### 5.3 - Ice Box Problems - chem30

```5.3 – ICE BOX
PROBLEMS
Unit 5: Equilibrium

As we already know, a chemical equation tells us
several pieces of information.
H2(g) + I2(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.
STUFF WE SHOULD KNOW
H2(g) + I2(g) ↔ 2HI (g)

For example, if [H2] changes by “x” mols/L, then:


[I2] changes by “x” mols/L, and
[HI] changes by “2x” mols/L.

However, [H2] and [I2] are decreasing

while [HI] is increasing.

Thus, we say that Δ[H2] = Δ[I2] = “

x”, and the Δ[HI] = “2x”.
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.
NEW-ISH STUFF…



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!
ICE BOX PROBLEMS

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.
ICE BOX PROBLEMS
ICE BOX PROBLEMS

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:
N2(g) + 3H2(g) ↔ 2 NH3(g)

If the initially [N2] = 0.96 M and [H2] = 0.72 M, and at equilibrium [NH3] =
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
ICE BOX PROBLEMS – TYPES OF
QUESTIONS
Type 2: Solving for equilibrium concentrations when Keq is given

Example 2: Initially, for the reaction below, [H2] = [I2] = 0.200 M.
H2(g) + I2(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”
ICE BOX PROBLEMS – TYPES OF
QUESTIONS
ICE BOX PROBLEMS – TYPES OF
QUESTIONS

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:
ax2 + 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,
PCl5(g) ↔ PCl3(g) + Cl2(g)

Calculate all three equilibrium concentrations if initially [PCl 5] = 1.00 M and
Keq = 16.0.
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