Transcript 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)
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This equation tells us how the number of
molecules or moles of each substance relate to
each other by looking at the balancing
coefficients.
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The balancing coefficients also tell us how the
concentration of each substance changes as a
reaction goes on.
STUFF WE SHOULD KNOW
ALREADY…
H2(g) + I2(g) ↔ 2HI (g)
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For example, if [H2] changes by “x” mols/L, then:
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[I2] changes by “x” mols/L, and
[HI] changes by “2x” mols/L.
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However, [H2] and [I2] are decreasing
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while [HI] is increasing.
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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…
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The ICE is an acronym for:
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I = Initial concentration
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C = Change in concentration
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E = Equilibrium concentration
An ICE box problem looks at the ENTIRE equilibrium chemical
reaction:
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The concentration of initial reactant (s) before they interact with each
other

The change in concentrations for each substance (both increasing
and decreasing)
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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
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The following process can be used to solve an ICE box problem
for a general reaction: A + B ↔ C
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1. Balance the equation.
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2. Set up the ICE box:
[A]
[B]
[C]
[Initial]
[Change]
[Equilibrium]
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3. Use information from the question to plug values into the ICE
box.
ICE BOX PROBLEMS
ICE BOX PROBLEMS
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Some hints you may want to consider:
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i. Initially, you should have no product.
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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?
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Therefore, the initial product box (es) can have a value of zero
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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!
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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
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iv. You can manipulate, use stoichiometry, or assign variables for the
CHANGE values ONLY.
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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.
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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:
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Example 1: Ammonia is created by the following process:
N2(g) + 3H2(g) ↔ 2 NH3(g)
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If the initially [N2] = 0.96 M and [H2] = 0.72 M, and at equilibrium [NH3] =
0.24 M what is the equilibrium constant?
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Use the information given to problem-solve for a CHANGE value
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Use stoichiometry to solve for the remaining CHANGE values
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Go where the questions asks
ICE BOX PROBLEMS – TYPES OF
QUESTIONS
Type 2: Solving for equilibrium concentrations when Keq is given
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Example 2: Initially, for the reaction below, [H2] = [I2] = 0.200 M.
H2(g) + I2(g) ↔ 2HI(g)
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Calculate all three equilibrium concentrations if Keq = 64.0.
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Since no equilibrium concentrations are given, you need to assign variables
for your CHANGE values
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Let the balancing coefficients assist you in assigning variables
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i.e.: if there is a balancing coefficient of “1”, assign that variable “x”
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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
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Type 3: Solving for equilibrium concentrations when Keq is given BUT you
have to use the quadratic formula
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When solving for x, if you are left with the following pattern:
ax2 + bx + c
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You will need to use the quadratic formula to solve for x
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This will give you two answers; always neglect the negative answer for “x”
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Remember, “x” represents concentration, so you can’t have a negative
concentration
Example 3: Given this equation,
PCl5(g) ↔ PCl3(g) + Cl2(g)
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Calculate all three equilibrium concentrations if initially [PCl 5] = 1.00 M and
Keq = 16.0.