D19.6 ICE Table
In many cases it is not necessary to know all equilibrium concentrations or partial pressures to calculate an equilibrium constant. Often chemical stoichiometry can be used to determine relationships among equilibrium concentrations or partial pressures. From stoichiometric relationships and experimental data, one can solve mathematically for the equilibrium constant. An ICE table (acronym for Initial concentration/partial pressure, Change in concentration/partial pressure, and Equilibrium concentration/partial pressure) helps organize such calculations. ICE tables also help to solve many other types of equilibrium problems.
For example, let’s determine the equilibrium constant for the reaction:
Experimentally we find that the 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 emphasizes that this is the I2 equilibrium concentration. Because an 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“. (In general x is multiplied by the stoichiometric coefficient, but in this case, all coefficients are 1 so we just have −x.).
To get 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} M | = | 6.61 × 10−4 M |
| x | = | 3.39 × 10−4 |
and then calculate the equilibrium constant K:
![\begin{array}{r cl} K & = & \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}}{(6.61\;\times\;10^{-4})(6.61\;\times\;10^{-4})} \\[1em] & = & 776 \end{array}](https://wisc.pb.unizin.org/app/uploads/quicklatex/quicklatex.com-a193107d146cdf287af17540e6ecee3f_l3.png "Rendered by QuickLaTeX.com")
Exercise: Concentration Changes During Reactions
Activity: Calculating Equilibrium Partial Pressures I
Activity: Calculating Equilibrium Partial Pressures II
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