# Position of equilibrium

> IB Chemistry SL · Reactivity 3: Equilibrium and Organic Chemistry
> Source: https://www.owlsprep.com/study/ib-chemistry-sl-u6-position-of-equilibrium/

This module explains how to quantify the position of equilibrium, predict the direction a reaction shifts to reach equilibrium, and how changing reaction conditions alters equilibrium position for IB Chemistry SL.

**Prerequisites:** [Fundamentals of chemical equilibrium](https://www.owlsprep.com/study/ib-chemistry-sl-u6-introduction-to-equilibrium/)

## Learning objectives

- Distinguish between reaction quotient Q and equilibrium constant K
- Predict the direction of net reaction from Q vs K comparison
- Apply Le Chatelier's principle to predict equilibrium shifts
- Relate K magnitude and changes to equilibrium position

## What is the Position of Equilibrium?

**Position of equilibrium** — The relative concentrations of reactants and products in a system at dynamic equilibrium, where forward and reverse reaction rates are equal.

*Example:* If $K > 1$, position of equilibrium lies right (favours products); if $K < 1$, it lies left (favours reactants).

The position of equilibrium describes how far a reaction goes towards products before it reaches dynamic equilibrium, it does not describe how fast the reaction reaches equilibrium. A reaction can reach equilibrium quickly but still favour reactants.

**Worked example:** For the reaction $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$, $K_c = 1.7 \times 10^{-4}$ at 500 K. Comment on the position of equilibrium.

1. Write the equilibrium constant expression:
2. $$K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$$
3. Compare $K_c$ to 1: $1.7 \times 10^{-4} < 1$.
4. When $K_c < 1$, the denominator (reactant concentrations) is larger than the numerator (product concentrations).
5. Conclusion: The position of equilibrium lies far to the left, favouring reactants, so very little ammonia forms at equilibrium.

## Comparing Q and K to Predict Reaction Direction

**Reaction Quotient** — A value calculated using the same ratio as $K$, but using current concentrations of reactants and products at any point in the reaction (not just at equilibrium).

*Notation:* Q

To predict which direction a reaction will shift to reach equilibrium, compare $Q$ (current ratio) to $K$ (equilibrium ratio):

- $Q < K$: Reaction shifts right to make more products
- $Q > K$: Reaction shifts left to make more reactants
- $Q = K$: System is at equilibrium, no net change

**Worked example:** For the reaction $2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$, $K_c = 4.2$ at 600 °C. Current concentrations are $[\text{SO}_2] = 0.10$ M, $[\text{O}_2] = 0.20$ M, $[\text{SO}_3] = 0.30$ M. Predict the direction of reaction.

1. Write the expression for Q:
2. $$Q = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]}$$
3. Substitute current concentrations and calculate Q:
4. $$Q = \frac{(0.30)^2}{(0.10)^2 (0.20)} = 45$$
5. Compare Q to $K_c$: $45 > 4.2$, so $Q > K_c$.
6. Conclusion: The product concentration is too high, so the reaction shifts left to consume products and form more reactants to reach equilibrium.

## Le Chatelier's Principle for Equilibrium Shifts

**Le Chatelier's Principle** — If a change in condition is applied to a system at equilibrium, the position of equilibrium shifts in the direction that counteracts the applied change.

- Concentration: Adding a substance shifts equilibrium to consume the added substance; removing shifts to produce more of it.
- Pressure (gases only): Increasing pressure shifts to the side with fewer moles of gas; decreasing shifts to more moles of gas.
- Temperature: Increasing temperature shifts in the endothermic direction to absorb added heat; decreasing shifts in exothermic direction.

> **warning**
>
> Catalysts do not change the position of equilibrium or the value of $K$. They only speed up the rate at which equilibrium is reached.

**Worked example:** Predict how the position of equilibrium shifts when temperature is increased for this exothermic reaction: $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \quad \Delta H = -92 \text{ kJ mol}^{-1}$

1. Identify the endothermic direction: forward reaction is exothermic (releases heat), so reverse reaction is endothermic (absorbs heat).
2. Apply Le Chatelier's principle: increasing temperature adds heat, so equilibrium shifts to absorb the extra heat.
3. Conclusion: Equilibrium shifts left (towards reactants), so ammonia concentration decreases, and reactant concentrations increase.

## Effect of Changes on Equilibrium Constant K

Only changes in temperature alter the value of the equilibrium constant $K$. Changes in concentration, pressure, or adding a catalyst do not change $K$, because they do not alter the thermodynamics of the reaction, only the position of equilibrium shifts temporarily.

For exothermic forward reactions: increasing temperature decreases $K$ (equilibrium shifts left, so product ratio falls). For endothermic forward reactions: increasing temperature increases $K$ (equilibrium shifts right, product ratio rises).

**Worked example:** For the endothermic decomposition of calcium carbonate: $\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \quad \Delta H = +178 \text{ kJ mol}^{-1}$, what happens to $K_p$ when temperature is decreased?

1. Forward reaction is endothermic, so decreasing temperature removes heat. Equilibrium shifts in the exothermic direction (reverse) to counteract the change.
2. $K_p$ for this reaction is equal to the partial pressure of $\text{CO}_2$, because solids are excluded from equilibrium expressions: $K_p = P_{\text{CO}_2}$.
3. Shifting left reduces the partial pressure of $\text{CO}_2$, so $K_p$ decreases.
4. Conclusion: Decreasing temperature for an endothermic reaction decreases the value of $K$.

## Common pitfalls

- **Wrong:** Claiming adding a catalyst shifts the position of equilibrium
  - Why it fails: Catalysts lower activation energy of both forward and reverse reactions equally, so they do not change equilibrium concentrations
  - Correct: Always state catalysts have no effect on position of equilibrium or the value of K
- **Wrong:** Predicting all pressure changes shift equilibrium for gaseous reactions
  - Why it fails: If total moles of gas are equal on both sides, changing pressure changes all concentrations equally, so Q stays equal to K
  - Correct: First count moles of gas on each side; if equal, no shift occurs after pressure change
- **Wrong:** Reversing the direction of shift when comparing Q and K
  - Why it fails: If Q < K, the product ratio is too small, so reaction shifts right to make more products, not left
  - Correct: Remember the rule: $Q < K \rightarrow$ shift right, $Q > K \rightarrow$ shift left
- **Wrong:** Claiming changing concentration or pressure changes K
  - Why it fails: K is only temperature dependent for a given reaction, other changes do not alter its value
  - Correct: Only temperature changes change K; concentration/pressure changes only shift equilibrium
- **Wrong:** Mixing up K change for exothermic reactions when temperature increases
  - Why it fails: Increasing temperature shifts exothermic reactions left, which lowers the product/reactant ratio, so K decreases, not increases
  - Correct: Recall: increasing T increases K for endothermic, decreases K for exothermic

## Cheatsheet

| Change | Effect on equilibrium position | Effect on K |
| --- | --- | --- |
| Increase [reactant] | Shifts right | No change |
| Increase pressure (fewer moles right) | Shifts right | No change |
| Increase pressure (equal moles both sides) | No shift | No change |
| Increase T (exothermic forward) | Shifts left | Decreases K |
| Increase T (endothermic forward) | Shifts right | Increases K |
| Add catalyst | No shift | No change |

## What's next

Mastering position of equilibrium is a core foundation for all subsequent equilibrium topics in IB Chemistry SL, which make up a large portion of exam marks. You will apply the same Q vs K comparison and Le Chatelier's principle to acid-base equilibria, the most heavily tested sub-topic in this unit. This concept also extends to solubility equilibria, where you predict precipitation formation using the same reasoning. Building a strong understanding here will simplify all later equilibrium topics.

- [Introduction to organic chemistry](https://www.owlsprep.com/study/ib-chemistry-sl-u6-introduction-to-organic-chemistry/)
- [Functional group chemistry](https://www.owlsprep.com/study/ib-chemistry-sl-u6-functional-group-chemistry/)
- [Organic reaction mechanisms](https://www.owlsprep.com/study/ib-chemistry-sl-u6-organic-reaction-mechanisms/)

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