Dynamic Equilibrium in Chemical Systems
(Assigned Readings pp. 424-436)
Closed System – a system that may exchange energy but not
matter with its surroundings (e.g. bottle of carbonated soft
drink)
Dynamic Equilibrium – a balance between forward and
reverse processes occurring at the same rate.
Equilibrium is symbolized by combining a forward () and a
reverse () arrow into a single symbol as follows:
or
or
When an equation is written with double arrows to show that
the change occurs both ways, the left-to-right change is called
the forward reaction, and the right-to-left change is called
the reverse reaction.
Three Types of Equilibrium
1. Solubility Equilibrium – a dynamic equilibrium
between a solute and a solvent in a saturated solution in
a closed system
Example: A saturated aqueous solution of iodine
, the rate of the dissolving process is equal to
the rate of the crystallizing process.
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2. Phase Equilibrium – a dynamic equilibrium between
different physical states of a pure substance in a closed
system.
Example 1: Evaporation/condensation equilibrium
Example 2: Solid/liquid equilibrium, usually established
at the freezing/melting point
3. Chemical Reaction Equilibrium – a dynamic
equilibrium between reactants and products of a
chemical reaction in a closed system
Example: Production of calcium oxide (lime) by heating
calcium carbonate (limestone)
Open system:
Closed system:
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ICE Tables
Previous chemistry courses, stoichiometric calculations were
straightforward when reactions proceed to completion (called
quantitative reactions – where virtually all of the limiting
reagent is consumed)
When reversible reactions achieve equilibrium before all of
the reactants become products, the stoichiometry needs more
thought and organization (using ICE Tables)
I – Initial
C – Change
E – Equilibrium
For systems composed of aqueous solutions or gases,
I, means initial concentrations of reactants and products
(before reaction)
C, stands for the change in the concentrations of reactants and
products between the start and the point at which equilibrium
is achieved
E, stands for the concentrations of reactants and products at
equilibrium
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Calculating Concentrations at Equilibrium
Consider the following equation for the formation of
hydrogen fluoride from its elements at SATP:
If the reaction begins with 1.00 mol/L concentrations of H2(g)
and F2(g) and no HF(g), calculate the concentrations of H2(g) and
HF(g) at equilibrium if the equilibrium concentrations of F2(g)
is measured to be 0.24 mol/L.
List all given information:
[H2(g)]initial = 1.00 mol/L
[F2(g)]initial = 1.00 mol/L
[HF(g)]initial = 0.00 mol/L
[F2(g)]equilibrium = 0.24 mol/L
Table 1: ICE Table for the Reaction of H2(g) and F2(g)
H2(g)
+
F2(g)
2 HF(g)
Initial
concentration
(mol/L)
Change in
concentration
(mol/L)
Equilibrium
concentration
(mol/L)
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The balanced chemical equation indicates that the change
occurs in a 1:1:2 molar ratio, but we do NOT know what
amount of the reactants is converted to product.
We choose the variable “x” to represent changes in the
concentrations of reactants and products, with the coefficients
of “x” corresponding to the coefficients in the balanced
equation.
Table 1: ICE Table for the Reaction of H2(g) and F2(g)
H2(g)
Initial
concentration
(mol/L)
Change in
concentration
(mol/L)
Equilibrium
concentration
(mol/L)
1.00
+
F2(g)
2 HF(g)
1.00
0.00
-x
1.00 – x
-x
1.00 –x
+ 2x
2x
Knowing that [F2(g)]equilibrium = 0.24 mol/L, we can determine
the value of x.
1.00 mol/L – x = 0.24 mol/L
- x = 0.24 mol/L – 1.00 mol/L
- x = - 0.76 mol/L
x = 0.76 mol/L
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Now use the value of x to calculate the equilibrium
concentrations of the other two entities.
[H2(g)] = 1.00 mol/L – x
= 1.00 mol/L – 0.76 mol/L
= 0.24 mol/L
[HF(g)] = 2x
= 2(0.76 mol/L)
= 1.52 mol/L
The equilibrium concentrations of H2(g) and HF(g) are
0.24 mol/L and 1.52 mol/L.
Learning Tip
Some problems do not provide you with M or mol/L. Instead
they specify size of container and moles of reactant and
product and you need to calculate molar concentration.
n
C
v
C = molar concentration (mol/L)
n = moles (mol)
v = volume (L)
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