Direction of reversible reactions

AP Chemistry· AP Chemistry CED — Equilibrium· 14 min read

1. Core Concept: Net Direction of Reversible Reactions★★☆☆☆⏱ 3 min

Reversible reactions proceed simultaneously in both forward (reactants converting to products) and reverse (products converting to reactants) directions. Net direction describes which overall change will occur when a system is not at equilibrium, or after a disturbance to an existing equilibrium. At equilibrium, forward and reverse reaction rates are equal, so there is no net change in concentrations.

📘 Definition

Net Reaction Direction

shift right = net forward, shift left = net reverse

The overall net change in concentrations of reactants and products that occurs as a non-equilibrium system progresses toward a new equilibrium state.

Example:

A system with too few products will have a net forward direction (shift right) to produce more products.

2. Quantitative Prediction: Comparing Q and K★★★☆☆⏱ 4 min

The most rigorous method to determine net direction is comparing the reaction quotient ( $Q$ ), calculated from current non-equilibrium concentrations, to the equilibrium constant $K$ , which is fixed at a given temperature. $Q$ uses the exact same expression form as $K$ , only substituting current instead of equilibrium values.

aA + bB \rightleftharpoons cC + dD \\ Q = \frac{[C]^c[D]^d}{[A]^a[B]^b}

If $Q < K$ : Current product ratio is too low → net forward (shift right)
If $Q = K$ : System is at equilibrium → no net change
If $Q > K$ : Current product ratio is too high → net reverse (shift left)

📐 Worked Example

For the gas-phase reaction $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$ , $K_{p} = 0.36$ at 25°C. The current partial pressures are $P_{N_{2} O_{4}} = 0.20$ atm, $P_{N O_{2}} = 0.15$ atm. Predict the net direction as the system approaches equilibrium.

1
Write the correct $Q_{p}$ expression matching the balanced equation:
2
$Q_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}}$
3
Substitute current partial pressures and calculate $Q_{p}$ :
4
$Q_p = \frac{(0.15)^2}{0.20} = 0.1125$
5
Compare $Q_{p}$ to $K_{p}$ and apply the Q vs K rule: $0.1125 < 0.36$ , so $Q < K$ . Net reaction proceeds forward (shifts right) to produce more $N O_{2}$ .

Exam tip:

Always confirm your expression matches the balanced reaction stoichiometry

3. Qualitative Prediction: Le Chatelier's Principle★★★☆☆⏱ 4 min

Le Chatelier's principle is a qualitative rule that predicts how an equilibrium system shifts after a disturbance. It states that when a system at equilibrium is disturbed by a change to concentration, pressure/volume, or temperature, the system will shift in the net direction that counteracts the disturbance.

Concentration changes: Adding a reactant/removing a product shifts right; adding a product/removing a reactant shifts left. $K$ never changes.
Pressure/volume changes (gases only): Increasing pressure (decreasing volume) shifts to the side with fewer moles of gas; decreasing pressure (increasing volume) shifts to the side with more moles of gas. Equal moles → no shift. $K$ never changes.
Temperature changes: Only temperature changes alter the value of $K$ . For endothermic ( $Δ H > 0$ , heat = reactant): increasing T shifts right, $K$ increases. For exothermic ( $Δ H < 0$ , heat = product): increasing T shifts left, $K$ decreases.

📐 Worked Example

The reaction $2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$ $Δ H = - 198$ kJ/mol is at equilibrium. Predict the net shift after each disturbance: (a) add more $O_{2}$ , (b) increase temperature, (c) decrease total pressure by increasing container volume.

1
(a) $O_{2}$ is a reactant. Adding a reactant causes the system to shift right (forward) to consume the added $O_{2}$ , per Le Chatelier's principle.
2
(b) $Δ H$ is negative, so the reaction is exothermic, and heat is a product of the forward reaction. Increasing temperature adds heat, so the system shifts left (reverse) to consume the excess heat.
3
(c) Count moles of gas: 3 moles on the reactant side, 2 moles on the product side. Decreasing pressure shifts to the side with more moles of gas, so net shift left (reverse).

4. Special Cases: Inert Gas Pressure Disturbances★★★★☆⏱ 3 min

A pressure change only causes an equilibrium shift if it changes the partial pressures of reactant and product gases. Two commonly tested special cases involve adding inert (unreactive) gases that do not participate in the reaction:

Inert gas at constant volume: Total pressure increases, but partial pressures of reacting gases do not change → $Q$ remains equal to $K$ → no shift.
Inert gas at constant total pressure: Volume must increase to keep total pressure constant, which decreases partial pressures of all reacting gases → shift follows the mole rule (to the side with more moles of gas).

📐 Worked Example

The reaction $H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)$ is at equilibrium in a rigid 1.0 L container at 25°C. Predict the net shift when 0.5 mol of argon (inert) is added at constant temperature.

1
The container is rigid, so volume remains constant after adding argon.
2
Moles and partial pressures of all reacting gases ( $H_{2}$ , $I_{2}$ , $H I$ ) do not change, so $Q$ is still equal to $K$ .
3
Conclusion: the system stays at equilibrium, no net shift occurs.

5. AP Style Worked Practice Problems★★★★☆⏱ 6 min

📐 Worked Example

Multiple Choice: For the reaction $A g C l (s) ⇌ A g^{+} (a q) + C l^{-} (a q)$ , $K_{s p} = 1.8 \times 1 0^{- 10}$ at 25°C. A solution has $[A g^{+}] = 1.0 \times 1 0^{- 5}$ M and $[C l^{-}] = 2.0 \times 1 0^{- 5}$ M. Which correctly describes the net reaction direction?

1
Calculate $Q$ (ion product, equivalent to reaction quotient for solubility):
2
$Q = [Ag^+][Cl^-] = (1.0 \times 10^{-5})(2.0 \times 10^{-5}) = 2.0 \times 10^{-10}$
3
Compare $Q$ to $K_{s p}$ : $2.0 \times 1 0^{- 10} > 1.8 \times 1 0^{- 10}$ , so $Q > K$ . For this reaction, forward is dissolution of solid AgCl to ions, reverse is precipitation of solid AgCl. $Q > K$ means too many ions, so net reverse reaction (precipitation) occurs. Correct answer: A.

📐 Worked Example

Free Response: The Haber process reaction is $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g) Δ H = - 92$ kJ/mol, initially at equilibrium. For each disturbance, predict shift, state if $K$ increases/decreases/unchanged, and justify: (a) add more $N_{2}$ at constant volume and temperature, (b) increase temperature at constant volume, (c) decrease container volume at constant temperature.

1
(a) Net shift: right (forward). $K$ stays the same. Justification: Adding $N_{2}$ (a reactant) disturbs equilibrium, so the system shifts right to consume the added $N_{2}$ per Le Chatelier's principle. Temperature is unchanged, so $K$ does not change.
2
(b) Net shift: left (reverse). $K$ decreases. Justification: The reaction is exothermic, so heat is a product. Increasing temperature adds heat, so the system shifts left to consume excess heat. Shifting left lowers the product/reactant ratio, so $K$ decreases.
3
(c) Net shift: right (forward). $K$ stays the same. Justification: Decreasing volume increases all partial pressures. There are 4 moles of gas on reactant side, 2 on product side, so the system shifts right to reduce total moles of gas, counteracting the pressure increase. Temperature is unchanged, so $K$ does not change.

6. Common Pitfalls

Wrong move:

Claiming $Q < K$ means the system shifts left

Why:

Confusion over the product-to-reactant ratio leads to flipped direction predictions

Correct move:

Use the mnemonic: Q smaller than K = need more Products = shift Right (Q<K→P→R) to remember the rule correctly

Wrong move:

Claiming any disturbance that causes a shift also changes the value of $K$

Why:

Confusing temperature changes with concentration/pressure changes, which all cause shifts but do not alter K

Correct move:

Memorize that only temperature changes change the value of $K$ ; all other disturbances leave K unchanged

Wrong move:

Counting moles of solids/liquids when predicting shifts from pressure changes

Why:

Students incorrectly count all moles instead of only gaseous moles when applying the pressure shift rule

Correct move:

Only count moles of gaseous reactants and products when predicting shifts from pressure/volume changes

Wrong move:

Claiming adding an inert gas always causes an equilibrium shift

Why:

Students forget that constant volume inert gas addition does not change partial pressures of reacting species

Correct move:

For any inert gas addition, first confirm if volume is constant (no shift) or pressure is constant (shift follows mole count of gaseous species)

Wrong move:

Claiming increasing temperature shifts an exothermic reaction to the right

Why:

Students forget that heat is a product for exothermic forward reactions

Correct move:

Always add heat to the reaction equation (e.g. reactants → products + heat for exothermic) before applying Le Chatelier's principle

7. Quick Reference Cheatsheet

Category	Rule / Relationship	Key Notes
Reaction Quotient	$Q = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}$	Products in numerator, uses current non-equilibrium concentrations, same form as $K$
Q vs K Direction Rule	$Q < K$ : Shift right (forward) $Q = K$ : At equilibrium $Q > K$ : Shift left (reverse)	Quantitative prediction for all reversible systems
Concentration Change (Le Chatelier)	Add reactant/remove product → shift right Add product/remove reactant → shift left	$K$ does not change
Pressure/Volume Change (Le Chatelier)	Increase P/decrease V → shift to fewer moles of gas Decrease P/increase V → shift to more moles of gas	Count only gaseous moles. Equal moles → no shift. $K$ does not change
Temperature Change (Le Chatelier)	Endothermic ( $Δ H > 0$ ): increase T → shift right, $K$ increases Exothermic ( $Δ H < 0$ ): increase T → shift left, $K$ decreases	Only temperature change changes $K$
Inert Gas Addition	Constant volume → no shift Constant pressure → volume increases, shift follows mole rule	$K$ does not change. Only partial pressure changes cause shifts
Solution Dilution	Dilution → shift to more moles of dissolved solute	Same logic as gas volume change. Equal moles → no shift. $K$ does not change

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Q vs K direction prediction
2022 · FRQ
Le Chatelier shift justification

Going deeper

unit overviewAP Chemistry Unit 7 Equilibrium OverviewParent unit for this topic

What's Next

This topic is the foundational prerequisite for all further equilibrium topics in AP Chemistry. Next, you will apply direction prediction rules to build ICE tables for calculating equilibrium concentrations, which requires correctly assigning signs to concentration changes based on the predicted net direction. Without mastering direction prediction here, you will get incorrect signs for concentration changes, leading to wrong equilibrium results and lost points on both MCQ and FRQ questions. This topic also underpins all subsequent equilibrium applications including acid-base equilibrium, solubility equilibrium, and reaction thermodynamics, where you will regularly use Q vs K comparisons.