Equilibrium
In Chemical Reaction
N2 + 3H2
2NH3
The double arrow tells us that
this reaction can go in both directions:
N2 + 3H2
2NH3
1) Reactants react to become products,
N2 + 3H2
2NH3
(‘forward’ reaction)
N2 + 3H2
2NH3
1) Reactants react to become products,
N2 + 3H2
2NH3
(‘forward’ reaction)
while simultaneously,
2) Products react to become reactants
N2 + 3H2
2NH3
(‘reverse’ reaction)
N2 + 3H2
2NH3
In a closed system,
where no reactants, products, or energy
can be added to or removed from the reaction,
a reversible reaction will reach equilibrium.
N2 + 3H2
2NH3
At equilibrium, the rate of the forward reaction becomes
equal to the rate of the reverse reaction, and so, like our
escalator metaphor, the two sides, reactants and products,
will have constant amounts, even though the reactions
continue to occur.
N2 + 3H2
2NH3
However (like the metaphor), the equilibrium
amounts of reactants and products are usually
not equal, they just remain unchanged.
N2 + 3H2
2NH3
N2 + 3H2
2NH3
N2 + 3H2
2NH3
N2 + 3H2
2NH3
reverse
forward
reverse
forward
reverse
forward
reverse
forward
reverse
forward
reverse
forward
reverse
forward
reverse
forward
reverse
forward
reverse
forward
reverse
forward
etc!
the reactions go on continuously in both directions.
concentration
Changes in the concentrations of the reactants
and products can be graphed; the graph
indicates when equilibrium has been reached.
time
concentration
For N2 + 3H2 2NH3,
suppose you begin with the following:
N2 = 1 M, H2 = 1 M, and NH3 = 0 M
time
concentration
For N2 + 3H2 2NH3,
suppose you begin with the following:
N2 = 1 M, H2 = 1 M, and NH3 = 0 M
time
concentration
For N2 + 3H2 2NH3,
suppose you begin with the following:
N2 = 1 M, H2 = 1 M, and NH3 = 0 M
time
concentration
For N2 + 3H2 2NH3,
suppose you begin with the following:
N2 = 1 M, H2 = 1 M, and NH3 = 0 M
time
concentration
For N2 + 3H2 2NH3,
suppose you begin with the following:
N2 = 1 M, H2 = 1 M, and NH3 = 0 M
time
concentration
For N2 + 3H2 2NH3,
suppose you begin with the following:
N2 = 1 M, H2 = 1 M, and NH3 = 0 M
time
concentration
For N2 + 3H2 2NH3,
suppose you begin with the following:
N2 = 1 M, H2 = 1 M, and NH3 = 0 M
time
concentration
For N2 + 3H2 2NH3,
suppose you begin with the following:
N2 = 1 M, H2 = 1 M, and NH3 = 0 M
time
concentration
For N2 + 3H2 2NH3,
suppose you begin with the following:
N2 = 1 M, H2 = 1 M, and NH3 = 0 M
time
concentration
For N2 + 3H2 2NH3,
suppose you begin with the following:
N2 = 1 M, H2 = 1 M, and NH3 = 0 M
time
concentration
For N2 + 3H2 2NH3,
suppose you begin with the following:
N2 = 1 M, H2 = 1 M, and NH3 = 0 M
time
concentration
For N2 + 3H2 2NH3,
suppose you begin with the following:
N2 = 1 M, H2 = 1 M, and NH3 = 0 M
time
concentration
For N2 + 3H2 2NH3,
suppose you begin with the following:
N2 = 1 M, H2 = 1 M, and NH3 = 0 M
time
concentration
For N2 + 3H2 2NH3,
suppose you begin with the following:
N2 = 1 M, H2 = 1 M, and NH3 = 0 M
time
concentration
For N2 + 3H2 2NH3,
suppose you begin with the following:
N2 = 1 M, H2 = 1 M, and NH3 = 0 M
time
concentration
For N2 + 3H2 2NH3,
suppose you begin with the following:
N2 = 1 M, H2 = 1 M, and NH3 = 0 M
time
concentration
For N2 + 3H2 2NH3,
suppose you begin with the following:
N2 = 1 M, H2 = 1 M, and NH3 = 0 M
time
For N2 + 3H2 2NH3,
suppose you begin with the following:
N2 = 1 M, H2 = 1 M, and NH3 = 0 M
concentration
[N 2 ]
[H 2 ]
[NH3]
time
concentration
For N2 + 3H2 2NH3,
suppose you begin with the following:
N2 = 1 M, H2 = 1 M, and NH3 = 0 M
[N 2 ]
[H 2 ]
[NH3]
time
Question 3: at what point
has equilibrium been established?
concentration
For N2 + 3H2 2NH3,
suppose you begin with the following:
N2 = 1 M, H2 = 1 M, and NH3 = 0 M
[N 2 ]
[H 2 ]
[NH3]
time
Question 4: what does the graph tell you about the
concentration of each species once equilibrium is established?
concentration
For N2 + 3H2 2NH3,
suppose you begin with the following:
N2 = 1 M, H2 = 1 M, and NH3 = 0 M
[N 2 ]
[H 2 ]
[NH3]
time
Question 5: what might a rate vs time graph
look like for the above reaction?
rate
For N2 + 3H2
2NH3 and still beginning with
N2 = 1 M, H2 = 1 M, and NH3 = 0 M
time
Question 5: what might a rate vs time graph
look like for the above reaction?
rate
For N2 + 3H2
2NH3 and still beginning with
N2 = 1 M, H2 = 1 M, and NH3 = 0 M
time
rate
For N2 + 3H2
2NH3 and still beginning with
N2 = 1 M, H2 = 1 M, and NH3 = 0 M
time
Question 6: at what point
has equilibrium been established?
rate
concentration
N2
H2
NH3
time
time
Question 7: describe how the two graphs are related.
rate
concentration
N2
H2
NH3
time
time
Question 8: do either of the two graphs
indicate if Keq >1 or Keq <1?
Equilibrium: the extent of a reaction
In stoichiometry we talk about theoretical yields,
and the many reasons actual yields may be
lower.
Another critical reason actual yields may be
lower is the reversibility of chemical reactions:
some reactions may produce only 70% of the
product you may calculate they ought to
produce.
Equilibrium looks at the extent of a chemical
reaction.
58
The Concept of Equilibrium
 Consider colorless frozen N2O4. At room temperature, it
decomposes to brown NO2:
N2O4(g)  2NO2(g).
 At some time, the color stops changing and we have a mixture of
N2O4 and NO2.
 Chemical equilibrium is the point at which the rate of the forward
reaction is equal to the rate of the reverse reaction.
At that point, the concentrations of all species are constant.
 Using the collision model:
 as the amount of NO2 builds up, there is a chance that two NO2
molecules will collide to form N2O4.
 At the beginning of the reaction, there is no NO2 so the reverse
reaction (2NO2(g)  N2O4(g)) does not occur.
59
 As the substance warms it begins to decompose:
N2O4(g)  2NO2(g)
 When enough NO2 is formed, it can react to form N2O4:
2NO2(g)  N2O4(g).
 At equilibrium, as much N2O4 reacts to form NO2 as NO2
reacts to re-form N2O4
 The double arrow implies the process is dynamic.
N2O4(g)A
B 2NO2(g)
60
As the reaction progresses
– [A] decreases to a constant,
– [B] increases from zero to a constant.
– When [A] and [B] are constant,
equilibrium is achieved.
AA
B
B
61
• No matter the starting composition of reactants
and products, the same ratio of concentrations
is achieved at equilibrium.
• For a general reaction
aA + bB(g)
pP + qQ
the equilibrium constant expression is
Kc 
p
q
P Q
a
b
A B
where Kc is the equilibrium constant.
62
Kc is based on the molarities of reactants and
products at equilibrium.
We generally omit the units of the equilibrium
constant.
Note that the equilibrium constant expression
has products over reactants.
63
Write the equilibrium expression for
the following reaction:
N2(g) + 3H2(g)
2NH3(g)
64
The Equilibrium Constant
in Terms of Pressure
• If KP is the equilibrium constant for reactions
involving gases, we can write:
 
p
q
 PP  PQ
KP 
a
b
 PA   PB 
• KP is based on partial pressures measured in
atmospheres.
65
The Magnitude
of Equilibrium Constants
 Therefore, the larger K the more products are
present at equilibrium.
 Conversely, the smaller K the more reactants
are present at equilibrium.
 If K >> 1, then products dominate at
equilibrium and equilibrium lies to the right.
 If K << 1, then reactants dominate at
equilibrium and the equilibrium lies to the left.
66
An equilibrium can be approached from
any direction.
Example:
N2O4(g)
2NO2(g)

NO 2 2
Kc 
 0.212
N 2O4 
However,
2NO2(g)
N2O4(g)

N 2O4 
1
Kc 

 4.72
NO 2 2 0.212
The equilibrium constant for a reaction in
one direction is the reciprocal of the
equilibrium constant of the reaction in the
opposite direction.
Heterogeneous Equilibria
• When all reactants and products are in one
phase, the equilibrium is homogeneous.
• If one or more reactants or products are in a
different phase, the equilibrium is
heterogeneous.
• Consider:
CaCO3(s)
CaO(s) + CO2(g)
– experimentally, the amount of CO2 does not
seem to depend on the amounts of CaO and
CaCO3. Why?
Heterogeneous Equilibria
Heterogeneous Equilibria
• Neither density nor molar mass is a variable, the
concentrations of solids and pure liquids are
constant. (You can’t find the concentration of
something that isn’t a solution!)
• We ignore the concentrations of pure liquids and
pure solids in equilibrium constant expressions.
• The amount of CO2 formed will not depend greatly
on the amounts of CaO and CaCO3 present.
CaCO3(s)
CaO(s) + CO2(g)
Kc = [CO2]  Kp = [pCO2]
Applications of Equilibrium Constants
Predicting the Direction of Reaction
We define Q, the reaction quotient, for a reaction at
conditions NOT at equilibrium
aA + bB(g)
as
pP + qQ

P  p Qq
Q
Aa Bb
where [A], [B], [P], and [Q] are molarities at any time.
Q = K only at equilibrium.
72
Predicting the Direction of Reaction
• If Q > K then the reverse reaction must occur to
reach equilibrium (go left)
• If Q < K then the forward reaction must occur to
reach equilibrium (go right)
73
Note the moles into a 10.32 L vessel stuff ... calculate molarity.
Starting concentration of HI: 2.5 mol/10.32 L = 0.242 M
2 HI
Initial:
Change:
Equil:
0.242 M
-2x
0.242-2x
H2 + I2
0
0
+x
+x
x
x
[ H 2 ][I 2 ]
Keq 
[ HI ]2
[ x][x]
x2
3
Keq 


1
.
26
x
10
[0.242 2 x]2 [0.242 2 x]2
What we are asked for here is the equilibrium concentration of H2 ...
... otherwise known as x. So, we need to solve this beast for x.
And yes, it’s a quadratic equation. Doing a bit of rearranging:
x2
3

1
.
26
x
10
[0.242 2 x]2
x 2  1.26x103[0.242 2x]2
 1.26x103[0.0586 0.968x  4x 2 ]
 7.38x105  1.22x103 x  5.04x103 x 2
0.995x 2  1.22x103 x  7.38x105  0
 b  b 2  4ac
x
2a
x = 0.00802 or –0.00925
Since we are using this to
model a real, physical system,
we reject the negative root.
The [H2] at equil. is 0.00802 M.
This type of problem is typically tackled using the “three line”
approach:
2 NO + O2
2 NO2
Initial:
Change:
Equil.:
Approximating
If Keq is really small the reaction will not proceed to the right
very far, meaning the equilibrium concentrations will be
nearly the same as the initial concentrations of your
reactants.
0.20 – x is just about 0.20 is x is really dinky.
If the difference between Keq and initial concentrations is
around 3 orders of magnitude or more, go for it.
Otherwise, you have to use the quadratic.
77
Initial Concentration of I2: 0.50 mol/2.5L = 0.20 M
I2
Initial
change
equil:
2I
0.20
-x
0.20-x
0
+2x
2x
[ I ]2
Keq 
 2.94x1010
[I2 ]
[ 2 x ]2

 2.94x1010
[0.20  x]
More than 3
orders of mag.
between these
numbers. The
simplification will
work here.
With an equilibrium constant that small, whatever x is, it’s near dink, and 0.20
minus dink is 0.20 (like a million dollars minus a nickel is still a million
dollars).
0.20 – x is the same as 0.20
[2 x]2
 2.94x1010
0.20
x = 3.83 x 10-6 M
Initial Concentration of I2: 0.50 mol/2.5L = 0.20 M
I2
Initial:
Change:
equil:
0.20
-x
0.20-x
2I
0
+2x
2x
[ I ]2
Keq 
 0.209
[I 2 ]
[2 x]2

[0.20  x]
These are too close to
each other ...
0.20-x will not be
trivially close to 0.20
 0.209 here.
Looks like this one has to proceed through the quadratic ...
Le Châtelier’s Principle
Le Chatelier’s Principle: if you disturb an equilibrium,
it will shift to undo the disturbance.
Remember, in a system at equilibrium, come what
may, the concentrations will always arrange
themselves to multiply and divide in the Keq
equation to give the same number (at constant
temperature).
Change in Reactant or Product Concentrations
• Adding a reactant or product shifts the equilibrium
away from the increase.
• Removing a reactant or product shifts the equilibrium
towards the decrease.
• To optimize the amount of product at equilibrium, we
need to flood the reaction vessel with reactant and
continuously remove product (Le Châtelier).
• We illustrate the concept with the industrial
preparation of ammonia
N2(g) + 3H2(g)
2NH3(g)
• Consider the Haber process
N2(g) + 3H2(g)
2NH3(g)
• If H2 is added while the system is at equilibrium, the
system must respond to counteract the added H2 (by
Le Châtelier).
• That is, the system must consume the H2 and produce
products until a new equilibrium is established.
• Therefore, [H2] and [N2] will decrease and [NH3]
increases.
82
Change in Reactant or Product Concentrations
• The unreacted nitrogen and hydrogen are recycled
with the new N2 and H2 feed gas.
• The equilibrium amount of ammonia is optimized
because the product (NH3) is continually removed
and the reactants (N2 and H2) are continually being
added.
Effects of Volume and Pressure
• As volume is decreased pressure increases.
• Le Châtelier’s Principle: if pressure is increased the
system will shift to counteract the increase.a
83
• Consider the production of ammonia
N2(g) + 3H2(g)
2NH3(g)
• As the pressure increases, the amount of ammonia
present at equilibrium increases.
• As the temperature decreases, the amount of
ammonia at equilibrium increases.
• Le Châtelier’s Principle: if a system at equilibrium is
disturbed, the system will move in such a way as to
counteract the disturbance.
Change in Reactant or Product Concentrations
85
86
Effects of Volume and Pressure
• The system shifts to remove gases and
decrease pressure.
• An increase in pressure favors the direction
that has fewer moles of gas.
• In a reaction with the same number of
product and reactant moles of gas, pressure
has no effect.
2NO2(g)
• Consider N2O4(g)
87
N2O4(g)
2NO2(g)
Effects of Volume and Pressure
• An increase in pressure (by decreasing the
volume) favors the formation of colorless N2O4.
• The instant the pressure increases, the system
is not at equilibrium and the concentration of
both gases has increased.
• The system moves to reduce the number moles
of gas (i.e. the reverse reaction is favored).
• A new equilibrium is established in which the
mixture is lighter because colorless N2O4 is
favored.
88
Effect of Temperature Changes
• The equilibrium constant is temperature
dependent.
• For an endothermic reaction, H > 0 and heat
can be considered as a reactant.
• For an exothermic reaction, H < 0 and heat can
be considered as a product.
• Adding heat (i.e. heating the vessel) favors away
from the increase:
– if H > 0, adding heat favors the forward reaction,
– if H < 0, adding heat favors the reverse reaction.
Effect of Temperature Changes
• Removing heat (i.e. cooling the vessel), favors
towards the decrease:
– if H > 0, cooling favors the reverse reaction,
– if H < 0, cooling favors the forward reaction.
• Consider
Cr(H2O)6(aq) + 4Cl-(aq)
CoCl42-(aq) + 6H2O(l)
for which H > 0.
– Co(H2O)62+ is pale pink and CoCl42- is blue.
90
Penggabungan Rumus Tetapan Kesetimbangan
Jika diketahui:
N2(g) + O2(g)  2NO(g) Kc = 4,1 x 10-31
N2(g) + ½ O2(g)  N2O(g) Kc = 2,4 x 10-18
Bagaimana Kc reaksi:
N2O(g) + ½ O2(g)  2NO(g)
Kc = ?
Kita dapat menggabungkan persamaan diatas
N2(g) + O2(g)  2NO(g)
Kc = 4,1 x 10-31
N2O(g)  N2(g) + ½ O2(g)
Kc = 1/(2,4 x 10-18) = 4,2 x 1017
N2O(g) + ½ O2(g)  2NO(g) Kc = ?
[ NO ]2 [ N 2 ][O2 ]1 / 2
[ NO]2
x

 K c (net)
1/ 2
[ N 2 ][O2 ]
[ N 2 O]
[ N 2 O][O2 ]
K c (bersih)  K c (1) x K c (2)  4,1 x 1031 x 4,2 x 1017  1,7 x 1013
Tetapan kesetimbangan untuk reaksi bersih adalah hasil
kali tetapan kesetimbangan untuk reaksi-reaksi terpisah
yang digabungkan
Soal Latihan
1.Untuk reaksi NH3 ↔ ½ N2 + 3/2 H2
Kc = 5,2 x 10-5
pada 298 K. Berapakah nilai Kc pada 298 K untuk reaksi:
N2 + 3H2 ↔ 2NH3
2. Senyawa ClF3 disiapkan melalui 2 tahap reaksi fluorinasi
gas klor sebagai berikut
(i) Cl2(g) + F2(g)  ClF(g)
(ii) ClF(g) + F2(g)  ClF3(g)
– Seimbangkan masing-masing reaksi diatas dan tuliskan
reaksi overallnya!
– Buktikan bahwa Kc overall sama dengan hasil kali Kc
masing-masing tahap reaksi ?
Hubungan Tetapan Kesetimbangan Kc dan Kp
• Tetapan kesetimbangan dalam sistem gas dapat
dinyatakan berdasarkan tekanan parsial gas, bukan
konsentrasi molarnya
• Tetapan kesetimbangan yang ditulis dengan cara ini
dinamakan tetapan kesetimbangan tekanan parsial
dilambangkan Kp.
Misalkan suatu reaksi
2SO2(g) + O2(g)  2SO3(g) Kc = 2,8 x 102 pd 1000 K
[ SO3 ]2
Kc 
[ SO2 ]2 [O2 ]
Sesuai dengan hukum gas ideal, PV = nRT
[ SO]3 
nSO3
V

PSO3
RT
[ SO2 ] 
nSO2
V

PSO2
[O2 ] 
RT
nO2
V
Dengan mengganti suku-suku yang dilingkari dengan
konsentrasi dalam Kc akan diperoleh rumus;
Kc 
( PSO3 / RT ) 2
2
( PSO2 / RT ) ( PO2 / RT )

( PSO3 ) 2
2
( PSO2 ) ( PO2 )
x RT
Terlihat ada hubungan antara Kc dan Kp yaitu:
K c  K p x RT dan
Kc
Kp 
 K c ( RT ) 1
RT

PO2
RT
Jika penurunan yang sama dilakukan terhadap reaksi
umum:
aA(g) + bB(g) + …  gG(g) + hH(g) + …
Hasilnya menjadi
Kp = Kc (RT)n
Dimana n adalah selisih koefisien stoikiometri dari
gas hasil reaksi dan gas pereaksi yaitu n = (g+h+…)
– (a+b+…).
Dalam persamaan reaksi pembentukan gas SO3
diatas kita lihat bahwa n = -1
Soal Latihan
Hitunglah nilai Kp reaksi kesetimbangan berikut:
• PCl3(g) + Cl2(g)  PCl5(g) Kc = 1,67 (at 500 K)
• N2O4(g)  2NO2(g); Kc = 6,1 x 10-3 (298 K)
• N2(g) + 3H2(g)  2NH3(g) Kc = 2,4 x 10-3 (at 1000 K).
Kesetimbangan yang melibatkan cairan dan
padatan murni (Reaksi Heterogen)
• Persamaan tetapan kesetimbangan hanya
mengandung suku-suku yang konsentrasi atau
tekanan parsialnya berubah selama reaksi
berlangsung
• Atas dasar ini walaupun ikut bereaksi tapi karena
tidak berubah, maka padatan murni dan cairan
murni tidak diperhitungkan dalam persamaan
tetapan kesetimbangan.
C(s) + H2O(g)  CO(g) + H2(g)
CaCO3(s)  CaO(s) + CO2(g)
[CO][H 2 ]
Kc 
[ H 2 O]
Kc = [CO2(g)]
atau jika dituliskan dalam bentuk tekanan parsial
menjadi
Kp = PCO2
Kp = Kc(RT)
Latihan:
Calculate Kc for the following reaction
CaCO3(s)  CaO(s) + CO2(g)
Kp = 2,1 x 10-4
(at 1000 K)
Arti Nilai Tetapan Kesetimbangan
Soal Latihan
1)
Reaksi: CO(g) + H2O(g) ↔ CO2(g) + H2(g)
pada suhu 1100 K nilai Kc = 1,00. Sejumlah zat berikut
dicampur pada suhu tersebut dan dibiarkan bereaksi:
1,00 mol CO, 1,00 mol H2O, 2,00 mol CO2 dan 2,00
mol H2. Kearah mana reaksi akan berjalan dan
bagaimana komposisi akhirnya?
2)
Klorometana terbentuk melalui reaksi:
CH4(g) + Cl2(g)  CH3Cl(g) + HCl(g)
pada 1500 K, konstanta kesetimbangan Kp = 1,6 x
104. Didalam campuran reaksi terdapat P (CH4) = 0,13
atm, P(Cl2) = 0,035 atm, P(CH3Cl) = 0,24 atm dan
P(HCl) = 0,47 atm. Apakah reaksi diatas menuju
kearah pembentukan CH3Cl atau pembentukan CH4.
Study Check
• In a study of hydrogen halide decomposition; a researcher fills an
evacuated 2,00 L flask with 0,2 mol HI gas and allows the reaction
to proceed at 453oC.
2HI(g)  H2(g) + I2(g)
At equilibrium [HI] = 0.078 M, calculate Kc
• In study of the conversion of methane to other fuels a chemical
engineer mixes gaseous CH4 and H2O in a 0,32 L flask at 1200 K. at
equilibrium, the flask contains 0,26 mol CO; 0,091 mol H2 and 0,041
mol CH4. What is [H2O] at equilibrium? Kc = 0,26 for the equation
CH4(g) + H2O  CO(g) + 3H2(g)
• The decomposition of HI at low temperature was studied by injecting
2,50 mol HI into a 10,32-L vessel at 25oC. What is [H2] at equilibrium
for the reaction 2HI(g)  H2(g) + I2(g); Kc = 1.26 x 10-3?
Prinsip Le Chatelier
• Usaha untuk mengubah suhu, tekanan atau
konsentrasi pereaksi dalam suatu sistem dalam
keadaan setimbang merangsang terjadinya reaksi
yang mengembalikan kesetimbangan pada sistem
tersebut
Pengaruh perubahan Jumlah
spesies yang bereaksi
[ SO3 ] 2
Q
 Kc
2
[ SO2 ] [O2 ]
[ SO3 ] 2
Q
 Kc
2
[ SO2 ] [O2 ]
Kesetimbangan awal
Gangguan
Kesetimbangan akhir
Pengaruh Perubahan Tekanan
Jika tekanan pada
campuran
kesetimbangan yang
melibatkan gas
ditingkatkan reaksi
bersih akan
berlangsung kearah
yang mempunyai
jumlah mol gas lebih
kecil begitupun
sebaliknya
Pengaruh Gas Lembam (inert)
• Pengaruh tidaknya gas lembam tergantung pada
cara melibatkan gas tersebut
• Jika sejumlah gas helium ditambahkan pada
keadaan volume tetap, tekanan akan meningkat,
sehingga tekanan gas total akan meningkat. Tetapi
tekanan parsial gas-gas dalam kesetimbangan tetap
• Jika gas ditambahkan pada tekanan tetap, maka
volume akan bertambah. Pengaruhnya akan sama
dengan peningkatan volume akibat penambahan
tekanan eksternal.
• Gas lembam mempengaruhi keadaan
kesetimbangan hanya jika gas tersebut
mengakibatkan perubahan konsentrasi (atau tekanan
parsial) dari pereaksi-pereaksinya
Pengaruh Suhu
• Penambahan kalor akan menguntungkan reaksi
serap-panas (endoterm)
• Pengurangan kalor akan menguntungkan reaksi
lepas-panas (eksoterm)
• Peningkatan suhu suatu campuran kesetimbangan
menyebabkan pergeseran kearah reaksi endoterm.
Penurunan suhu menyebabkan pergeseran kearah
reaksi eksoterm
Pengaruh Suhu pada Kesetimbangan
• Umumnya tetapan kesetimbangan suatu reaksi
tergantung pada suhu
• Nilai Kp untuk reaksi oksidasi belerang dioksida
diperlihatkan pada tabel berikut
Hubungan pada tabel tersebut dapat dituliskan dengan:
 H o 1
log K 
 tetapan
2,303R T
Persamaan garis lurus
y =
m .x
+b
Dan jika ada dua keadaan yang berbeda kita dapat
menghubungkan dengan modifikasi sederhana hingga
diperoleh:
K2
H o  T2  T1 


log

K1 2,303R  T2T1 
• K2 dan K1 adalah tetapan kesetimbangan pada suhu
kelvin T2 dan T1. ∆Ho adalah entalpi (kalor) molar
standar dari reaksi. Nilai positif dan negatif untuk
parameter ini dimungkinkan dan diperlukan asumsi
bahwa ∆Ho tidak tergantung pada suhu
• Menurut prinsip Le Chatelier, jika ∆Ho > 0 (endoterm)
reaksi kedepan terjadi jika suhu ditingkatkan,
menyiratkan bahwa nilai K meningkat dengan suhu.
Jika ∆Ho < 0 (eksoterm) reaksi kebalikan terjadi jika
suhu ditingkatkan dan nilai K menurun dengan suhu
• Persamaan diatas menghasilkan nilai kuantitatif yang
sesuai dengan pengamatan kualitatif dari prinsip Le
Chatelier.
Soal Latihan
• Untuk reaksi N2O4(g)  2NO2(g), ∆Ho = +61,5 kJ/mol dan
Kp = 0,113 pada 298K
Berapa nilai Kp pada 0oC? (1,2x10-2)
Pada suhu berapa nilai Kp = 1,00 (326 K)
Pengaruh Katalis pada Kesetimbangan
• Katalis dalam reaksi reversibel dapat mempercepat
reaksi baik kekanan atau kekiri. Keadaan
kesetimbangan tercapai lebih cepat tetapi tidak
mengubah konstanta kesetimbangan dari spesiesspesies yang bereaksi.
• Peranan katalis adalah mengubah mekanisme reaksi
agar tercapai energi aktivasi yang lebih rendah.
• Keadaan kesetimbangan tidak bergantung pada
mekanisme reaksi
• Sehingga tetapan kesetimbangan yang diturunkan
secara kinetik tidak dipengaruhi oleh mekanisme
yang dipilih.
Terima Kasih
Selamat Belajar