Theory — Chemical Equilibrium

Dynamic Equilibrium

A reversible reaction reaches equilibrium when the forward and reverse reactions occur at the same rate. The concentrations then stay constant — not because the reaction stops, but because the two opposing rates are balanced.

Equilibrium Constants Kc and Kp

For aA + bB ⇌ cC + dD, the equilibrium constant in terms of concentration is Kc; in terms of partial pressures of gases it is Kp.

Kc and Kp Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ
Kp = (P_C)ᶜ(P_D)ᵈ / (P_A)ᵃ(P_B)ᵇ
Kp = Kc (RT)^Δn,  Δn = (c + d) − (a + b),  R = 0.08206 L·atm/mol·K
If Δn = 0 (equal gas moles each side), Kp = Kc

Reaction Quotient Q

The reaction quotient Q has the same form as Kc but uses the current (not necessarily equilibrium) concentrations. Comparing Q with K predicts the direction of change.

Q vs K Q < K → reaction shifts right (toward products)
Q = K → at equilibrium (no net change)
Q > K → reaction shifts left (toward reactants)
The system always moves to make Q approach K

Le Châtelier's Principle

If a system at equilibrium is disturbed, it shifts to partly oppose the change: adding a species shifts away from it, raising temperature shifts in the endothermic direction, and raising pressure shifts toward the side with fewer gas moles.

Dynamic

Forward and reverse rates equal; concentrations constant but the reaction continues.

Kc / Kp

Ratios of products to reactants at equilibrium; related by Kp = Kc(RT)^Δn.

Q vs K

Q<K shifts right, Q>K shifts left, Q=K is equilibrium.

QuantityExpression
Kc[products]/[reactants], each to its coefficient
Kppartial pressures, each to its coefficient
RelationKp = Kc(RT)^Δn
Qsame as Kc but with current concentrations

Apparatus

The equipment a real chemical-equilibrium experiment uses. In the simulation these are modelled for you, but the readings correspond to what each instrument would measure.

holds samples
Test tube rack
Holds equilibrium mixtures whose colour shifts with conditions.
0.42 Ameasures absorbance
Colorimeter
Measures concentration from solution colour to find K.
reagent solutions
Reagent bottles
Add stress (more reactant or product) to test Le Chatelier.
holds solutions
Beaker
Holds the equilibrium solution for warming or cooling.
heats and stirs
Hot plate
Changes temperature to shift the equilibrium.
measures temperature
Thermometer
Records the temperature at which K is measured.

Instructions — Running the Virtual Experiment

The simulation has four sections. In the calculation sections, work out the value yourself first, then check it against the simulation and record both. Include screenshots in your report.

Part 1 — Dynamic Equilibrium (Equilibrium tab)
1
Open Simulation → Equilibrium. Set the starting amounts and run the reaction. Watch the forward and reverse rates approach each other and the concentrations level off. Record the equilibrium concentrations and note that the rates become equal.
Part 2 — Kc & Kp (Kc / Kp tab)
1
Open Kc / Kp. Choose a reaction and enter the equilibrium concentrations. Calculate Kc yourself first, then check. Set the temperature and calculate Kp using Kp = Kc(RT)^Δn, and check. Record both calculated and lab values.
Part 3 — Reaction Quotient (Q vs K tab)
1
Open Q vs K. Enter current concentrations and a value of K. Predict the direction of shift yourself first (compare Q with K), then check with the simulation.
Part 4 — Le Châtelier's Principle (Le Châtelier tab)
1
Open Le Châtelier. Apply each stress (add/remove a species, change temperature, increase or decrease pressure) and record the direction the equilibrium shifts and the reason.

Simulation — Chemical Equilibrium

Equilibrium Virtual LabDynamic · Kc/Kp · Q vs K · Le Châtelier

H₂ + I₂ ⇌ 2 HI

At equilibrium
[H₂] = [I₂]
[HI]
Forward rate
Reverse rate
At equilibrium the two rates are equal.
H₂(g) + I₂(g) ⇌ 2 HI(g)
Δn = 2 − 2 = 0, so Kp = Kc
Kc = [HI]² / ([H₂][I₂])

Equilibrium data

Result
Kc— hidden
Δn— hidden
(RT)^Δn— hidden
Kp— hidden
Calculate Kc and Kp yourself, enter them, then check.
H₂(g) + I₂(g) ⇌ 2 HI(g)
Q
K
Enter values

Current state — H₂ + I₂ ⇌ 2HI

Result
Q— hidden
K— hidden
Direction— hidden
Work out Q = [HI]² / ([H₂][I₂]), predict the direction, then check.
N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g)
Forward reaction is exothermic (ΔH < 0)
reactants
products
No shift yet

Equilibrium

Apply a stress

Effect
Stress
Shift
Apply a stress to see which way the equilibrium shifts.

Team Questions

Question 1. At equilibrium, what is true about the forward and reverse reaction rates? (Answer in a phrase)
Question 2. For H₂ + I₂ ⇌ 2HI at equilibrium, [H₂] = 0.20, [I₂] = 0.20, [HI] = 1.60 M. Calculate Kc. (Type to 1 decimal)
Question 3. For the reaction in Q2, what is Δn (moles of gas products − reactants), and what does that make Kp compared with Kc?
Question 4. For N₂ + 3H₂ ⇌ 2NH₃, what is Δn? (Type a number, e.g. -2)
Question 5. If Q = 10 and K = 50 for a reaction, which way does it shift? (left or right)
Question 6. For N₂ + 3H₂ ⇌ 2NH₃ (exothermic), which way does the equilibrium shift if you raise the temperature? (left or right)
Question 7 — Challenge. For N₂O₄ ⇌ 2NO₂, which way does decreasing the pressure shift the equilibrium? (left or right — compare gas moles)

Example Lab Report

Sample report demonstrating the expected format. Include your equilibrium data, the Kc and Kp calculations, the Q-vs-K predictions, and your Le Châtelier observations, with labelled screenshots.

Chemical Equilibrium

Chemistry | Section: [Your Section] | Date: [Date]

Lab Members: [Names of all members present]

Purpose

To observe equilibrium as a dynamic balance of forward and reverse rates, to calculate Kc and Kp and relate them, to use the reaction quotient Q to predict the direction of change, and to apply Le Châtelier's principle.

Theory

At equilibrium the forward and reverse rates are equal. Kc and Kp are the equilibrium constants in terms of concentration and partial pressure, related by Kp = Kc(RT)^Δn. Comparing Q with K predicts the direction of net reaction.

Kc = [products]/[reactants] · Kp = Kc(RT)^Δn · Q vs K → direction

Results (worked example)

H₂ + I₂ ⇌ 2HI at equilibrium: [H₂] = 0.20, [I₂] = 0.20, [HI] = 1.60 M.

Kc = (1.60)² / (0.20 × 0.20) = 2.56 / 0.040 = 64.0
Δn = 2 − 2 = 0 → Kp = Kc(RT)⁰ = Kc = 64.0

N₂ + 3H₂ ⇌ 2NH₃ with Kc = 0.25 at 298 K: Δn = 2 − 4 = −2.

Kp = 0.25 × (0.08206 × 298)⁻² = 0.25 × (24.45)⁻² = 4.18 × 10⁻⁴

Q vs K: with K = 50 and Q = 10, Q < K so the reaction shifts right. Le Châtelier (N₂ + 3H₂ ⇌ 2NH₃): adding N₂ → right; heating → left; increasing pressure → right.

Analysis

The equilibrium concentrations stayed constant once the forward and reverse rates became equal, confirming a dynamic balance. The calculated Kc and Kp matched the simulation values, and Kp equalled Kc whenever Δn was zero. Q correctly predicted the direction of shift, and all Le Châtelier predictions agreed with the simulation.

Conclusion

Equilibrium is dynamic, characterised by constant Kc and Kp; the reaction quotient predicts the direction of change; and Le Châtelier's principle correctly anticipates how concentration, temperature, and pressure changes shift the position of equilibrium.

Practice Questions

Show all work. Use Kc = [products]/[reactants], Kp = Kc(RT)^Δn with R = 0.08206, and compare Q with K.

Question 1
For H₂ + I₂ ⇌ 2HI, equilibrium concentrations are [H₂] = 0.10, [I₂] = 0.40, [HI] = 1.60 M. Calculate Kc.
Hint: Kc = (1.60)²/(0.10 × 0.40) = 2.56/0.040 = 64.
Question 2
A gas reaction has Kc = 1.5 and Δn = +1 at 350 K. Calculate Kp.
Hint: Kp = 1.5 × (0.08206 × 350)¹ = 1.5 × 28.7 = 43.1.
Question 3
Explain why Kp = Kc when Δn = 0, using the relationship Kp = Kc(RT)^Δn.
Hint: any non-zero quantity raised to the power 0 equals 1.
Question 4
For a reaction with K = 25, the current quotient is Q = 60. In which direction will the reaction proceed, and why?
Hint: Q > K, so it shifts left (toward reactants) to reduce Q.
Question 5
For 2SO₂ + O₂ ⇌ 2SO₃ (exothermic), predict the shift when you (a) add O₂, (b) raise the temperature, and (c) increase the pressure.
Hint: (a) right, (b) left, (c) right (3 gas moles → 2).
Question 6 — Challenge
For N₂O₄ ⇌ 2NO₂, explain how increasing and decreasing the pressure each affect the position of equilibrium.
Hint: increasing pressure shifts toward fewer moles (left); decreasing shifts toward more moles (right).