ChemWiki: The Dynamic Chemistry E-textbook > Physical Chemistry > Equilibria > Le Châtelier's Principle > ICE Tables

An ICE (**I**nitial, **C**hange, **E**quilibrium) table is simple matrix formalism that used to simplify the calculations in reversible equilibrium reactions (often involving weak acids or weak bases). ICE is a simple acronym for the titles of the first column of the table.

**I**stands for**initial**concentration.**C**stands for the**change**in concentration.**E**is for the concentration when the reaction is at**equilibrium**.

ICE tables are composed of the concentrations of molecules in solution in different stages of a reaction, and are usually used to calculate the K, or equilibrium constant expression, of a reaction (in some instances, K may be given, and one or more of the concentrations in the table will be the unknown to be solved for). ICE tables automatically set up and organize the variables and constants needed when calculating the unknown.

ICE is a simple acronym for the titles of the first column of the table.

**I**stands for**initial**concentration. This is the concentration that the reaction starts out in.**C**stands for the**change**in concentration. This is the concentration change needed to get the reaction from start to equilibrium. It is the difference between the equilibrium and initial rows. The concentrations in this row are, unlike the other rows, expressed with either an appropriate positive (+) or negative (-) sign because this row represents change up or down (or no change).**E**is for the concentration when the reaction is at**equilibrium**. This is the summation of the initial and change rows. If you are looking for K_{c}, as you often will be at this stage of chemistry, this is the goal. Once you have the values in this row, you can easily obtain K_{c}by plugging them into the equation for K_{c}.

We will learn by doing a few examples. Setting up an actual ICE table will familiarize you with them better than abstract definitions, so let us consider an example of a reaction of general form:

Example 1 | ||||||||||||||||||||||||||||||||||||||||||||||||
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Using an ICE table to determine the \( K_c \) for the balanced general reaction: \[ 2X(g) \rightleftharpoons (g) + 4Z(g) \] where the capital letters represent the products and reactants. - This equation will be placed horizontally above the table, with each product and reactant having a separate column.
- This statement implies that there are no initial amounts of Y and Z. For the I row of the Y and Z columns, 0.000 mol will be entered.
- Notice that we are using amounts. The amounts can either be converted to concentrations before putting them into the ICE table or after the equilibrium amounts have been calculated. In this example, we will use the number of moles when filling in the ICE table (concentration can be calculated later).
- For the equilibrium row of X, 0.350 mol will be entered. As you will see, this and the initial amount will help us find the other unknown amounts in the ICE table.
Desired Unknown \[ K_c = ? \]
We know that the equilibrium constant expression can be expressed as products over reactants, with each to the power of their respective amounts: \[ K_c = \dfrac{[Y]^3[Z]^4}{[X]^2} \] Unfortunately, we do not know what the concentrations of Y and Z is in the equilibrium stage. To solve for them, we will construct the ICE table. This is the information we were given:
This is the first step in setting up the ICE table. As mentioned above, the ICE mnemonic is vertical and the equation heads the table horizontally, giving the rows and columns of the table, respectively. The numerical amounts were given. Any amount not directly given is yet unknown.
Notice that the equilibrium in this equation is shifted to the right, meaning that some amount of reactant will be taken away and some amount of product will be added (for the Change row). The change in amount \(x\) can be found easily by using algebra: \[ Equilibrium \; Amount = Initial \; Amount + Change \; in \; Amount \] Solving for the Change in the amount of 2X gives: \[ 0.350 \; mol - 0.500 \; mol = -0.150 \; mol \] The change in reactants and the balanced equation of the reaction is known, so we can calculate the change in products. Take a look at the stoichiometric constants. For every 2 mol of \(x\) that is reacted, 3 mol of Y and 4 mol of Z are produced. The relationship is as follows: \[ Change \; in \; Product = -\left(\dfrac{\text{Stoichiometric Coefficient of Product}}{\text{Stoichiometric Coefficient of Reactant}}\right)(\text{Change in Reactant}) \] \[ Change \; in \; Y = -\left(\dfrac{3}{2}\right)(-0.150 \; mol) \] \[ = +0.225 \; mol \] Try obtaining the change in Z with this method (the answer is already in the ICE table).
If the initial amounts of Y and/or Z were nonzero, then they would be added together with the change in amounts to determine equilibrium amounts. However, because there was no initial amount for those two products, the equilibrium is simply equal to the change in amount. \[ Equilibrium \; Amount = Initial \; Amount + Change \; in \; Amount \] \[ Equilibrium \; Amount \; of \; Y = 0.000 \; mol\; + 0.225 \; mol \] \[ = +0.225 \; mol \] Use the same method to find the equilibrium amount of Z. Now we must convert the equilibrium amounts to concentrations. Recall that the volume of the system we were given is 0.750 liters. \[ [Equilibrium \; Concentration \; of \; Substance] = \dfrac{Amount \; of \; Substance}{Volume \; of \; System}\] \[ [X] = \dfrac{0.350 \; mol}{0.750 \; L} = 0.467 \; M \] \[ [Y] = \dfrac{0.225 \; mol}{0.750 \; L} = 0.300 \; M \] \[ [Z] = \dfrac{0.300 \; mol}{0.750 \; L} = 0.400 \; M \] We now have the values required to solve the \( K_c \) equation: \[ K_c = \dfrac{[Y]^3[Z]^4}{[X]^2} \] \[ K_c = \dfrac{[0.300]^3[0.400]^4}{[0.467]^2} \] \[ K_c = 3.17 \times 10^{-3} \] This is our answer! |

Example 2: Using an ICE Table with Concentrations | ||||||||||||||||||||||||||||||||||||||||||||||||
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In this example we will use an ICE table to find the equilibrium concentration of the reactants and products. (This example will be less in depth than the previous but remember that the same concepts are applied here.) These are calculations often done for weak acid titrations. Find the concentration of A \[ HA (aq) + H_2O (l) \rightleftharpoons A^- (aq) + H_3O (aq)\] with \(HA (aq) = 0.150 M\) and \(K_a = 1.6 \times 10^{-2}\)
This equation is for a weak acid reaction in solution with water. The acid (HA) will dissociate into its conjugate base (\(A^
- In this table notice that you can shorten the leftmost column so it is easier to write out when you are doing a problem.
- The change in concentration is unknown, so the variable \(x\) is used to show the change. \(x\) is the same for both products and reactants because equal amounts of A
^{-}and H_{3}O^{+}are made when HA dissociates in water.
- To find the equilibrium amounts the I row and the C row are added. Now you can use these values and \(K_a\)
_{ }(the equilibrium constant for acids) to find the concentration*\(x\)*.
We know that \(K_a\) is the concentrations of the products divided by the concentrations of the reactants. Plugging in the values at equilibrium into the equation for \(K_a\) we get: \(K_a = \dfrac{x^2}{0.150-x} = 1.6*10^{-2}\) To find the concentration \(x\), we rearrange this equation to be quadratic (of the form \(Ax^2+Bx+C=0\)), then use the quadratic formula to find \(x\) \((1.6 \times 10^{-2})({0.150-x}) = {x^2}\) \(x^2+(1.6 \times 10^{-2})x-(0.150)(1.6 \times 10^{-2})= 0\) This is the typical form for a quadratic equation: \[Ax^{2}+Bx+C=0\] where - \(A = 1\)
- \(B = 1.6 \times 10^{-2}\)
- \(C =( -0.150)( 1.6 \times 10^{-2}) = -2.4 \times 10^{-3} \)
The quadratic formula will give two solutions (but only one physically) for \(x\): \[x = \dfrac{-B+\sqrt{B^2-4AC}}{2A}\] and \[x = \dfrac{-B-\sqrt{B^2-4AC}}{2A}\] Then you just plug in your values for \(A\), \(B\), and \(C\), and you find \(x\)! You will get two values for \(x\), the value that makes the most sense is correct. For example, if you plug them in and one gives you a negative concentration, that value is incorrect because you cannot have a negative concentration of a solution. Make sure not to forget, your concentration \(x\) can be used to calculate the equilibrium concentrations of each product and reactant (just plug in your value for \(x\) back into the ICE table in the E row). [Solution: \(x\) = 0.0416, -0.0576. \(x\) = 0.0416 makes chemical sense and is therefore the correct answer.] NOTE: For some problems like this, if \(x\) is significantly less than the value for \(K_a\) then the \(x\) of the reactants (in the denominator) can be omitted and the concentration for \(x\) should not be greatly affected. This will make calculations faster, no longer needing the quadratic formula to calculate \(x\). |

- Make sure the reversible equation is
**balanced**at the start of the problem; otherwise, you could end up with wrong amounts in the table (in our example, we were given a balanced equation…but do not assume the equation always is!). - The given data should be in
**amounts, concentrations, partial pressures**, or somehow able to be converted to such. If it is not, then an ICE table will not help solve the problem. - If the ICE table has the
**equilibrium in amounts, make sure to convert equilibrium values to concentrations**before plugging in to solve for K_{c}. - If the given data is in
**amounts or concentrations**, use the ICE table to find**K**. If the given data is in_{c}**partial pressures**, use the ICE table to find \(K_p\). If you desire to convert from one to the other, remember that \(K_p = K_c(RT)^{\Delta n_{gas}}\); it is simpler to use the ICE table with the appropriate givens and convert at the end of the problem. - Enter in
**known data first**, and that should enable you to calculate the unknown data in the table. - If ever you have
**a negative value in the Initial or Equilibrium rows, double check your work**! A negative concentration, amount, or partial pressure is physically impossible. Obviously, the Change row can contain a value that is a negative number. **Pay attention to the state of each reactant and product.**If a compound is a solid or a liquid, its concentrations are not relevant to the calculations. Only concentrations of gaseous and aqueous compounds are used.- In the 'change' row the values will usually be a variable, denoted by
*x*. It must first be understood which direction the equation is going to reach equilibrium (from left to right or from right to left). The value for 'change' in the 'from' direction of the reaction will be the opposite of*x*and the 'to' direction will be the positive of \(x\) (adding concentration to one side and take away an equal amount from the other side). - Know the
**direction of the reaction**. This knowledge will affect the Change row of the ICE table (for our example, we knew the reaction would proceed forward, as there was no initial products). Direction of reaction can be calculated using \(Q\), the reaction quotient, which is then compared to a known \(K\) value. - It is easiest to
**use****the same units**every time you use an ICE table (usually molarity is preferred). This will minimize confusion when calculating the equilibrium constants. ICE tables are usually used for weak acid or weak base reactions because all of the nature of these solutions. The amount of acid or base that will dissociate is unknown (for strong acids and strong bases it can be assumed that all of the acid or base will dissociate, meaning that the concentration of the strong acid or base is the same as its dissociated particles).

******"Partial pressure" may also be substituted for "concentrations" in the ICE table, if desired (i.e., if the concentrations are not known, K _{p} instead of K_{c} is desired, etc.). "Amount" is also acceptable (the ICE table may be done in amounts until the equilibrium amounts are found, after which they will be converted to concentrations). For simplicity, assume that the word "concentration" can be replaced with "partial pressure" or "amounts" throughout the non-example portions of this module.*

- 0.200 M acetic acid is added to water. What is the concentration of H
_{3}O^{+}in solution if K_{c}= 1.8*10^{-6}? - At 0.975 atm sodium hydroxide is added to water. What is the partial pressure of sodium ions? [Hint: Sodium hydroxide (NaOH) is a strong base.]
- If the initial concentration of NH
_{3}is 0.350 M and the concentration at equilibrium is 0.325 M, what is K_{c}for this reaction? - How do you find K
_{c}from K_{p}? - Fill in this ICE table:

Reaction: | [HA] | [A]^{-} | [H]_{3}O^{+} |

I | 0.650 mol | ? | ? |

C | ? | ? | ? |

E | 0.250 mol | ? | ? |

1. 5.99*10^{-4}

2. 0.975 atm Na^{+}

3. 1.92*10^{-3}

4. \(K_p = K_c(RT)^{\Delta n}\) then solve for K_{c}

5.

Reaction: | HA | A^{-} | H_{3}O^{+} |

I | 0.650 mol | 0.000 mol | 0.000 mol |

C | -0.400 mol | +0.400 mol | +0.400 mol |

E | 0.250 mol | 0.400 mol | 0.400 mol |

- Petrucci, et al. General Chemistry: Principles & Modern Applications; Ninth Edition. Pearson/Prentice Hall; Upper Saddle River, New Jersey 07.

- Alexander Shei (UCD), Aileen McDuff (UCD)

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