![\begin{array}{r cl} K_c & = & \dfrac{[\text{I}_3^{-}]_e}{[\text{I}_2]_e[\text{I}^{-}]_e} \\[1em] & = & \dfrac{x}{{\{ (1.00 \times {{10}^{ - 3}})-x\} \{ (1.00 \times {{10}^{ - 3}})-x\} }} \\[1em] & = & \dfrac{3.39\;\times\;10^{-4}\;M}{(6.61\;\times\;10^{-4}\;M)(6.61\;\times\;10^{-4}\;M)} \\[1em] & = & 776 \end{array}](https://wisc.pb.unizin.org/app/uploads/quicklatex/quicklatex.com-4cf4ff4c085c43a7071470453cb18200_l3.png)
An ICE table (acronym for Initial concentration, Change in concentration, and Equilibrium concentration) is an useful methodology for calculating an equilibrium constant from experimental data. ICE tables also help to solve many other types of equilibrium problems.
An ICE table begins with the balanced reaction equation, using the reactants and products as column headings. The second row lists the initial concentrations of the reactants and products; these can usually be obtained from experimental data based on the assumption that no reaction has yet taken place. The third row is the change in concentration that occurs as the system proceeds toward equilibrium; this row is derived from the stoichiometry of the reaction. The last row is the sum of the above two rows, which yields the equilibrium concentrations.
For example, let’s determine the equilibrium constant for the reaction:
I2(aq) + I – (aq) ⇌ I3 – (aq)A solution initially has [I2]0 = [I – ]0 = 1.000 × 10 −3 M and no triiodide ions, [I3 – ]0 = 0. (The subscript “0” in […]0 clarifies that these are initial (at time = 0) concentrations.) When the reaction has reached equilibrium, we find that the equilibrium concentration of I2 is [I2]e = 6.61 × 10 −4 M. Again, the subscript “e” in [I2]e provides additional clarification that this is I2 equilibrium concentration. Because ICE table contains concentrations of reaction species at different stages of the reaction, inclusion of these subscripts can be helpful for keeping track of which concentration is which.
We can use an ICE table to help us determine the equilibrium constant for the reaction. First, write the balanced reaction at the top of the table. The next row (yellow) gives the initial concentrations.
| I2(aq) | + | I‾(aq) | ⇋ | I3‾(aq) |
| Initial concentration (M) | 1.000 × 10 -3 | 1.000 × 10 -3 | 0 | |
| Change in concentration (M) | –x | –x | +x | |
| Equilibrium concentration (M) | (1.000 × 10 -3 ) – x | (1.000 × 10 -3 ) – x | x |
In the “Change in concentration” row (green), the smallest change is represented by x and the mathematical sign indicates the direction of change: the sign is positive when the concentration increases as the reaction proceeds toward equilibrium, and negative when the concentration decreases. Here, because the reaction starts with no I3 – , [I2] must decrease as the reaction proceeds, so its (unknown) change in concentration is “-x“. Furthermore, x is multiplied by the stoichiometric coefficient, but in this case, all coefficients are 1.
In the “Equilibrium concentration” row (blue), we sum each column, initial concentration + change in concentration, to obtain the expression for equilibrium concentration for each reaction species.
Using the information that [I2]e = 6.61 × 10 −4 M, we can solve for x:
| [I2]e = (1.000 × 10 −3 ) – x | = | 6.61 × 10 −4 M |
| x | = | 3.39 × 10 −4 M |
and then calculate the equilibrium constant Kc:
Exercise: Concentration Changes During Reactions
