The Equilibrium Constant: Quick Revision

A brief note on The Equilibrium Constant. Read it before a test. Try each question at the end before you open its answer.

Equilibrium

• In a closed system, forward and backward processes reach equal rates. The concentrations then stay constant.
• Equilibrium is dynamic, and the same mixture is reached from either side.
• Physical examples: liquid and vapour, a saturated solution, a gas dissolved in a liquid (Henry's law).

The constants

• $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$: $K_c = \dfrac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}$, constant at a given temperature.
• $K_p = K_c(RT)^{\Delta n}$, with $R = 0.0831$ L bar K$^{-1}$ mol$^{-1}$ and $\Delta n$ the change in moles of gas.
• Pure solids and liquids are left out.
• Reverse the equation: $1/K$. Multiply it by n: $K^n$.
• K above about 1000: products favoured. Below about 0.001: reactants favoured.
• $\Delta G^\circ = -2.303\,RT\log K$.

Q and the direction

• $Q_c < K_c$: forward. $Q_c > K_c$: backward. $Q_c = K_c$: at equilibrium.

Le Chatelier's principle

• A disturbed equilibrium shifts to undo part of the change.
• Add a reactant or remove a product: forward.
• Higher pressure: towards fewer moles of gas.
• Heating: towards the endothermic direction. Only temperature changes K.
• A catalyst: equilibrium sooner, no shift.

A worked example

For $\text{A} + \text{B} \rightleftharpoons \text{C} + \text{D}$, $K_c = 9.0$. Start with 1.0 mol/L each of A and B. Find the equilibrium mixture.

1. ICE. At equilibrium, $[\text{A}] = [\text{B}] = 1.0 - x$ and $[\text{C}] = [\text{D}] = x$.
2. The equation. $\dfrac{x^2}{(1.0 - x)^2} = 9.0$, so $\dfrac{x}{1.0 - x} = 3.0$.
3. Solve. $x = 3.0 - 3.0x$, so $x = 0.75$ mol/L.
4. Answer. $[\text{A}] = [\text{B}] = 0.25$ mol/L and $[\text{C}] = [\text{D}] = 0.75$ mol/L.

Practice with new numbers

A program to try

The program works through the example above.

Mistakes to avoid

• Thinking the reactions stop at equilibrium.
• Putting solids or pure liquids into K.
• Using starting amounts in K instead of equilibrium amounts.
• Saying a catalyst gives more product.
• Saying a pressure change alters K. Only temperature does.

Check yourself

1. Write $K_p$ for $\text{C(s)} + \text{H}_2\text{O(g)} \rightleftharpoons \text{CO(g)} + \text{H}_2\text{(g)}$.

2. A reaction has $K_c = 0.20$, and a mixture has $Q_c = 0.050$. Which way does it go?

3. Does halving the volume shift $\text{H}_2 + \text{Br}_2 \rightleftharpoons 2\text{HBr}$ (all gases)?

4. An endothermic reaction is cooled. What happens to K?