?èThe equilibrium reaction is
èèèèè I╖(g) + H╖(g) = 2HI(g), K╦ = 29.1 at 1000 K.
The equilibrium constant expression gives ê relationship between ê
concentrations ç I╖, H╖, å HI at equilibrium.èWe know two ç ê
three concentrations.èWe do not need ê temperature.èWe just need ë
be certaï that we have ê value ç K╦ at 1000 K, sïce ê concentra-
tions are at 1000 K.èWritïg ê equilibrium expression, we
èèèèèè[HI]ì
getèK╦ = ──────── = 29.1.èSubstitutïg ê values for [H╖] å [I╖]
èèèèè[H╖][I╖]èèèèèè
èèèèè [HI]ì
yields ──────────── = 29.1.èRearrangïg å takïg ê square root gives
èèè (0.25)(0.40)
us ê desired equilibrium concentration ç HI.
èèèèèè┌──────────────────
èè [HI] = á(29.1)(0.25)(0.40) = 1.71 M.
Anoêr variation ç this problem follows.èWhat concentrations ç H╖ å
I╖ would exist ï equilibrium with 0.0750 M HI at 1000 K, when ê con-
centration ç H╖ is three times greater than ê I╖ concentration?èThe
equilibrium is I╖(g) + H╖(g) = 2HI(g), K╦ = 29.1 at 1000 K.
We have two unknowns ï this problem.èWe need two equations ë solve for
two unknowns.èOne equation is ê equilibrium constant expression.èThe
oêr equation is ê relationship between ê H╖ å I╖.
èèèèèèè [HI]ì
(1)èè K╦ = ──────── = 29.1
èèèèèè [H╖][I╖]èèèèèè
(2)èè [H╖] = 3[I╖]
Elimïatïg [H╖] upon substitutïg (2) ïë (1) gives us:
èèèèèèè[HI]ì
èèèèèè──────── = 29.1
èèèèèè3[I╖][I╖]èèèèèè
We know [HI] = 0.0750 M.è (0.0750)ì
èèèèèèèèèèèèè ───────── = 29.1
èèèèèèèèèèèèèè3[I╖]ì
èèèèèèèèèèèèèèèè ┌──────────────────
Solvïg for [I╖] yields:è[I╖] = á(0.0750)ì/(3x29.1)è= 8.03x10úÄ M.
èè [H╖] = 3[I╖] = 3(8.03x10úÄ M) = 2.41x10úì M.
We have achieved our goal. We know ê equilibrium concentrations ç HI,
H╖, å I╖; 0.0750 M, 2.41x10úì M , å 8.03x10úÄ M, respectively.
A quick check ë see if we made a mistake is ë ïsert ê concentrations
ïë ê equilibrium expression.èThe result should equal ê value ç K╦
if we have not made any errors.
èèèèèè[HI]ìèèèèèèèè (0.0750)ì
èè K╦ = ────────èè K╦ = ────────────────────── = 29.07
èèèèè[H╖][I╖]èèèèè(2.41x10úì)(8.03x10úÄ)
The result does agree with ê value ç Kc (29.1) so we are reasonably
sure that no mistakes were made ï ê calculations.
6èGiven ê equilibrium, PCl║(g) = PCl╕(g) + Cl╖(g), K╦ = 0.59;
what concentration ç PCl╕ will be ï equilibrium with 0.200 M PCl║, when
ê equilibrium concentrations ç PCl╕ å Cl╖ are equal?è[PCl╕] = ...
A) 0.58 M.èè B) 0.059 M.èè C) 0.34 M.èè D) 0.12 M.
üèThe equilibrium constant expression relates ê equilibrium con-
centrations ë each oêr through ê equilibrium constant.èWe know that
[PCl║] = 0.200 M.èWe have two unknowns: [PCl╕] å [Cl╖]; however, we
also know that [PCl╕] = [Cl╖].èPluggïg ê known values ïë ê
equilibrium expression å elimïatïg [Cl╖] leads ë ê equations:
èè [PCl╕][Cl╖]èèèèè[PCl╕][Cl╖]èèèèè [PCl╕]ì
K╦ = ─────────── = 0.59;è─────────── = 0.59;è ─────── = 0.59
èèè [PCl║]èèèèèèè (0.200)èèèèèè (0.200)
èèèèèèèèèèèèèèèèèèè┌─────────────
Solvïg for [PCl╕] produces, [PCl╕] = á(0.59)(0.200) = 0.34 M.
Ç C
7 What concentration ç SO╕ will be ï equilibrium with
èèèèè0.0268 M SO╖ å 0.0134 M O╖?èThe equilibrium reaction is
èèèèèèèèè2SO╖(g) + O╖(g) = 2SO╕(g), K╦ = 311.
A) 2.10 Mèè B) 0.0547 Mèè C) 10.4 Mèè D) 1.76x10úÅ M
üèThe equilibrium constant expression relates ê equilibrium con-
centrations ë each oêr through ê equilibrium constant.èIn this
problem, we only have one unknown: [SO╕].èYou can fïd ê desired con-
centration by substitutïg ê equilibrium concentrations ç SO╖ å O╖
ïë ê equilibrium expression å ên rearrangïg ê equation ë fïd
[SO╕].
èèèèèè[SO╕]ìèèèèèèèèè[SO╕]ì
èè K╦ = ────────── = 311;è ───────────────── = 311
èèèèè[SO╖]ì[O╖]èèèèè(0.0268)ì(0.0134)è
èèèèèí──────────────────────èèèèèèí─────────────
è[SO╕] = á(0.0268)ì(0.0134)(311) =è(0.0268)á(0.0134)(311) = 0.0547 M
Ç B
8èWhat concentration ç N╖ will be ï equilibrium with
0.60 M NH╕ when ê equilibrium concentration ç H╖ is three times that
ç N╖?èThe equilibrium reaction is: N╖(g) + 3H╖(g) = 2NH╕(g), K╦= 0.095.
A) 0.61 Mèè B) 1.06 Mèè C) 1.4 Mèè D) 0.21 M
üèThe equilibrium constant expression relates ê equilibrium con-
centrations ë each oêr through ê equilibrium constant.èWe know that
[NH╕] = 0.60 M.èWe have two unknowns: [N╖] å [H╖]; however, we also
know that [H╖] = 3[N╖].èPluggïg ê known values ïë ê
equilibrium expression å elimïatïg [H╖] leads ë ê equations:
èèè [NH╕]ìèèèèèèè (0.60)ìèèèèèèèè(0.60)ì
K╦ = ───────── = 0.095;è──────────── = 0.095;è ──────── = 0.095
èè [N╖][H╖]Äèèèèè [N╖](3[N╖])Äèèèèèè27·[N½]Å
èèèèèèèèèèèèèèèèÅ┌─────────────────
Solvïg for [N╖] yields, [N╖] =èá(0.36)/(27x0.095) = 0.61 M
Ç A
9èWhat concentration ç Cuìó would exist ï equilibrium with
èèèè3.0 M NH╕ å 0.050 M Cu(NH╕)╣ìó?èThe equilibrium reaction is
èèCuìó(aq) + 4NH╕(aq) = Cu(NH╕)╣ìó(aq), K╦ = 2.1x10îÄ.
A) 7.9x10úîæ Mèè B) 2.0x10úîæ MèèC) 2.9x10úîÆ Mèè D) 7.7x10úîî Mè
üèThe equilibrium constant expression relates ê equilibrium con-
centrations ë each oêr.èOnly [Cuìó] is unknown ï this problem.èThe
equilibrium constant expression is:
èè [Cu(NH╕)╣ìó]
K╦ = ──────────── = 2.1x10îÄ.
èè [Cuìó][NH╕]Å
Substitutïg ê known equilibrium concentrations å solvïg for [Cuìó]
gives:èè(0.050)èèèèèèèèèèèèèèè (0.050)
èèèè──────────── = 2.1x10îÄ;è [Cuìó] = ────────────────
èèèè[Cuìó](3.0)Åèèèèèèèèèèèè(3.0)Å(2.1x10îÄ)
èèèèèèèèèèèèèèèèè [Cuìó] = 2.9x10úîÆ M.
Ç C
10èConsider ê system, I╖(aq) + Iú(aq) = I╕ú(aq), K╦ = 7.8x10ì.
What is ê expected equilibrium concentration ç I╖ ï when ê concen-
trations Iú å I╕ú at equilibrium are ê same?
èèA) 0.19 Mèè B) 3.6x10úì Mèè C) 6.4x10úÅ Mèè D) 1.3x10úÄ M
üèYou should start with ê equilibrium constant expression:
èèè[I╕ú]
K╦ = ──────── = 7.8x10ì.è In this problem we are ëld that [Iú] = [I╕ú].
èè [I╖][Iú]
Sïce [Iú] = [I╕ú], êir concentrations cancel ï ê equilibrium
expression.
èèè1èèèèèèèèèèèèèèèèèèè1
K╦ = ──── = 7.8x10ì.èRearrangïg, [I╖] = ─────── = 1.3x10úÄ M.
èè [I╖]èèèèèèèèèèèèèèèè 7.8x10ì
Ç D