Using an ICE Table

equilibrium reaction of the deprotonation of alcohol

Core Concepts

In this tutorial, we will learn about the ICE table for chemistry: a method of completing calculations in equilibrium reactions, either to find the concentrations of reactants and products, or to find the value of the equilibrium constant.

Topics Covered in Other Articles

Review: Equilibrium Constant & Le Chatelier’s Principle

In order to use an ICE table correctly, you need to know how to calculate the equilibrium constant, K. K is a measure of the extent of the reaction, whether the position of equilibrium favors the reactants or products. For a reaction aA + bB \rightleftarrows cC + dD, then we calculate the equilibrium constant using the equation:

     \begin{gather*} K = \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{a}} \end{gather*}


Le Chatelier’s principle states that if a system in a state of dynamic equilibrium is disturbed by a change to its conditions, then the position of equilibrium will shift to counteract the change. For example, if more of one reactant is added to the reaction, then the position of equilibrium will shift towards the products.

solution which you can analyze with an ice table

What is an ICE table?

ICE stands for Initial, Change, Equilibrium. An ICE table is a tool used to calculate the changing concentrations of reactants and products in (dynamic) equilibrium reactions. This method first lists the concentrations of both reactants and products, before any changes occur. This is the initial stage. Then, the change is listed, in the form of addition or subtraction of a specific concentration. Alternately, the addition or subtraction of an unknown amount is listed (in the form of + or – x M), and the value of x is solved for. Finally, the equilibrium concentration is listed, which is the initial concentration after it has undergone the change described.

What is the change in an ICE table?

If the described change is unspecified, then the equilibrium concentration is listed in terms of x, and the equilibrium constant is used to solve for the variable. If the described change is specific, then the equilibrium concentration is listed as a concrete number, and it is used to solve for the equilibrium constant. We will explore both examples below.

Le Chatelier’s principle is especially relevant to ICE tables because the change in the ICE table represents the shift in the position of equilibrium: that means that if the change shows an increase in concentration of reactants, then there will be a subsequent decrease in the concentration of products. Vice versa as well: if the change shows a decrease in concentration of reactants, then there will be a subsequent increase in the concentration of products. This subsequent increase/decrease shows the position of equilibrium shifting to counteract the change to the reactants.

Additionally, Le Chatelier’s principle is relevant because the coefficient of the reactant or product affects the change to its concentration. For example, if a reactant has a coefficient of 1, then it changes by +x, but if a reactant has a coefficient of 2 or more, then it changes by +2x.

Examples of an ICE table + how to use them

ICE table #1: Find the equilibrium concentrations from the K value.

     \begin{gather*} {A (\text{aq}) + 2B (\text{aq}) \rightleftarrows C (\text{aq})}  \\ {K = 0.4} \end{gather*}

Initial0.5 \text{M}0.5 \text{M}0.0 \text{M}
Equilibrium0.5 - x \text{M}0.5 - 2x \text{M}0 + x \text{M}

Now, we solve for x:

     \begin{align*} {K &=\frac{[C]}{[A][B]^{2}}} \\  {K &= \frac{x}{(0.5-x)(0.5-2x)^{2}} = \frac{x}{(0.5-x)\left(4x^{2}-2x+0.25\right)}} \\ {K &= \frac{x}{2x^{2}-x+0.125-4x^{3}+2x^{2}-0.25x} = \frac{x}{-4x^{3}-1.25x+0.125}} \\ {0.4 &= \frac{x}{-4x^{3}-1.25x+0.125}} \\ {(0.4)\left(-4x^{3}-1.25x+0.125\right)&=x} \\ {-1.6x^{3}-0.5x+0.05&=x} \\ {-1.6x^{3}-1.5x+0.05&=x} \\ {0.03329&=x} \end{align*}

Now, we recalculate the equilibrium concentrations in the ICE table, using the newly found x value:

Initial0.5 \text{M}0.5 \text{M}0.0 \text{M}
Equilibrium0.5 - 0.03329 \text{M}0.5 - 0.06658 \text{M}0 + 0.03329 \text{M}

ICE table #2: Find the K value from known equilibrium concentrations.

     \begin{gather*} HA (\text{aq}) + H_{2}O (\text{l}) \leftrightarrows H_{3}O^{+} (\text{aq}) + A^{-} (\text{aq}) \end{gather*}


Initial1.000 \text{M}0.500 \text{M}0.200 \text{M}
Change- 0.150 \text{M}+ 0.150 \text{M}+ 0.150 \text{M}
Equilibrium0.850 \text{M}0.650 \text{M}0.350 \text{M}

     \begin{align*} {K&=\frac{[A^{-}][H_{3}O^{+}]}{[HA]}} \\ {K&=\frac{(0.350\text{M})(0.650\text{M})}{(0.850\text{M})}=\frac{(0.2275\text{M})}{(0.850\text{M})}} \\ {K&=0.268} \end{align*}

NOTE: these are purely theoretical examples, neither the concentrations nor K values are taken from real-life applications

ICE Table Practice Problems

Problem 1

Nitrogen monoxide forms through the following reaction:

     \begin{gather*} {N_{2} + O_{2} \rightleftarrows 2NO} \\ {K=4.2\cdot 10^{-8}} \end{gather*}

With initial concentrations  0.085\text{M}N_{2} and  0.038\text{M}O_{2}, what is the equilibrium concentration of nitrogen monoxide? Assume small x.

Problem 2

Consider the following reaction and initial concentrations:

     \begin{align*} {A + B &\rightleftarrows 2C} \end{align*}

     \begin{align*}  {[A]_{0} &= 0.50\text{M}} \\ {[B]_{0} &= 0.25\text{M}} \\ {[C]_{0} &= 1.0\text{M}} \end{align*}

At equilibrium, A has the following concentration:

     \begin{gather*} [A]_{eq} = 0.4875\text{M} \end{gather*}

What is the equilibrium concentration?

ICE Table Practice Problem Solutions

1:  1.2 \cdot 10^{-5} \text{M} \: \: NO

2:  K = 9.1

Further Reading