Write equilibrium constant expressions, interpret their magnitude, and calculate equilibrium concentrations using an ICE table

What this dot point is asking

For a general reaction $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$, the equilibrium constant in terms of concentration is

Products go on top, reactants on the bottom, and each concentration is raised to the power of its stoichiometric coefficient. Concentrations are in $\text{mol L}^{-1}$ and are the equilibrium values, not the starting values.

Two rules about what to include. Pure solids and pure liquids are left out of the expression because their concentration is effectively constant. For example, for $\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)$ the expression is just $K_c = [\text{CO}_2]$. Water as a solvent in dilute aqueous reactions is also omitted.

Interpreting the magnitude of K

The size of $K_c$ tells you where the position of equilibrium lies.

• $K_c \gg 1$ (large): the equilibrium mixture is mostly products; the forward reaction is favoured.
• $K_c \ll 1$ (small): the equilibrium mixture is mostly reactants; the reverse reaction is favoured.
• $K_c \approx 1$: significant amounts of both reactants and products are present.

$K_c$ depends only on temperature. It does not change when you alter concentrations, pressure, or add a catalyst, but it does change when the temperature changes.

The reaction quotient Q

The reaction quotient $Q$ has exactly the same form as $K_c$ but uses the concentrations at any moment, not necessarily at equilibrium. Comparing $Q$ to $K_c$ predicts the direction of net reaction.

• $Q < K_c$: too few products, so the reaction proceeds forwards (to the right).
• $Q > K_c$: too many products, so the reaction proceeds backwards (to the left).
• $Q = K_c$: the system is already at equilibrium; no net change.

Worked example: an ICE table

At a certain temperature, $1.00\ \text{mol}$ of $\text{H}_2$ and $1.00\ \text{mol}$ of $\text{I}_2$ are placed in a $2.00\ \text{L}$ flask. The reaction is

At equilibrium $1.56\ \text{mol}$ of $\text{HI}$ has formed. Find $K_c$.

Initial concentrations: $[\text{H}_2] = [\text{I}_2] = 1.00 / 2.00 = 0.500\ \text{mol L}^{-1}$, $[\text{HI}] = 0$.

Set up an ICE (Initial, Change, Equilibrium) table in concentration. The HI formed is $1.56 / 2.00 = 0.780\ \text{mol L}^{-1}$, so the change in HI is $+0.780$. From the 1:1:2 stoichiometry, $\text{H}_2$ and $\text{I}_2$ each change by $-0.390$.

• $[\text{H}_2]$ at equilibrium $= 0.500 - 0.390 = 0.110\ \text{mol L}^{-1}$
• $[\text{I}_2]$ at equilibrium $= 0.500 - 0.390 = 0.110\ \text{mol L}^{-1}$
• $[\text{HI}]$ at equilibrium $= 0.780\ \text{mol L}^{-1}$

Now substitute:

Because the moles of gas are equal on both sides, the volume cancels and you could have used moles directly here, but always convert to concentration as a habit so you do not slip on reactions where the volumes do not cancel.

Note that $K_c$ for this reaction has no units; in WACE you generally state $K_c$ as a number. Always quote the temperature, because $K_c$ is temperature-dependent.