Topic 7.4 Calculating the Equilibrium Constant: calculate the value of an equilibrium constant from equilibrium concentrations or pressures, using an ICE table where initial and equilibrium data are mixed.

What this topic is asking

The College Board (Topic 7.4) wants you to calculate the value of an equilibrium constant from equilibrium concentrations or pressures, using an ICE table when you are given initial amounts and one equilibrium value. This is the workhorse calculation of the unit: turning measured data into a value of $K$.

Direct calculation from equilibrium values

This is the simplest case: read off the equilibrium concentrations, plug them in, and compute. The arithmetic is straightforward; the only care needed is raising each concentration to its correct coefficient and keeping products over reactants.

The ICE table

The ICE table is needed when only the initial amounts and one equilibrium value are given. Because the changes all relate through the coefficients, one unknown $x$ describes the whole reaction. Solving for $x$ from the known equilibrium value fills in every entry, and the equilibrium row then goes into the $K$ expression.

Using the stoichiometry

The coefficients control the change row. For $\text{A} + 2\text{B} \rightleftharpoons 3\text{C}$, if A changes by $-x$ then B changes by $-2x$ and C by $+3x$. This proportionality is just conservation of atoms expressed in the change row. Getting the multiples and signs right is the most error-prone step, so always read them straight from the balanced equation.

It is worth being deliberate about which species you let define $x$. If you are given the equilibrium concentration of one product, set the change in that species equal to its coefficient times $x$ and solve for $x$ first; every other change then follows from the coefficients. If instead you are given how much of a reactant was consumed, that consumption is the change for the reactant, and you scale it by the coefficient ratios to get the others. The single unknown $x$ ties the whole table together, which is the great economy of the ICE method: no matter how many species are involved, one number describes the entire reaction's progress, and conservation of mass guarantees the changes stay in the ratio of the coefficients.

Try this

Q1. For $\text{A} \rightleftharpoons \text{B}$, a flask starts with $[\text{A}] = 0.50$ M; at equilibrium $[\text{B}] = 0.20$ M. Calculate $K_c$. [3 points]

• Cue. $[\text{A}]_\text{eq} = 0.50 - 0.20 = 0.30$ M; $K_c = \dfrac{0.20}{0.30} = 0.67$.

Q2. In an ICE table for $2\text{A} \rightleftharpoons \text{B}$, write the change in $[\text{A}]$ if $[\text{B}]$ changes by $+x$. [1 point]

• Cue. $-2x$ (two A consumed per B formed).