How to use the simulator
The system starts at equilibrium with relative amounts N2 = 1.00, H2 = 1.80 and NH3 = 1.00, drawn as three bars with their values printed on top. Each button applies one stress, the bars animate, a banner reports Shift RIGHT (toward products) or Shift LEFT (toward reactants), and a message box gives the reason.
- Reaction is Exothermic / Endothermic: a toggle that sets the sign of . The reaction line updates to say whether heat is a product () or a reactant (). Exothermic is the default and is the real sign for ammonia synthesis.
- + N2, + H2, + NH3: inject more of one gas. Its bar jumps first, then all three bars adjust as the system responds.
- Raise T and Lower T: change the temperature. The response depends on the toggle.
- ↓ Volume and ↑ Volume: compress or expand the container.
- Reset: return to the starting equilibrium.
Read the bars in two stages. The first jump is the stress itself; the slower change that follows is the system's response. For the volume buttons all three bars jump together by the same factor, which stands for the sudden change in every concentration and partial pressure (the y-axis says mol, but compressing a container changes concentrations, not the number of moles present). Each press applies a fixed-size shift, so the simulator shows direction, not a calculated new equilibrium position. Predict the direction before you click, then compare with the banner.
The key ideas
A system at equilibrium has . A stress either changes Q (concentration or volume changes) or changes K itself (temperature changes). The system then reacts in whichever direction brings Q back to K. For this reaction
- Adding a reactant lowers Q below K, so the forward reaction runs: shift right. Adding a product raises Q above K: shift left. Removing a species does the opposite.
- Temperature is the only stress that changes K. Treat heat as a product for an exothermic reaction and as a reactant for an endothermic one. Heating an exothermic system shifts it left and lowers K; heating an endothermic system shifts it right and raises K.
- Volume: decreasing the volume raises every partial pressure, and the system shifts toward the side with fewer moles of gas. Here that is 2 mol of NH3 versus 4 mol of reactant gas, so compression shifts right.
- Not modelled here but tested: adding a catalyst changes no equilibrium amounts (it only speeds up reaching equilibrium), and adding an inert gas at constant volume changes no partial pressures, so neither causes a shift.
Worked example
Question. At equilibrium, extra NH3 is injected into the container at constant temperature and volume. Predict the shift and justify it in terms of Q and K.
Answer. Raising [NH3] increases the numerator of Q, so . The reverse reaction now runs faster than the forward reaction until Q falls back to K, so the system shifts left: NH3 is partly consumed and N2 and H2 increase. K does not change because the temperature did not change.
Check it in the simulator. Press Reset, then + NH3. The NH3 bar jumps from 1.00 to 2.20. The system then shifts left by 0.28 units of reaction: N2 rises by 0.28 to 1.28, H2 rises by to 2.64, and NH3 falls by to 1.64. The changes follow the 1 : 3 : 2 coefficients, and NH3 ends higher than its original 1.00: the shift only partly undoes the stress.
A second case. Press Reset and then ↓ Volume. All bars jump by a factor of 1.25 (to 1.25, 2.25, 1.25), then the system shifts right to 0.97, 1.41 and 1.81, consuming 4 moles of gas for every 2 it makes. Now switch to Endothermic and press Raise T: the shift is to the right, the reverse of what the exothermic setting gives.
Common mistakes on the AP exam
- Saying K changes after a concentration or volume change. Only temperature changes K. Concentration and volume stresses change Q.
- Justifying with "the system wants to" language alone. A full-credit justification compares Q with K, or rates of the forward and reverse reactions.
- Counting all moles, not gas moles. For pressure and volume stresses, count only gaseous species on each side. If both sides have equal moles of gas, a volume change causes no shift.
- Adding a solid or pure liquid. Extra solid or liquid does not appear in Q, so it causes no shift.
- Thinking the stress is fully reversed. After adding a product, the new equilibrium still has more of that product than before the stress.
When the AP exam uses this
Le Chatelier questions appear in Unit 7 (Topics 7.9 and 7.10) and return in acid-base chemistry, solubility and buffers. Expect particle diagrams before and after a stress, concentration-versus-time graphs where you must identify when a species was added, and free-response prompts asking you to predict a shift and justify it with Q and K.