1. Chapter 14: Chemical Equilibrium
CHE 124: General Chemistry II
Dr. Jerome Williams, Ph.D.
Saint Leo University

2. Overview The Equilibrium Condition: A Second Example
Law of Mass Action: Equilibrium Constants
Homogeneous Equilibrium

3. H2(g) + I2(g) Û 2 HI(g)
At time 0, there are only reactants in the mixture, so only the forward reaction can take place
[H2] = 8, [I2] = 8, [HI] = 0
At time 16, there are both reactants and products in the mixture, so both the forward reaction and reverse reaction can take place
[H2] = 6, [I2] = 6, [HI] = 4
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4. H2(g) + I2(g) Û 2 HI(g)
At time 32, there are now more products than reactants in the mixture − the forward reaction has slowed down as the reactants run out, and the reverse reaction accelerated
[H2] = 4, [I2] = 4, [HI] = 8
At time 48, the amounts of products and reactants in the mixture haven’t changed – the forward and reverse reactions are proceeding at the same rate – it has reached equilibrium
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5. H2(g) + I2(g) Û 2 HI(g)
Because the reactant concentrations are decreasing, the forward reaction rate slows down
And because the product concentration is increasing, the reverse reaction rate speeds up
As the reaction proceeds, the [H2] and [I2] decrease, and the [HI] increases
Because the [HI] at equilibrium is larger than the [H2] or [I2], we say the position of equilibrium favors products
At equilibrium, the forward reaction rate is the same as the reverse reaction rate
Once equilibrium is established, the concentrations no longer change
Equilibrium established
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6. Equilibrium  Equal
The rates of the forward and reverse reactions are equal at equilibrium
But that does not mean the concentrations of reactants and products are equal
Some reactions reach equilibrium only after almost all the reactant molecules are consumed – we say the position of equilibrium favors the products
Other reactions reach equilibrium when only a small percentage of the reactant molecules are consumed – we say the position of equilibrium favors the reactants
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7. An Analogy: Population Changes
However, after a time, emigration will occur in
both directions at the same rate, leading to
populations in Country A and Country B that are
constant, though not necessarily equal
When Country A citizens feel overcrowded, some will emigrate to Country B
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8. Equilibrium Constant
Even though the concentrations of reactants and products are not equal at equilibrium, there is a relationship between them
The relationship between the chemical equation and the concentrations of reactants and products is called the Law of Mass Action
For the general equation aA + bB Û cC + dD, the Law of Mass Action gives the relationship below
the lowercase letters represent the coefficients of the balanced chemical equation
always products over reactants
K is called the equilibrium constant
unitless
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9. Writing Equilibrium Constant Expressions
For the reaction aA(aq) + bB(aq) Û cC(aq) + dD(aq) the equilibrium constant expression is
So for the reaction
2 N2O5(g) Û 4 NO2(g) + O2(g)
the equilibrium constant
expression is
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10. 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
the position of equilibrium favors products
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
the position of equilibrium favors reactants
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11. A Large Equilibrium Constant
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12. A Small Equilibrium Constant
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13. Practice – Write the equilibrium constant expressions, K, and predict the position of equilibrium for the following
2 SO2(g) + O2(g) Û 2 SO3(g) K = 8 x 1025
N2(g) + 2 O2(g) Û 2 NO2(g) K = 3 x 10−17
favors products
favors reactants
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14. Relationships between K and Chemical Equations
When the reaction is written backward, the equilibrium constant is inverted
For the reaction aA + bB Û cC + dD
the equilibrium constant expression is
For the reaction cC + dD Û aA + bB
the equilibrium constant expression is
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15. Relationships between K and Chemical Equations
When the coefficients of an equation are multiplied by a factor, the equilibrium constant is raised to that factor
For the reaction aA + bB Û cC
the equilibrium constant expression is
For the reaction 2aA + 2bB Û 2cC
the equilibrium constant expression is
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16. Relationships between K and Chemical Equations
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 Û bB and (2) bB Û cC the equilibrium constant expressions are
For the reaction aA Û cC
the equilibrium constant expression is
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17. Example 14.2: Compute the equilibrium constant at 25 °C for the reaction NH3(g)  0.5 N2(g) H2(g)
Given:
Find:
for N2(g) + 3 H2(g) Û 2 NH3(g), K = 3.7 x 108 at 25 °C
K for NH3(g)  0.5N2(g) + 1.5H2(g), at 25 °C
Conceptual Plan:
Relationships:
Kbackward = 1/Kforward, Knew = Koldn K’ K
Solution:
N2(g) + 3 H2(g) Û 2 NH3(g) K1 = 3.7 x 108
2 NH3(g) Û N2(g) + 3 H2(g)
NH3(g) Û 0.5 N2(g) H2(g)
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18. Practice – When the reaction A(aq) Û 2 B(aq) reaches equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M. When the reaction 2 B(aq) Û Z(aq) reaches equilibrium [B] = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M. Calculate the equilibrium constant for each of these reactions and the equilibrium constant for the reaction 3 Z(aq) Û 3 A(aq)
Krxn 1 = 1.6 x 104
Krxn 2 = 5.0 x 10−1
2 B Û A K3 = 1/Krxn 1 = 6.25 x 10−5
Z Û 2 B K4 = 1/Krxn 2 = 2
Z Û A K3K4 = 1.25 x 10−4
3 Z Û 3 A (K3K4)3 = 2.0 x 10−12
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