# 8: ICE Tables (Worksheet)

- Page ID
- 79393

Name: ______________________________

Section: _____________________________

Student ID#:__________________________

Work in groups on these problems. You should try to answer the questions without accessing the Internet.

When a system is at equilibrium, the Law of Mass action is ALWAYS satisfied (i.e., \(Q=K\)). However, the actual concentrations (or pressures) under this equilibrium can be any number, just as long as the respective concentrations satisfy the numerical value of \(K\). The gist of this worksheet is to identify how a nonequilibrium system moves toward equilibrium.

### Q1: Two Component System

Consider the general reaction

\[2 A \rightleftharpoons B \nonumber \]

In the plot below, draw the relationship that represents all possible combinations of \([A]\) and \([B]\) that satisfy the Law of Mass Action with \(K_c=1\).

- What is the dimensionality of the relationship you drew (point=0, curve=line=1, surface=area=2)?
- Does each point(s) you drew describe a possible equilibrium configuration?
- If so, what distinguishes one point from another?
- If not, draw the relationship you can draw for only equilibrium condition(s).

- Draw three points on the plot that are under non-equilibrium conditions. Why did you specifically identify those points?

### Q2: Three Component System

A reaction is represented by this equation: \(\ce{A}(aq)+\ce{2B}(aq)⇌\ce{2C}(aq) \) with \(K_c=1×10^3\)

- Write the mathematical expression for the equilibrium constant.
- Using concentrations ≤1
*M*, make up two sets of concentrations that describe a mixture of A, B, and C at equilibrium.

## ICE Tables: The Quest for Equilibrium

An ICE (**I**nitial, **C**hange, **E**quilibrium) table is simple matrix formalism that used to simplify the calculations in reversible equilibrium reactions. 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 row contains the initial concentrations of products and reactants.**C**stands for the**change**in concentration. This is the concentration change required for the reaction to reach 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 and a variable; this is because this row represents an increase or decrease (or no change) in concentration.**E**is for the concentration when the reaction is at**equilibrium**. This is the summation of the initial and change rows. Once this row is completed, its contents can be plugged into the equilibrium constant equation to solve for \(K_c\).

### Example: The Haber-Bosch Reaction (Identifying the Equilibrium Constant)

If 1.5 moles of \(N_2(g)\) were added to 3.50 moles of \(H_2 (g)\) in a \(1 L\) vessel at 700 °C. The possible Haber-Bosch reaction occurs

\[ 3H_2(g) + N_2(g) \rightleftharpoons 2NH_3(g) \nonumber \]

Let's say an experimentalist is able to determine the equilibrium concentration of \(NH_3\) at 0.540 M. Determine the equilibrium constant (\(K_c\)) for this reaction (at 700 °C).

To solve this, we use an ICE table that relates equilibrium and non-equilibrium conditions via the Law of Mass Action and the stoichiometry of the **balanced **reaction. Let's start by constructing one, by entering the balanced species into the top line:

ICE Table |
\(3H_2(g)\) |
\(N_2\) |
\(\rightleftharpoons\) |
\(2NH_3\) |
---|---|---|---|---|

Initial |
- | |||

Change |
- | |||

Equilibrium |
- |

Now we insert the initial concentrations into the next line (the I in ICE). We can work an ICE table in terms of moles, concentrations (any), or pressures. Since this is a 1 L vessel, we can stick to molarity for now.

ICE Table |
\(3H_2(g)\) |
\(N_2\) |
\(\rightleftharpoons\) |
\(2NH_3\) |
---|---|---|---|---|

Initial |
3.5 | 1.5 | - | 0 |

Change |
- | |||

Equilibrium |
- |

Now we insert the change of concentrations when the reaction moves toward equilibrium into the next line (the C in ICE). This is dictated **exclusively **by the stoichiometric coefficients in the balance equation (always make sure the reaction is balanced).

ICE Table |
\(3H_2(g)\) |
\(N_2\) |
\(\rightleftharpoons\) |
\(2NH_3\) |
---|---|---|---|---|

Initial |
3.5 | 1.5 | - | 0 |

Change |
-3x | -x | - | +2x |

Equilibrium |
- |

Lastly, we insert the equilibrium concentration taken from the top two rows (the E in ICE). This dictated exclusively by the top two lines (although you may have more information from the problem, ignore it when constructing the ICE table).

ICE Table |
\(3H_2(g)\) |
\(N_2\) |
\(\rightleftharpoons\) |
\(2NH_3\) |
---|---|---|---|---|

Initial |
3.5 | 1.5 | - | 0 |

Change |
-3x | -x | - | +2x |

Equilibrium |
3.5 -3x | 1.5 -x | - | 2x |

Now we can take the Law of Mass Action to construct the equilibrium constant for this system

\[ K_c = \dfrac{[NH_3]^2}{[H_2]^3[N_2]^1} \label{Ex1.1} \]

We can now insert the values for the equilibrated populations from the ICE table into the Equation \(\ref{Ex1.1}\) to get

\[ K_c = \dfrac{(2x)^2}{(3.5 -3x)^3 (1.5 -x)^1} \label{Ex1.2} \]

To solve this equation, we need to know the value of \(x\), which we don't know explicitly. However, we do know from the initial information and he ICE table that

\[ [NH_3] =0.540 = 2x \label{Ex1.3} \]

or

\[x = \dfrac{0.540}{2} = 0.270 \label{Ex1.4} \]

This tells us how far the system has to evolve to equilibrate from the original non-equlibrium conditions. Plug \(x\) from Equation \(\ref{Ex1.4}\) into \(K_c\) expression from Equation \(\ref{Ex1.2}\) results in

\[ K_c = \dfrac{(2 \times 0.270)^2}{(3.5 -3 \times 0.270)^3 (1.5 - 0.270)^1} = 0.0122 \label{Ex1.5} \]

### Q3

The above ICE mathematics relate equilibrium concentrations to initial non-equilibrium concentrations based off of the specific equilibrium constant and the balanced stoichiometric coefficients. Many possible question can be asked that move from one to the other. When calculating concentrations (equilibrated or not), what reality checks can you do to ensure the predicted concentrations from such a calculations are reasonable?

### Q4: ICE Table with Concentrations

A mixture consisting of 0.201 mol \(H_2\), and 0.201 mol of \(I_2\) is brought to equilibrium at 500°C, in a 4.5 L flask. What are the equilibrium amounts of \(H_2\), \(I_2\), and \(HI\)? \(K_c = 50.2\) at 500°C. Is the Video Solution correct?

### Q5: ICE Table with Pressures

For the reversible reaction between \(CO\) and \(CO_2\) in the gas phase

\[2CO(g)+O_2(g) \rightleftharpoons 2CO_2(g) \nonumber \]

If 0.352 mol of \(CO\) were added to 0.067 mol of \(CO_2\) in a 3 L flask at 668 K, how many moles of \(O_2(g)\) will be present at equilibrium? The \(K_c\) of this reaction is \(1.2 \times 10^3\) at 668 K.

### Q6: Heterogeneous Reaction

Equilibrium is reached in a 2.5 L flask for the following reaction

\[ \ce{CaCO_3(s)} \rightleftharpoons CaO(s)+ CO_2(g) \nonumber \]

with \(K_c=2.68 \times 10^{-3}\) at 800 °C.

- How many moles of \(CaCO_3(s)\) is present at equilibrium after 0.715 mols \(CaCO_3(g)\) is introduced into an empty flask?
- What is the pressure of \(CO_2\) under equilibrium?