Chemical Equilibrium – Chemical Equilibrium & ICE Tables
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Chemical Equilibrium
Most chemical reactions are quantitative. That is, 100% of the limiting reagent reacts to form products.
However, some chemical reactions are reversible. This is understandable given that the Transition State Theory
predicts that the activated complex formed in a chemical reaction can form either reactants or products. In
quantitative reactions, the reverse reaction is so unlikely to occur, that the reaction can be considered to be
irreversible. However, if the rate of the reverse reaction is significant enough, then both the forward and
reverse reaction can occur simultaneously in a reaction mixture.
In order to understand what is happening in a chemical equilibrium, let us consider the chemical reaction of
catalytic methanation wherein carbon monoxide is reacted with hydrogen gas in the presence of a catalyst
(usually an industry secret) to form methane and water (steam):
CO(g) + 3H2(g) → CH4(g) + H2O(g)
Let us use numbers that make for some easy calculation. Since the ratio of carbon monoxide to hydrogen is 1:3,
let us suppose you put 1.000 mol CO and 3.000 mol H2 into a 1.00 L vessel at 1200 K (927oC). At first, the
forward rate of reaction is fast, but as the reaction progresses, the concentration of the reactants will decrease,
so the forward rate of reaction will decrease.
This reaction is reversible. Therefore, we have:
CH4(g) + H2O(g) → CO(g) + 3H2(g)
Initially, the amount of methane and steam is zero. Therefore, the initial rate of the reverse reaction is zero.
However, as the forward reaction progress, the concentration of methane and steam increases. Therefore, the
reverse rate of reaction will increase.
Eventually, the rate of the forward reaction and the rate of the reverse reaction will be equal.
At this point, the concentrations of both the reactants and products will be constant.
ICE Tables
Once the reaction reaches equilibrium, you only need one substance to determine the amount of all the others.
For example, the amount of water produced by catalytic methanation can be determined by rapidly cooling the
vessel to condense the water, and then measuring the volume.
If we know the amount of one of the substances at equilibrium, how do we calculate the rest?
The most important tool you will use for calculation equilibrium problems is the ICE table. ICE stands for Initial,
Change, and Equilibrium. Using algebra, we assign the change in concentration of each substance the variable x,
since their changes are all multiples based on the molar coefficient. Let’s look at an example:
Sample Question
1.000 mol CO and 3.000 mol H2 are placed in a 10.00 L vessel at 927oC and allowed to come to equilibrium. The
amount of water at equilibrium is determined to be 0.0387 mol/L. What is the molar composition of the
equilibrium mixture? (i.e. how many moles of CO, H2, and CH4 are there?)
To determine this, set up a table to determine the initial (I), change (C), and equilibrium (E) values (or, an ICE
table).
Reactants
Products
Units = Mol/L
CO(g)
H2(g)
CH4(g)
H2O(g)
Initial
0.1000
0.3000
0
0
Change
-x
-3x
+x
+x
Equilibrium
0.1000-x
0.3000-3x
x
x
* notice that the amount of CO and H2 has been calculated for the molar concentration using C = n/V
Since the amount of water at equilibrium is 0.0387 mol/L, then we know that x = 0.0387 mol/L. Therefore, we
can calculate the remaining concentrations:
Equilibrium amount CO = (0.1000 – x) mol/L
= (.1000 – 0.0387) mol/L
= 0.0613 mol/L
Equilibrium amount H2 = (.3000 – 3x) mol/L
= [.3000 – 3(0.0387)] mol/L
= 0.1839 mol/L
Equilibrium amount CH4 = x mol = 0.0387 mol/L
Recall that by rearranging C = n/V, we get n = CV, and given that the volume of the container is 10.00 L, then
multiply each concentration by 10.00 L.
Therefore, the amounts of substances in the equilibrium mixture are 0.613 mol CO, 1.839 mol H2, 0.387 mol CH4
and 0.387 mol H2O
Homework:
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