Chapter 15 – Chemical Equilibrium
Equilibrium
To be in equilibrium is to be in a state of balance:
- Static Equilibrium (nothing happens; e.g. a tug of war).
- Dynamic Equilibrium (lots of things happen, but the net effect is zero; e.g. cars traveling).
Chemical Equilibrium
It occurs when a reaction and its reverse reaction proceed at the same rate.
15.1 The Concept of Equilibrium
Many reactions “go to completion”, we predict the amount of product formed using stoichiometry.
Many reactions (particularly organic reactions) DO NOT go to completion, but seem to stop before the reaction
is complete. E.g.
𝐶𝑂(𝑔) + 3𝐻2 (𝑔) ⇌ 𝐶𝐻4 (𝑔) + 𝐻2 𝑂(𝑔)
If we mix 1 mol CO + 3 mol H2, we would expect to get 1 mol CH4 and 1 mol of H2O. However, we get:
1.8 mol H2, 0.6 mol CO, 0.4 mol CH4, 0.4 mol H2O (at 1200 K in a 10.0L container), which is a mixture of
reactants and products in a dynamic equilibrium.
Once the two reactions (forward and reverse) get to the same rate, the concentrations do not change (again, this
is a dynamic equilibrium).
𝑁2 𝑂4 (𝑔) ⇌ 2𝑁𝑂2 (𝑔)
15.2 The Equilibrium Constant
15.3 Understanding and Working with Equilibrium Constants
Equilibrium constants depend on the reaction (its stoichiometry, not its mechanism) and T.
𝐶𝑂(𝑔) + 3𝐻2 (𝑔) ⇌ 𝐶𝐻4 (𝑔) + 𝐻2 𝑂(𝑔)
𝐾𝐶 =
𝐾=
[𝐶𝐻4 ][𝐻2 𝑂]
[𝐶𝑂][𝐻2 ]3
;
[𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠]
[𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠]
𝐾𝑃 =
𝑃𝐶𝐻4 𝑃𝐻2 𝑂
𝑃𝐶𝑂 𝑃𝐻32
;
;
𝑚𝑜𝑙
[ ] ⟹ 𝐶𝑜𝑛𝑐. 𝑖𝑛 𝑀 (
)
𝐿
𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑠 𝑏𝑒𝑐𝑜𝑚𝑒 𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑠.
𝑃 ⟹ 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑖𝑛 𝑎𝑡𝑚.
𝐾𝑃 ≠ 𝐾𝐶 𝑖𝑛 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 ; 𝐾𝑃 = 𝐾𝐶 (𝑅𝑇)∆𝑛
∆𝑛 = ∆(#𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑔𝑎𝑠𝑒𝑠) ; 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 − 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠
1
Fossum-Reyes
Taking:
𝐶𝑂(𝑔) + 3𝐻2 (𝑔) ⇌ 𝐶𝐻4 (𝑔) + 𝐻2 𝑂(𝑔)
For the above reaction: ∆𝑛𝑔𝑎𝑠 = 2 − 4 = −2
𝐾𝐶
𝐿𝑎𝑡𝑚
𝐾𝑃 = 𝐾𝐶 (𝑅𝑇)−2 =
; 𝑅 = 0.08206
; 𝑇[𝐾𝑒𝑙𝑣𝑖𝑛]
2
(𝑅𝑇)
𝑚𝑜𝑙𝐾
If: ∆𝑛𝑔𝑎𝑠 = 0 ; 𝐾𝑃 = 𝐾𝐶 .
Equilibrium constants are reported without units since they are related not only to the kinetics, but also to the
thermodynamics of the reaction.
Equilibrium constants derived from thermodynamic measurements have no units, so, it is customary to write all
types of equilibrium constants without units.
Nevertheless, for gases we use P and for solutes we use M units (could be mixed units). Solids, liquids and
solvents do not appear in K expressions (their concentrations are constant).
Changing the amount of a solid or liquid has no effect on the equilibrium, as long as there is some present.
𝐹𝑜𝑟: 𝐻3 𝑂+(𝑎𝑞) + 𝐻𝐶𝑂3−
𝐾𝑒𝑞 =
(𝑎𝑞)
⇌ 2𝐻2 𝑂 (𝑙) + 𝐶𝑂2
(𝑔)
𝑃𝐶𝑂2
[𝐻3 𝑂+ ][𝐻𝐶𝑂3− ]
Q. Write the Keq expression for:
2𝑁𝑂2
(𝑔)
+ 7𝐻2
(𝑔)
⇌ 2𝑁𝐻3
(𝑔)
+ 4𝐻2 𝑂 (𝑙)
Meaning of K
If the system has reached equilibrium and the M’s/P’s are substituted into the K expression, the result is a
constant; conversely, if the values are plugged in and the result equals K, then the system is at equilibrium.
There are infinite possibilities for the equilibrium mixture.
𝑁2 𝑂4 (𝑔) ⇌ 2𝑁𝑂2 (𝑔)
What Does the Value of Keq Imply?
When the value of Keq >> 1, we know that when the reaction reaches equilibrium there will be many more
product molecules present than reactant molecules.
When the value of Keq << 1, we know that when the reaction reaches equilibrium there will be many more
reactant molecules present than product molecules.
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Fossum-Reyes
Q. Write the equilibrium constant expressions, KC, and predict the position of equilibrium for the following:
2 SO2(g) + O2(g) ⇌ 2 SO3(g) K = 8 x 1025
N2(g) + 2 O2(g) ⇌ 2 NO2(g) K = 3 x 10−17
Manipulating Equilibrium Constants
When the reaction is written backward, the equilibrium constant is inverted.
For the reaction aA + bB ⇌ cC + dD
the equilibrium constant expression is:
𝐾𝑓𝑜𝑟𝑤𝑎𝑟𝑑 =
For the reaction cC + dD ⇌ aA + bB
the equilibrium constant expression is:
[𝐶]𝑐 [𝐷]𝑑
[𝐴]𝑎 [𝐵]𝑏
𝐾𝑏𝑎𝑐𝑘𝑤𝑎𝑟𝑑 =
∴ 𝐾𝑏𝑎𝑐𝑘𝑤𝑎𝑟𝑑 =
[𝐴]𝑎 [𝐵]𝑏
[𝐶]𝑐 [𝐷]𝑑
1
𝐾𝑓𝑜𝑟𝑤𝑎𝑟𝑑
When the coefficients of an equation are multiplied by a factor, the equilibrium constant is raised to that factor.
For the reaction aA + bB ⇌ cC
the equilibrium constant expression is:
𝐾𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙
For the reaction 2aA + 2bB ⇌ 2cC
the equilibrium constant expression is:
[𝐶]𝑐
=
[𝐴]𝑎 [𝐵]𝑏
∴ 𝐾𝑛𝑒𝑤 = (𝐾𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 )
𝑛
[𝐶]2𝑐
𝐾𝑛𝑒𝑤 =
[𝐴]2𝑎 [𝐵]2𝑏
2
[𝐶]𝑐
= ( 𝑎 𝑏)
[𝐴] [𝐵]
When you add equations to get a new equation, the equilibrium constant of the new equation is the product of
the equilibrium constants of the old equations
For the reactions (1) aA ⇌ bB and (2) bB ⇌ cC
the equilibrium constant expressions are:
𝐾1 =
[𝐵]𝑏
[𝐴]𝑎
;
𝐾2 =
[𝐶]𝑐
[𝐵]𝑏
∴ 𝐾𝑛𝑒𝑤 = 𝐾1 × 𝐾2
For the reaction aA ⇌ cC
the equilibrium constant expression is:
[𝐶]𝑐
[𝐴]𝑎
𝑏
[𝐵]
[𝐶]𝑐
=
×
[𝐴]𝑎 [𝐵]𝑏
𝐾𝑛𝑒𝑤 =
3
Fossum-Reyes
Q. For N2(g) + 3 H2(g) ⇌ 2 NH3(g), K = 3.7 x 108 at 25 °C; find K for:
NH3(g) ⇌ 0.5N2(g) + 1.5H2(g), at 25 °C.
Q. Using:
𝐻2 + 𝐵𝑟2 ⇌ 2𝐻𝐵𝑟
𝐻2 ⇌ 2𝐻
𝐵𝑟2 ⇌ 2𝐵𝑟
𝐾𝑝 = 7.9 × 1011
𝐾𝑝 = 4.8 × 10−41
𝐾𝑝 = 2.2 × 10−15
Calculate KP for: 𝐻 + 𝐵𝑟 ⇌ 𝐻𝐵𝑟
15.4 Heterogeneous Equilibria
Again: Solids, liquids and solvents do not appear in K expressions (their concentrations do not change);
however, they must be present for equilibrium to occur.
𝐶𝑎𝐶𝑂3 (𝑠) ⇌ 𝐶𝑎𝑂(𝑠) + 𝐶𝑂2 ; 𝐾𝐶 = [𝐶𝑂2 ] 𝑎𝑛𝑑 𝐾𝑃 = 𝑃𝐶𝑂2
𝐻2 𝑂 (𝑙) + 𝐶𝑂32− (𝑎𝑞) ⇌ 𝑂𝐻 − (𝑎𝑞) + 𝐻𝐶𝑂3− (𝑎𝑞) ; 𝐾𝐶 =
[𝑂𝐻 − ][𝐻𝐶𝑂3− ]
[𝐶𝑂32− ]
15.5 Calculating Equilibrium Constants
1. Easy: Given all equilibrium amounts, just substitute the values into the K expression and solve it.
2. Given initial amounts and one equilibrium amount, calculate K. You have to set up a chart.
4
Fossum-Reyes
Q. 2𝐻2 𝑆 (𝑔) ⇌ 2𝐻2 (𝑔) + 𝑆2 (𝑔)
Start with 0.100 mol H2S in a 10.0 L vessel at 1132°C. At equilibrium, there’s 0.0285 mol H2 present. What is
the value of KC at this temperature?
(𝐾𝐶 = 2.3 × 10−4 )
15.6 Applications of Equilibrium Constants
If a system is not in equilibrium, the reaction will go in whichever direction needed (→ or ←) to get to
equilibrium.
Reaction quotient: Q
Q has the same form as K, but actual concentrations are used instead of equilibrium concentrations.
Predicting the Direction of Reaction
If Q > K, the reaction will proceed fastest in the reverse direction
–
the [products] will decrease and [reactants] will increase
If Q < K, the reaction will proceed fastest in the forward direction
–
the [products] will increase and [reactants] will decrease
If Q = K, the reaction is at equilibrium
–
the [products] and [reactants] will not change
If a reaction mixture contains just reactants, Q = 0, and the reaction will proceed in the forward direction
If a reaction mixture contains just products, Q = ∞, and the reaction will proceed in the reverse direction
Note: if any product or reactant has a concentration of 0, the reaction must proceed to make some.
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Fossum-Reyes
Q. 𝐶𝑂2 (𝑔) + 𝐶 (𝑠) ⇌ 2𝐶𝑂(𝑔) ; 𝐾𝐶 = 1.17 @ 1000℃
If 0.0015 mol CO2 and 0.10 mol CO are combined with C(s) in a 10.0 L vessel @ 1000°C, will more CO form?
Calculating Equilibrium Concentrations
1. Easy – given K and some eq. concentrations (or P’s), calculate the missing concentration (or P).
2. Given K and initial concentrations (or P’s), find KC or KP: Set up a chart.
Q. 𝐻2 (𝑔) + 𝐼2 (𝑔) ⇌ 2𝐻𝐼(𝑔) ; 𝐾𝐶 = 49.7 @ 458℃
Start with 0.500 moles of each (H2 and i2) in a 2.00 L flask. Calculate the equilibrium concentrations.
For some problems, we will need to solve a quadratic equation:
𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 ;
𝑥=
−𝑏 ± √𝑏 2 − 4𝑎𝑐
2𝑎
2 solutions: find both, but keep the one that makes sense physically.
6
Fossum-Reyes
Q. 𝑃𝐶𝑙5 (𝑔) ⇌ 𝑃𝐶𝑙3 (𝑔) + 𝐶𝑙2 (𝑔) ; 𝐾𝐶 = 0.0211 @ 160℃
If initial [PCl5] is 1.00M, what is the equilibrium composition at 160°C.
Approximations to Simplify the Math
When the equilibrium constant is very small, the position of equilibrium favors the reactants; therefore, if we
have relatively large initial concentrations of reactants, the reactant concentration will not change significantly
when it reaches equilibrium. So:
[X]equilibrium = ([X]initial  ax)  [X]initial
Example: 1.00 − 𝑥 ; 𝑖𝑓 𝑥 = 0.0001, 𝑡ℎ𝑒𝑛:
1.00 − 0.0001 = 1.00 to 2 or 3 sig. figs.
we are approximating the equilibrium concentration of reactant to be the same as the initial concentration
assuming the reaction is proceeding forward
We can check our approximation by comparing the approximate value of x to the initial concentration:
If the approximate value of x is less than 5% of the initial concentration, the approximation is valid.
𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑒 𝑥
𝑖𝑓
× 100 < 5% 𝑣𝑎𝑙𝑖𝑑 𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑖𝑜𝑛
𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛
7
Fossum-Reyes
Q. For the reaction I2(g) ⇌ 2 I(g) the value of Kc = 3.76 x 10−5 at 1000 K. If 1.00 mole of I2 is placed into a
2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?
(use the simplifying assumption to solve for x)
Q. For the reaction 2 H2S(g) ⇌ 2 H2(g) + S2(g) at 800 °C, Kc = 1.67 x 10−7. If a 0.500 L flask initially
containing 1.25 x 10−4 mol H2S is heated to 800 °C, find the equilibrium concentrations.
8
Fossum-Reyes
For reactions with large K values, two steps are required (two charts)
1. Assume the reaction goes to completion.
2. Consider the reverse reaction at equilibrium.
Example:
𝐴 (𝑔) + 3𝐵 (𝑔) ⇌ 𝐶 (𝑔) + 2𝐷 (𝑔)
𝐾𝑃 = 2.9 × 107
Mix 0.100 atm A and 0.200 atm B. Determine all pressures at equilibrium.
1. The reaction goes to completion, so, the LR gets used up. Find the new P’s.
2. Consider equilibrium in the reverse direction.
A
3B
C
2D
Initially
0.100 atm
0.200 atm
0
0
Change
-X
-3X
+X
+2X
0.100 – X
0.200 – 3X
X
2X
End (not Eq!)
𝐿. 𝑅. ? 𝑛𝑒𝑒𝑑:
3𝐵
0.2𝐵 2𝐵
; ℎ𝑎𝑣𝑒:
=
; 𝐵 𝑖𝑠 𝐿. 𝑅. ;
𝐴
0.1𝐴 1𝐴
𝐴 = 0.100 − 0.0666̅7 = 0.033̅3𝑎𝑡𝑚 ;
0.200 − 3𝑥 = 0 ; 𝑥 =
0.200
= 0.0666̅7𝑎𝑡𝑚
3
𝐶 = 𝑥 = 0.0666̅7𝑎𝑡𝑚 ; 𝐷 = 2𝑥 = 0.133̅3𝑎𝑡𝑚 ; 𝐵 = 0
We now consider the reverse reaction:
𝐶 (𝑔) + 2𝐷 (𝑔) ⇌ 𝐴 (𝑔) + 3𝐵 (𝑔)
C
;
𝐾𝑃 =
Initially
0.0666̅7
2D
0.133̅3𝑎𝑡𝑚
Change
-Y
0.0666̅7 – Y
-2Y
0.133̅3 – 2Y
Equilibrium
1
= 3. 4̅5 × 10−8
2.9 × 107
A
0.033̅3𝑎𝑡𝑚
3B
0
+Y
+3Y
0.033̅3𝑎𝑡𝑚+Y
3Y
K is small. The reactions does not proceed much. Y is small compared to other numbers.
0.033̅3 + 𝑌 ≈ 0.033̅3 ; 0.133̅3 − 2𝑌 ≈ 0.133̅3
We can neglect Y.
𝐾𝑃 =
𝑃𝐴 𝑃𝐵3
(0.033̅3 + 𝑦)(3𝑦)3
=
= 3.45 × 10−8
𝑃𝐶 𝑃𝐷2 (0.0666̅7 − 𝑦)(0.133̅3 + 2𝑦)2
Reduces to:
(0.033̅3)(3𝑦)3
= 3.45 × 10−8 ; (3𝑦)3 = 27𝑦 3
(0.0666̅7)(0.133̅3)2
3
𝑦=√
(3.45 × 10−8 )(0.0666̅7)(0.133̅3)2
= 3. 5̅68 × 10−4 𝑎𝑡𝑚
(0.033̅3)(27)
Equilibrium Amounts (from table)
𝑃𝐴 = 0.033 𝑎𝑡𝑚 ; 𝑃𝐶 = 0.0667 𝑎𝑡𝑚
𝑃𝐵 = 0.0011 𝑎𝑡𝑚 ; 𝑃𝐷 = 0.133 𝑎𝑡𝑚
9
Fossum-Reyes
15.7 Le Châtelier’s Principle
Once at equilibrium (when the concentrations of all the reactants and products remain the same), what happens
if the conditions are changed?
The concentrations of all the chemicals will change until equilibrium is restored. The new concentrations will
be different, but the equilibrium constant will be the same… Unless…
…the Temperature is changed.
Le Châtelier's Principle guides us in predicting the effect various changes in conditions have on the position of
equilibrium.
If a system at equilibrium is disturbed, the position of equilibrium will shift to minimize the disturbance. (A new
equilibrium will ensue.)
1. Changing Concentrations
This does not change the value of K.
Equilibrium shifts away from side with added chemicals or toward side with removed chemicals…
If something is added to the mixture, the reaction shifts to use up some.
If something is removed from the mixture, the reactions shifts to create more.
Adding/removing solids or liquids has no effect (as long as there’s still some present.)
Q. 𝐻2 (𝑔) + 𝐼2 (𝑔) ⇌ 2𝐻𝐼(𝑔)
Which way will the equilibrium shift if you…:
a. Add 𝐻𝐼
b. Add 𝐼2
c. Remove 𝐻2
d. Add NaOH
2. Changing Pressure/Volume
If VContainer is decreased, P increases.
The equilibrium shifts so as to decrease P: towards the side with fewer moles of gas.
𝐶𝑂 (𝑔) + 3𝐻2 (𝑔) ⇌ 𝐶𝐻4 (𝑔) + 𝐻2 𝑂 (𝑔)
If 𝑉 ↓, 𝑃 ↑, Eq. will shift → to right, toward fewer moles of gas (takes up less space…)
Note: adding an inert gas won’t change partial P’s; therefore, there is no shift in equilibrium.
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Fossum-Reyes
3. Changing Temperature
When we change the temperature, the value of K also changes.
If ∆𝐻(+)(endothermic), if 𝑇 ↑, 𝐾 ↑ or 𝑇 ↓, 𝐾 ↓
If ∆𝐻(−)(exothermic), if 𝑇 ↑, 𝐾 ↓ or 𝑇 ↓, 𝐾 ↑
In other words, if T is increased, the reaction will go in the endothermic direction; if T is decreased, the reaction
will go in the exothermic direction.
Q. 2𝑆𝑂2 (𝑔) + 𝑂2 (𝑔) ⇌ 2𝑆𝑂3 (𝑔) ∆𝐻 = −198 𝑘𝐽
If 𝑇 ↑, which way does this eq. shift?
Q. 2𝐶𝑂2 (𝑔) ⇌ 2𝐶𝑂(𝑔) + 𝑂2 (𝑔) ∆𝐻 = 566 𝑘𝐽
What conditions of P and T would give the best yield of 𝐶𝑂(𝑔)?
Be careful! Language…
What is the difference between saying that: the equilibrium “shifts to the left” and the equilibrium “lies to the
left”?
“shifts to the left”
It means the reaction
will go ←.
𝑄>𝐾
“lies to the left”
It means at eq. we
have mostly reactants:
𝐾𝑒𝑞 ≪ 1
The Effect of Catalysts
Catalysts increase the rate of the forward and reverse reactions by the same factor; therefore, catalysts do not
affect the position of equilibrium (it only is achieved faster).
Kinetic vs Thermodynamic Control
Changing conditions to affect the reaction rates.
Adjusting the conditions to ensure that only the desired product or products are present in significant
concentrations at equilibrium.
11
Fossum-Reyes