Unit 5: Equilibrium


As we already know, a chemical equation tells us several pieces of information.

H
2
(g) + I
2
(g) ↔ 2HI (g)
This equation tells us how the number of molecules or moles of each substance relate to each other by looking at the balancing coefficients.

The balancing coefficients also tell us how the concentration of each substance changes as a reaction goes on.




H
2
(g) + I
2
(g) ↔ 2HI (g)
For example, if [H
2
] changes by “x” mols/L, then:

[I
2
] changes by
mols/L, and

[HI] changes by
mols/L.
However, [H
2
] and [I
2
] are decreasing while [HI] is increasing.

Thus, we say that Δ[H
2
] = Δ[I
2
] =
, and the Δ[HI] =

Note that the negatives mean decrease (or losing) and a positive value means gaining.

This knowledge of changing concentrations is key for an ICE box problem.



The ICE is an acronym for:

I = Initial concentration


C = Change in concentration
E = Equilibrium concentration
An ICE box problem looks at the ENTIRE equilibrium chemical reaction:



The concentration of initial reactant (s) before they interact with each other
The change in concentrations for each substance (both increasing and decreasing)
The final concentrations of each substance once equilibrium has been established

Therefore you will be working with three (3) sets of values in each
ICE box question – so make sure to be thorough and show all your work; it’s easy to make simple mistakes!


The following process can be used to solve an ICE box problem for a general reaction: A + B ↔ C

1. Balance the equation.

2. Set up the ICE box:
[A] [B] [C]
[Initial]
[Change]
[Equilibrium]

3. Use information from the question to plug values into the ICE box.


Some hints you may want to consider:




 i. Initially, you should have no product.

Think about it, if the substances haven’t even touched or interacted with each other, how can any product be formed at the start of the reaction?

Therefore, the initial product box (es) can have a value of zero ii. All of the values in the ICE box represent concentrations (molarity, moles/L). If the information given is not expressed in molarity, make sure you convert it to molarity first! iii. Read the information given in the question very carefully – it will often give away many of the values for you to place in the ice box iv. You can manipulate, use stoichiometry, or assign variables for the
CHANGE values ONLY.

You can NOT do this for your initial or equilibrium values v. Information given in the question will always be for the INITIAL or
EQUILIBRIUM concentrations.

The CHANGE values will never be explicitly given to you, rather, you will have to problem-solve to figure them out


Type 1: Solving for Keq when initial and equilibrium conditions are given:
Example 1: Ammonia is created by the following process:

N
2
(g) + 3H
2
(g) ↔ 2 NH
3
(g)
If the initially [N
2
] = 0.96 M and [H
2
] = 0.72 M, and at equilibrium [NH
3
] =
0.24 M what is the equilibrium constant?

Use the information given to problem-solve for a CHANGE value

Use stoichiometry to solve for the remaining CHANGE values

Go where the questions asks

Type 2: Solving for equilibrium concentrations when Keq is given


Example 2: Initially, for the reaction below, [H
2
] = [I
2
] = 0.200 M.
H
2
(g) + I
2
(g) ↔ 2HI(g)
Calculate all three equilibrium concentrations if Keq = 64.0.

Since no equilibrium concentrations are given, you need to assign variables for your CHANGE values

Let the balancing coefficients assist you in assigning variables

 i.e.: if there is a balancing coefficient of “1”, assign that variable “x” i.e.: if there is a balancing coefficient of “2”, assign that variable “2x”


Type 3: Solving for equilibrium concentrations when Keq is given BUT you have to use the quadratic formula

When solving for x, if you are left with the following pattern:


 ax 2 + bx + c
You will need to use the quadratic formula to solve for x
This will give you two answers; always neglect the negative answer for “x”
Remember, “x” represents concentration, so you can’t have a negative concentration

Example 3: Given this equation,
PCl
5
(g) ↔ PCl
3
(g) + Cl
2
(g)

Calculate all three equilibrium concentrations if initially [PCl
Keq = 16.0.
5
] = 1.00 M and