CHEMICAL EQUILIBRIUM
Many chemical reactions run to completion
The forward reaction is the only reaction that can
take place when a product escapes from the
reaction vessel
However, if the products stay in the reaction
vessel, the reverse reaction may also take place
1A-1 (of 14)
H2 (g) + Cl2 (g)
2HCl (g)
CHEMICAL EQUILBRIUM – When forward and reverse
reactions are proceeding at equal rates in a system
1A-2 (of 14)
X
X
X
X
X
X
X
X
X
H2
X
X
X
X
X
X
X
X
X
Cl2
X
X
X
X
X
X
X
X
X
X
X
X
HCl
EQUILBRIUM CONSTANT (Keq) – The ratio of product of the equilibrium
product concentrations to the product of the equilibrium reactant
concentrations
H2 (g) + Cl2 (g) ⇆ 2HCl (g)
Keq =
[HCl]2
___________
[H2] [Cl2]
=
[12]2
_______
=
16
[3] [3]
EQUILBRIUM CONSTANT EXPRESSION
[ ] = concentrations, in mol/L
1A-3 (of 14)
aA + bB ⇆ cC + dD
Keq = [C]c[D]d
___________
[A]a[B]b
Write the equilibrium constant expression for
N2 (g) + 3H2 (g) ⇆ 2NH3 (g)
Keq =
[NH3]2
___________
[N2] [H2]3
1A-4 (of 14)
aA + bB ⇆ cC + dD
Keq = [C]c[D]d
___________
[A]a[B]b
If Keq > 1
There will be more products than reactants in the
container when the reaction reaches equilibrium
If Keq < 1
There will be more reactants than products in the
container when the reaction reaches equilibrium
1A-5 (of 14)
A
B
A
B
A
C
A
C
D
B
D B
A
A
B B
A
C
B
D
A
B
A
C
B
D
A
C
B
D
CALCULATING AN EQUILIBRIUM CONSTANT FROM EQUILIBRIUM
CONCENTRATIONS
2H2 (g) + O2 (g) ⇆ 2H2O (g)
Keq = 9.0 M-1
A 2.0 L flask contains 2.0 moles hydrogen, 2.0 moles oxygen, and 6.0 moles water
vapor at equilibrium. Find Keq for the reaction.
[H2]eq
= 2.0 mol / 2.0 L = 1.0 M H2
[O2]eq
= 2.0 mol / 2.0 L = 1.0 M O2
Keq =
1A-6 (of 14)
___________
[H2]2[O2]
[H2O]eq = 6.0 mol / 2.0 L = 3.0 M H2O
Because the Keq > 1 there will be more
products than reactants in the container
when the reaction reaches equilibrium
[H2O]2
=
(3.0 M)2
__________________
(1.0 M)2(1.0 M)
= 9.0 M-1
2H2 (g) + O2 (g) ⇆ 2H2O (g)
For nonequilibrium conditions
Q =
[H2O]2
___________
[H2]2[O2]
where Q = REACTION QUOTIENT
1A-7 (of 14)
Keq = 9.0 M-1
Keq = 9.0 M-1
2H2 (g) + O2 (g) ⇆ 2H2O (g)
Describe the system in which [H2] = [O2] = [H2O] = 1.0 M
Q =
[H2O]2
___________
[H2]2[O2]
=
(1.0 M)2
___________________
=
1.0 M-1
(1.0 M)2(1.0 M)
if Q < Keq, the forward reaction is spontaneous
1A-8 (of 14)
Keq = 9.0 M-1
2H2 (g) + O2 (g) ⇆ 2H2O (g)
Describe the system in which [H2] = 1.0 M, [O2] = 2.0 M, and [H2O] = 6.0 M
Q =
[H2O]2
___________
[H2]2[O2]
=
(6.0 M)2
___________________
=
18 M-1
(1.0 M)2(2.0 M)
if Q > Keq, the reverse reaction is spontaneous
1A-9 (of 14)
RELATED EQUILIBRIUM CONSTANTS
2N2 (g) + O2 (g) ⇆ 2N2O (g)
Keq =
[N2O]2
___________
= 100. M-1
[N2]2[O2]
Find the equilibrium constant for:
Keq΄ =
[N2]2[O2]
____________
[N2O]2
=
1
[N2O]2
___________
2N2O (g) ⇆ 2N2 (g) + O2 (g)
=
1
Keq
=
1
=
0.0100 M
100. M-1
[N2]2[O2]
When a reaction is reversed, its equilibrium constant is the reciprocal of the
forward reaction’s equilibrium constant
1A-10 (of 14)
RELATED EQUILIBRIUM CONSTANTS
2N2 (g) + O2 (g) ⇆ 2N2O (g)
Keq =
[N2O]2
___________
= 100. M-1
[N2]2[O2]
Find the equilibrium constant for:
Keq΄ =
[N2O]4
____________
[N2]4[O2]2
=
[N2O]2
___________
4N2 (g) + 2O2 (g) ⇆ 4N2O (g)
2
= Keq2
=
(100. M-1)2
_
= 10,000 M-2
[N2]2[O2]
When a reaction is multiplied by an integer, its equilibrium constant is raised to
that integer as a power
1A-11 (of 14)
RELATED EQUILIBRIUM CONSTANTS
2N2 (g) + O2 (g) ⇆ 2N2O (g)
Keq =
[N2O]2
___________
= 100. M-1
[N2]2[O2]
Find the equilibrium constant for:
Keq΄ =
[N2O]
___________
[N2][O2]½
=
[N2O]2
___________
N2 (g) + ½O2 (g) ⇆ N2O (g)
½
= Keq½
=
(100. M-1)½
= 10.0 M-½
[N2]2[O2]
When a reaction is multiplied by a fraction, its equilibrium constant is raised to
that fraction as a power
1A-12 (of 14)
EQUILIBRIA INVOLVING SOLIDS OR LIQUIDS
C (s) + CO2 (g) ⇆ 2CO (g)
Gases (and dissolved solutes) have variable concentrations
Variables appear in equilibrium constant expressions
1A-13 (of 14)
EQUILIBRIA INVOLVING SOLIDS OR LIQUIDS
C (s) + CO2 (g) ⇆ 2CO (g)
Solids (and liquids) have constant concentrations
Constants do not appear in equilibrium constant expressions
Keq =
[CO]2
_________
[CO2]
1A-14 (of 14)
TYPES OF EQUILIBRIUM CONSTANTS
There are 2 types of equilibrium constants:
(1) Kc
(what we have been writing)
(2) Kp
concentration units of MOLARITY are used in the expression
pressure units of ATMOSPHERES are used in the expression
H2 (g) + Cl2 (g) ⇆ 2HCl (g)
Kc =
[HCl]2
____________
[H2] [Cl2]
1B-1 (of 16)
Kp =
pHCl2
__________
pH2 pCl2
CALCULATING AN EQUILIBRIUM CONSTANT FROM EQUILIBRIUM PRESSURES
H2 (g) + I2 (g) ⇆ 2HI (g)
A mixture at equilibrium is 0.40 atm H2, 0.30 atm I2, and 0.20 atm HI. Calculate the
Kp for the above reaction.
Kp =
pHI2
________
pH2pI2
1B-2 (of 16)
=
(0.20 atm)2
_________________________
(0.40 atm)(0.30 atm)
=
0.33
H2 (g) + I2 (g) ⇆ 2HI (g)
If a new flask at the same T is charged with 0.20 atm H2, 0.30 atm I2, and 0.40 atm
HI, is the forward or reverse reaction spontaneous?
Q =
pHI2
________
pH2pI2
=
(0.40 atm)2
_________________________
(0.20 atm)(0.30 atm)
Q > Kp
 the reverse reaction is spontaneous
1B-3 (of 16)
=
2.7
Write the Kp and the Kc expression for
2SO2 (g) + O2 (g) ⇆ 2SO3 (g)
Kp =
pSO32
___________
pSO22 pO2
1B-4 (of 16)
Kc =
[SO3]2
______________
[SO2]2 [O2]
Write the Kp and the Kc expression for
2SO2 (g) + O2 (g) ⇆ 2SO3 (g)
Numerical values for Kc and Kp are usually different
Kp =
pSO32
___________
Kp =
p
V
p
Kp =
= nRT
______
_________________________
([SO2]RT)2 ([O2]RT)
pSO22 pO2
pV = nRT
([SO3]RT)2
Kp =
[SO3]2
(RT)2
[SO2]2 [O2]
(RT)2(RT)
______________ x _____________
Kc
x
RT
= MRT
Kp =
1B-5 (of 16)
1
____
Kc
x
(RT)-1
Kp
= Kc(RT)Δn
Δn = moles of products in Keq – moles of reactants in Keq
2SO2 (g) + O2 (g) ⇆ 2SO3 (g)
The Kc for the above reaction is 100. M-1 at 27ºC. Calculate Kp.
Δn = 2 – 3 = -1
Kp
= Kc(RT)Δn
= (100. L/mol) [(0.08206 Latm/molK)(300.2 K)]-1
= 4.06 atm-1
1B-6 (of 16)
Write the Kc and the Kp expression for
NH4HS (s) ⇆ NH3 (g) + H2S (g)
Kc = [NH3][H2S]
Kp = pNH3pH2S
The Kc for the above reaction is 0.10 M2 at 2ºC. Calculate Kp.
Δn = 2 – 0 = 2
Kp
= Kc(RT)Δn
= (0.10 mol2/L2) [(0.08206 Latm/molK)(275.2 K)]2
= 51 atm2
1B-7 (of 16)
4HCl (g) + O2 (g) ⇆ 2Cl2 (g) + 2H2O (l)
The Kp for the above reaction is 50. atm-3 at 500. K. Calculate Kc.
Δn = 2 – 5 = -3
Kp
= Kc(RT)Δn
Kp
= Kc
_________
(RT)Δn
(50. atm-3) [(0.08206 Latm/molK)(500. K)]-(-3)
=
3.5 x 106 L3/mol3
=
3.5 x 106 M-3
=
1B-8 (of 16)
= Kp(RT)-Δn
2HCl (g) ⇆ H2 (g) + Cl2 (g)
The Kc for the above reaction is 25 at 20.ºC. Calculate Kp.
Δn = 2 – 2 = 0
Kp = Kc(RT)Δn
= (25) [(0.08206 Latm/molK)(293.2 K)]0
= 25
1B-9 (of 16)
CALCULATING EQUILIBRIUM CONCENTRATIONS AND PRESSURES
1B-10 (of 16)
CALCULATING EQUILIBRIUM CONCENTRATIONS AND PRESSURES
H2 (g) + I2 (g) ⇆ 2HI (g)
Kp = 36.0
A tank is charged with 0.200 atm H2 and 0.200 atm I2. Find the equilibrium
pressure of HI.
H2 (g)
Initial atm’s
Change in atm’s
Final atm’s
0.200
- 0.200
0
+
I2 (g)
0.200
- 0.200
0
But the reaction does not go to completion!
1B-11 (of 16)
⇆
2HI (g)
0
+ 0.400
0.400
Ratios from the
balanced equation
CALCULATING EQUILIBRIUM CONCENTRATIONS AND PRESSURES
H2 (g) + I2 (g) ⇆ 2HI (g)
Kp = 36.0
A tank is charged with 0.200 atm H2 and 0.200 atm I2. Find the equilibrium
pressure of HI.
H2 (g)
Initial atm’s
Change in atm’s
Equilibrium atm’s
Kp =
pHI2
________
pH2pI2
1B-12 (of 16)
0.200
-x
0.200 - x
+
I2 (g)
0.200
-x
0.200 - x
36.0 =
⇆
2HI (g)
0
+ 2x
2x
(2x)2
______________
(0.200 – x)2
Ratios from the
balanced equation
CALCULATING EQUILIBRIUM CONCENTRATIONS AND PRESSURES
H2 (g) + I2 (g) ⇆ 2HI (g)
Kp = 36.0
A tank is charged with 0.200 atm H2 and 0.200 atm I2. Find the equilibrium
pressure of HI.
36.0 =
(2x)2
_______________
6.00
=
(0.200 – x)2
0.200 – x
2x
1.20 – 6.00x =
2x
0.150 =
pHI (eq) = 2x
= 2(0.150 atm)
= 0.300 atm
(0.200 – x) 6.00 =
1.20 =
1B-13 (of 16)
2x
_____________
8.00x
x
PF5 (g) ⇆ PF3 (g) + F2 (g)
Kp = 15.0
If a tank is initially charged with only 0.200 atm PF5, find the equilibrium pressure
of PF5.
PF5 (g) ⇆ PF3 (g) + F2 (g)
Initial atm’s
Change in atm’s
Equilibrium atm’s
Kp = pPF3pF2
_________
pPF5
1B-14 (of 16)
0.200
-x
0.200 - x
0
+x
x
15.0 =
0
+x
x
x2
____________
0.200 – x
PF5 (g) ⇆ PF3 (g) + F2 (g)
Kp = 15.0
If a tank is initially charged with only 0.200 atm PF5, find the equilibrium pressure
of PF5.
15.0 =
pPF5 (eq) = 0.200 – x
x2
____________
= 0.200 atm – 0.197 atm
0.200 – x
(0.200 – x) 15.0 =
x2
3.00 – 15.0x =
x2
= 0.003 atm
0 =
x2 + 15.0x – 3.00
x =
-15.0 ±
15.02 - 4(1)(-3.00)
________________________________________
2(1)
x =
1B-15 (of 16)
0.197
PF5 (g) ⇆ PF3 (g) + F2 (g)
Kp = 15.0
If a tank is initially charged with only 0.200 atm PF5, find the equilibrium pressure
of PF5.
What is the percent dissociation of PF5?
% dissociation =
amount dissociated
__________________________
x 100
=
original amount
x
__________
0.200 M
=
0.197 M x 100
___________
0.200 M
=
1B-16 (of 16)
x 100
98.5%
H2 (g) + Cl2 (g) ⇆ 2HCl (g)
Kc = 75
If 0.10 moles of HCl are placed in a 2.0 L flask, what will be the concentration of
the HCl at equilibrium?
[HCl]in = 0.10 mol / 2.0 L = 0.050 M HCl
H2 (g)
Initial M’s
Change in M’s
Equilibrium M’s
Kc =
[HCl]2
___________
[H2] [Cl2]
1B-17 (of 16)
0
+x
x
+
Cl2 (g)
0
+x
x
⇆
2HCl (g)
0.050
- 2x
0.050 - 2x
75 = (0.050 – 2x)2
________________
x2
H2 (g) + Cl2 (g) ⇆ 2HCl (g)
Kc = 75
If 0.10 moles of HCl are placed in a 2.0 L flask, what will be the concentration of
the HCl at equilibrium?
75 = (0.050 – 2x)2
________________
x2
8.66 = 0.050 – 2x
_____________
x
8.66x = 0.050 – 2x
10.66x = 0.050
x = 0.00469
1B-18 (of 16)
[HCl]eq =
0.050 – 2x
= 0.050 M – 2(0.00469 M)
= 0.041 M
H2 (g) + Cl2 (g) ⇆ 2HCl (g)
Kc = 75
If 0.10 moles of HCl are placed in a 2.0 L flask, what will be the concentration of
the HCl at equilibrium?
H2 (g)
Initial M’s
Change in M’s
Equilibrium M’s
0
+x
x
+
Cl2 (g)
0
+x
x
⇆
2HCl (g)
0.050
- 2x
0.050 - 2x
What is the percent dissociation of HCl?
% dissociation =
2x
__________
0.050 M
1B-19 (of 16)
x 100
=
2(0.0047 M) x 100 = 19%
_______________
0.050 M
2HF (g) ⇆ H2 (g) + F2 (g)
Kc = 0.100 at 10.ºC
A 3.00 L flask is charged with 0.150 moles each of HF, H2, and F2. Find the
equilibrium concentration of HF.
[each]in = 0.150 mol / 3.00 L = 0.0500 M
⇆
2HF (g)
Initial M’s
Change in M’s
Equilibrium M’s
Q = [H2][F2]
__________
[HF]2
1C-1 (of 14)
0.0500
+ 2x
0.0500 + 2x
=
(0.0500)2
___________
(0.0500)2
H2 (g)
0.0500
-x
0.0500 - x
= 1.00
+
F2 (g)
0.0500
-x
0.0500 - x
Q > Kc  the reverse reaction
is spontaneous
2HF (g) ⇆ H2 (g) + F2 (g)
Kc = 0.100 at 10.ºC
A 3.00 L flask is charged with 0.150 moles each of HF, H2, and F2. Find the
equilibrium concentration of HF.
Kc = [H2][F2]
__________
[HF]2
0.100 =
(0.0500 – x)2
_________________
(0.0500 + 2x)2
x = 0.02094
1C-2 (of 14)
[HF]eq = 0.0500 + 2x
=
0.0500 M + 2(0.02094 M)
= 0.0919 M
N2 (g) + O2 (g) ⇆ 2NO (g)
At equilibrium a flask has 0.250 atm N2, 0.250 atm O2, and 0.400 atm NO.
If 0.100 atm of NO is added to the flask, find the new equilibrium pressure of NO.
Kp =
pNO2
________
pN2pO2
1C-3 (of 14)
=
(0.400 atm)2
________________
(0.250 atm)2
=
2.560
N2 (g) + O2 (g) ⇆ 2NO (g)
At equilibrium a flask has 0.250 atm N2, 0.250 atm O2, and 0.400 atm NO.
If 0.100 atm of NO is added to the flask, find the new equilibrium pressure of NO.
N2 (g)
Equilibrium
Initial atm’s (w/ added NO)
Change in atm’s
Equilibrium atm’s
0.250
0.250
+x
0.250 + x
+
O2 (g)
0.250
0.250
+x
0.250 + x
[NO]w/added = 0.400 atm + 0.100 atm = 0.500 atm NO
1C-4 (of 14)
⇆
2NO (g)
0.400
0.500
- 2x
0.500 - 2x
N2 (g) + O2 (g) ⇆ 2NO (g)
At equilibrium a flask has 0.250 atm N2, 0.250 atm O2, and 0.400 atm NO.
If 0.100 atm of NO is added to the flask, find the new equilibrium pressure of NO.
Kp =
pNO2
_________
pN2pO2
2.560 = (0.500 – 2x)2
_______________
(0.250 + x)2
x = 0.02778 atm
1C-5 (of 14)
pNO (eq) = 0.500 – 2x
= 0.500 atm – 2(0.02778 atm)
= 0.444 atm
Kc = 1.6 x 109 at 250°C
2ClF (g) ⇆ Cl2 (g) + F2 (g)
If a tank is charged with 0.40 M ClF, find the equilibrium concentration of ClF at
250°C.
2ClF (g)
Initial M’s
Change in M’s
Equilibrium M’s
Kc = [Cl2][F2]
__________
[ClF]2
1C-6 (of 14)
0.40
- 2x
0.40 - 2x
⇆
Cl2 (g)
0
+x
x
1.6 x 109 =
+
F2 (g)
0
+x
x
x2
______________
(0.40 – 2x)2
2ClF (g) ⇆ Cl2 (g) + F2 (g)
Kc = 1.6 x 109 at 250°C
If a tank is charged with 0.40 M ClF, find the equilibrium concentration of ClF at
250°C.
1.6 x 109
=
x2
______________
4.0 x 104 =
(0.40 – 2x)2
0.40 - 2x
4.0 x 104 (0.40 – 2x) =
x
1.6 x 104 – 8.0 x 104x =
x
1.6 x 104 =
0.200 =
1C-7 (of 14)
x
___________
8.0001 x 104x
x
Kc = 1.6 x 109 at 250°C
2ClF (g) ⇆ Cl2 (g) + F2 (g)
If a tank is charged with 0.40 M ClF, find the equilibrium concentration of ClF at
250°C.
2ClF (g)
Initial M’s
Change in M’s
Equilibrium M’s
⇆
0.40
- 2x
0.40 - 2x
Cl2 (g)
0
+x
x
+
F2 (g)
0
+x
x
[ClF]eq = 0.40 M – 2x
= 0.40 M – 2(0.200 M) = 0 M ??
1C-8 (of 14)
Kc = 1.6 x 109 at 250°C
2ClF (g) ⇆ Cl2 (g) + F2 (g)
If a tank is charged with 0.40 M ClF, find the equilibrium concentration of ClF at
250°C.
2ClF (g)
Initial M’s
Change in M’s
Equilibrium M’s
0.40
- 2x
0.40 - 2x
⇆
Cl2 (g)
+
0
+x
x
F2 (g)
0
+x
x
A large Kc means mostly products at equilibrium
∴ if this reaction goes in the forward direction, x will be large relative to 0.40
Must make the reaction go in the reverse direction to make x small
∴ first make the reaction goes to completion
1C-9 (of 14)
Kc = 1.6 x 109 at 250°C
2ClF (g) ⇆ Cl2 (g) + F2 (g)
If a tank is charged with 0.40 M ClF, find the equilibrium concentration of ClF at
250°C.
2ClF (g)
Initial M’s
Change in M’s
New Initial M’s
Change in M’s
Equilibrium M’s
1C-10 (of 14)
0.40
- 0.40
0
+ 2x
2x
⇆
Cl2 (g)
0
+ 0.20
0.20
-x
0.20 - x
+
F2 (g)
0
+ 0.20
0.20
-x
0.20 -x
make all ClF react away
2ClF (g) ⇆ Cl2 (g) + F2 (g)
Kc = 1.6 x 109 at 250°C
If a tank is charged with 0.40 M ClF, find the equilibrium concentration of ClF at
250°C.
Kc = [Cl2][F2]
___________
1.6 x 109 =
[ClF]2
= 2(2.50 x 10-6 M)
(0.20 – x)2
= 5.0 x 10-6 M
_____________
(2x)2
x = 2.50 x 10-6
1C-11 (of 14)
[ClF]eq = 2x
Kc = 1.5 x 106 at 150°C
2NO (g) + O2 (g) ⇆ 2NO2 (g)
If a tank is charged with 0.30 M NO and 0.30 M O2, find the equilibrium
concentration of NO at 150°C.
2NO (g)
Initial M’s
Change in M’s
Equilibrium M’s
0.30
- 2x
0.30 - 2x
+
O2 (g)
⇆
0.30
-x
0.30 - x
2NO2 (g)
0
+ 2x
2x
A large Kc means mostly products at equilibrium
Reaction goes in the forward direction so x will be large relative to 0.30
Need x to be small
∴ first make the reaction goes to completion
1C-12 (of 14)
Kc = 1.5 x 106 at 150°C
2NO (g) + O2 (g) ⇆ 2NO2 (g)
If a tank is charged with 0.30 M NO and 0.30 M O2, find the equilibrium
concentration of NO at 150°C.
2NO (g)
Initial M’s
Change in M’s
New Initial M’s
Change in M’s
Equilibrium M’s
1C-13 (of 14)
0.30
- 0.30
0
+ 2x
2x
+
O2 (g)
0.30
- 0.15
0.15
+x
0.15 + x
⇆
2NO2 (g)
0
+ 0.30
0.30
- 2x
0.30 - 2x
make all of one
reactant react away
2NO (g) + O2 (g) ⇆ 2NO2 (g)
Kc = 1.5 x 106 at 150°C
If a tank is charged with 0.30 M NO and 0.30 M O2, find the equilibrium
concentration of NO at 150°C.
Kc =
[NO2]2
____________
[NO]2[O2]
1.5 x 106 =
(0.30 – 2x)2
_________________
(2x)2(0.15 + x)
1.5 x 106 =
(0.30)2
_____________
(2x)2(0.15)
x
=
1C-14 (of 14)
3.2 x 10-4
[NO]eq = 2x = 2(3.16 x 10-4 M)
= 6.3 x 10-4 M
If you know x is very small, you may ignore it
when it is added to or subtracted from an
initial molarity or pressure
CALCULATING EQUILIBRIUM CONSTANTS
2NH3 (g) ⇆ N2 (g) + 3H2 (g)
2.0 moles of NH3 are placed in a 1.0 L container. At equilibrium 1.0 moles of NH3
remain. Find Kc for the above reaction.
[NH3]in = 2.0 mol / 1.0 L = 2.0 M NH3
2NH3 (g)
Initial M’s
Change in M’s
Equilibrium M’s
Kc = [N2][H2]3
___________
[NH3]2
1D-1 (of 17)
2.0
- 2x
2.0 - 2x
Kc =
⇆
N2 (g)
0
+x
x
(x)(3x)3
_____________
(2.0 – 2x)2
[NH3]eq = 1.0 mol / 1.0 L = 1.0 M NH3
+
3H2 (g)
0
+ 3x
3x
equals 2.0 - 2x
CALCULATING EQUILIBRIUM CONSTANTS
2NH3 (g) ⇆ N2 (g) + 3H2 (g)
2.0 moles of NH3 are placed in a 1.0 L container. At equilibrium 1.0 moles of NH3
remain. Find Kc for the above reaction.
[NH3]eq = 1.0 M = 2.0 M – 2x
Kc = [N2][H2]3
___________
[NH3]2
-1.0 M = -2x
0.50 M = x
Kc =
(x)(3x)3
_____________
(2.0 – 2x)2
Kc = (0.50 M)(1.5 M)3 = 1.7 M2
___________________
(1.0 M)2
1D-2 (of 17)
SF4 (g) ⇆ SF2 (g) + F2 (g)
A container has an initial SF4 concentration of 0.308 M. When the container reaches
equilibrium, the SF4 is found to be 35.0% dissociated. Calculate Kc.
SF4 (g)
Initial M’s
Change in M’s
Equilibrium M’s
Kc = [SF2][F2]
___________
[SF4]
⇆
0.308
-x
0.308 - x
Kc =
SF2 (g)
0
+x
x
x2
_____________
(0.308 – x)
+
F2 (g)
0
+x
x
35.0 =
x
_______
0.308
0.1078 = x
1D-3 (of 17)
(100)
SF4 (g) ⇆ SF2 (g) + F2 (g)
A container has an initial SF4 concentration of 0.308 M. When the container reaches
equilibrium, the SF4 is found to be 35.0% dissociated. Calculate Kc.
Kc = [SF2][F2]
___________
[SF4]
Kc =
x2
_____________
(0.308 – x)
1D-4 (of 17)
=
(0.1078 M)2
________________________
(0.308 – 0.1078 M)
=
(0.1078 M)2
_______________
(0.2002 M)
= 0.0580 M
NH4Cl (s) ⇆ NH3 (g) + HCl (g)
1.00 g NH4Cl is placed in a flask and the total pressure at equilibrium is 0.600 atm.
Find Kp for the above reaction.
NH4Cl (s)
Initial atm’s
Change in atm’s
Equilibrium atm’s
⇆
NH3 (g)
HCl (g)
0
0
+x
x
+x
x
Kp = pNH3pHCl
Kp = x2 = (0.300 atm)2 = 0.0900 atm2
1D-5 (of 17)
+
ptotal = pNH3 + pHCl
0.600 atm =
x
0.300 atm =
x
+
x
2N2O (g) ⇆ 2N2 (g) + O2 (g)
A flask is charged with 0.450 atm N2O and when equilibrium is reached the total
pressure is 0.500 atm. Find Kp for the above reaction.
2N2O (g)
Initial atm’s
Equilibrium atm’s
0.450
- 2x
0.450 – 2x
Kp = pN22pO2
(2x)2 x
Change in atm’s
__________
pN2O2
1D-6 (of 17)
=
________________
(0.450 – 2x)2
⇆
2N2 (g)
+
0
+ 2x
2x
O2 (g)
0
+x
x
ptotal
=
pN2O + pN2 + pO2
0.500 atm = (0.450 – 2x) + (2x) + (x)
0.500 atm =
0.450 + x
0.050 atm =
x
2N2O (g) ⇆ 2N2 (g) + O2 (g)
A flask is charged with 0.450 atm N2O and when equilibrium is reached the total
pressure is 0.500 atm. Find Kp for the above reaction.
Kp = pN22pO2
__________
pN2O2
Kp =
(2x)2 x
________________
(0.450 – 2x)2
= 4.1 x 10-3 atm
1D-7 (of 17)
= (2 x 0.050 atm)2(0.050 atm)
___________________________________
(0.450 – [2 x 0.050] atm)2
= (0.10 atm)2(0.050 atm)
_____________________________
(0.350 atm)2
AB
Initial Molecules/L
Equilibrium Molecules/L
Initial Molecules/L
Equilibrium Molecules/L
Keq = [A][B]
_________
[AB]
= (20)2
_______
100
80
0
80
=
⇆
A
0
20
100
20
+
B
0
20
100
20
5
(80)
The same ratio of products to reactants (Keq) is always achieved
However, if starting at equivalent stoichiometric points, the same numerical
values of A, B, and AB will always be achieved
1D-8 (of 17)
AB
Equilibrium Molecules/L
Add AB Molecules/L
Equilibrium Molecules/L
Keq = [A][B] =
(25)2
________
_______
[AB]
(125)
80
130
125
=
⇆
A
20
20
25
+
B
20
20
25
5
The same ratio of products to reactants (Keq) is always achieved
However, if starting at different stoichiometric points, the numerical values of
A, B, and AB can be different
1D-8 (of 17)
LE CHATELIER’S PRINCIPLE – If a system is changed so that it is no longer at
equilibrium, either the forward or reverse reaction will become spontaneous until
the system reaches equilibrium again
If a stress is applied to a system at equilibrium, the equilibrium will shift to relieve
the stress
1D-10 (of 17)
H2 (g) + I2 (g) ⇆ 2HI (g)
H2 is added
I2 is added
H2 is removed
HI is added
He is added
Shift Direction
Right
Right
Left
Left
No Change
[HI]new eq
Increase
Increase
Decrease
Increase
Same
H2 is added
I2 is added
H2 is removed
HI is added
Shift Direction
Right
Right
Left
Left
[I2]new eq
Decrease
Increase
Increase
Increase
1D-10 (of 16)
Only changes in temperature can change the numerical value of the Keq
ENDOTHERMIC – A process that absorbs energy
EXOTHERMIC – A process that releases energy
Energy can be treated as a reactant (for endothermic reactions) or a product
(for exothermic reactions) to predict shifts in equilibrium
1D-12 (of 17)
Endothermic:
energy
Equilibrium Molecules
Equilibrium at Higher T
Keq = [A][B]
_________
[AB]
= (30)2
______
(70)
For endothermic reactions:
T ↑, Keq ↑
T ↓, Keq ↓
1D-13 (of 17)
+
AB
80
70
= 13
⇆
A
20
30
+
B
20
30
H2 (g) + I2 (g) ⇆ 2HI (g) + energy
CHANGE IN ENTHALPY (ΔH) – The energy change during a chemical reaction
ΔH positive – energy absorbed, endothermic
ΔH negative – energy released, exothermic
The ΔH for this reaction is -10.2 kJ/ 2 mol HI
1D-14 (of 17)
H2 (g) + I2 (g) ⇆ 2HI (g) + energy
T increased
T decreased
Shift Direction
Left
Right
For exothermic reactions:
T ↑, Keq ↓
T ↓, Keq ↑
1D-15 (of 17)
[HI]new eq
Decrease
Increase
AB
Equilibrium Molecules/L
Change Volume from 1 L to 0.5 L
Equilibrium Molecules/L
Q = [A][B]
_________
[AB]
=
(40)2
_______
(160)
=
10
80
160
171
⇆
A
+
B
20
40
29
20
40
29
Keq = [A][B]
=
_________
[AB]
(29)2
_______
=
(171)
Decrease in volume: the reaction that produces the lesser number of gas
molecules will be spontaneous
Increase in volume: the reaction that produces the greater number of gas
molecules will be spontaneous
1D-16 (of 17)
5
4RbCl (s) + O2 (g) ⇆ 2Cl2 (g) + 2Rb2O (s)
Volume increased
Volume decreased
1D-17 (of 17)
Shift Direction
Right
Left