🔄 The Reaction Quotient Q

Part 1 of 7 — Same Expression as K, but at Any Time

The reaction quotient $Q$ has the exact same mathematical form as the equilibrium constant $K$, but it uses current concentrations (or pressures) rather than equilibrium values. It tells us where the system is relative to equilibrium.

Defining Q

For the general reaction:

$aA + bB \rightleftharpoons cC + dD$

$Q_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \quad \text{(using current concentrations)}$

$Q_p = \frac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b} \quad \text{(using current pressures)}$

Key Distinction

Q at Special Times

• At $t = 0$ (only reactants): $Q = 0$ (numerator = 0)
• At equilibrium: $Q = K$
• If only products present: $Q = \infty$ (denominator = 0)

Calculating Q

Example

For: $\text{N}_2(g) + 3\,\text{H}_2(g) \rightleftharpoons 2\,\text{NH}_3(g)$, $K_c = 0.50$ at 400°C

Current concentrations: $[\text{N}_2] = 1.0$ M, $[\text{H}_2] = 2.0$ M, $[\text{NH}_3] = 3.0$ M

$Q_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} = \frac{(3.0)^2}{(1.0)(2.0)^3} = \frac{9.0}{8.0} = 1.125$

Since $Q_c = 1.125 > K_c = 0.50$:

• The system has too many products relative to equilibrium
• The reaction will shift to the left (toward reactants) to reach equilibrium

The Rules for Solids and Liquids

Just like with $K$, pure solids and pure liquids are excluded from the $Q$ expression.

Understanding Q 🎯

Calculating Q 🧮

For the reaction: $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\,\text{HI}(g)$, $K_c = 50$ at 448°C.

Current concentrations: $[\text{H}_2] = 0.10$ M, $[\text{I}_2] = 0.10$ M, $[\text{HI}] = 0.50$ M

1. Calculate $Q_c$. (Enter as a whole number)

2. Is $Q > K$, $Q < K$, or $Q = K$? (Enter "Q > K", "Q < K", or "Q = K")

3. If you start with only reactants and no products, what is the initial value of Q? (Enter as a number)

Q Concepts 🔍

Exit Quiz — Reaction Quotient ✅

🔄 Comparing Q and K

Part 2 of 7 — Predicting the Direction of Shift

By comparing Q to K, you can predict exactly which direction a reaction will shift to reach equilibrium. This is one of the most powerful tools in equilibrium chemistry.

The Three Cases

Case 1: $Q < K$ — Shift Right (→)

$Q < K \implies \frac{\text{products}}{\text{reactants}} < \text{equilibrium ratio}$

• There are too few products (or too many reactants)
• The system shifts right (forward) to make more products
• Q increases until $Q = K$

Case 2: $Q > K$ — Shift Left (←)

$Q > K \implies \frac{\text{products}}{\text{reactants}} > \text{equilibrium ratio}$

• There are too many products (or too few reactants)
• The system shifts left (reverse) to make more reactants
• Q decreases until $Q = K$

Case 3: $Q = K$ — At Equilibrium

$Q = K \implies \text{system is at equilibrium}$

• No net change occurs
• Forward and reverse rates are equal

Memory Aid

Think of Q as "chasing" K:

• $Q < K$: Q needs to increase → more products → shift right
• $Q > K$: Q needs to decrease → more reactants → shift left

Visual Summary

$\underbrace{Q = 0}_{\text{pure reactants}} \quad \xleftarrow{\text{shift right}} \quad Q < K \quad \longrightarrow \quad \underbrace{Q = K}_{\text{EQUILIBRIUM}} \quad \longleftarrow \quad Q > K \quad \xrightarrow{\text{shift left}} \quad \underbrace{Q = \infty}_{\text{pure products}}$

Worked Example

$\text{CO}(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}_2(g) + \text{H}_2(g)$, $K_c = 5.0$ at 700 K

Given: $[\text{CO}] = 0.10$, $[\text{H}_2\text{O}] = 0.10$, $[\text{CO}_2] = 0.20$, $[\text{H}_2] = 0.20$ M

$Q_c = \frac{[\text{CO}_2][\text{H}_2]}{[\text{CO}][\text{H}_2\text{O}]} = \frac{(0.20)(0.20)}{(0.10)(0.10)} = \frac{0.04}{0.01} = 4.0$

Since $Q = 4.0 < K = 5.0$:

• The system shifts right to produce more CO₂ and H₂
• $[\text{CO}]$ and $[\text{H}_2\text{O}]$ will decrease
• $[\text{CO}_2]$ and $[\text{H}_2]$ will increase

Predicting the Direction of Shift 🎯

Q vs K Calculations 🧮

For: $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\,\text{NO}_2(g)$, $K_c = 0.36$ at 100°C

Current state: $[\text{N}_2\text{O}_4] = 0.50$ M, $[\text{NO}_2] = 0.20$ M

1. Calculate $Q_c$. (Enter as a decimal to 2 places)

2. Does the reaction shift right or left? (Enter "right" or "left")

3. At equilibrium, will $[\text{NO}_2]$ be higher or lower than 0.20 M? (Enter "higher" or "lower")

Round all answers to 3 significant figures.

Q vs K — Quick Concepts 🔍

Exit Quiz — Comparing Q and K ✅

🔄 Le Chatelier's Principle — Concentration Changes

Part 3 of 7 — How the System Responds to Stress

Le Chatelier's Principle states: When a system at equilibrium is subjected to a stress, the system will shift in the direction that partially relieves that stress. In this part, we focus on concentration changes.

Adding or Removing Species

Adding Reactant → Shift Right

For: $\text{N}_2(g) + 3\,\text{H}_2(g) \rightleftharpoons 2\,\text{NH}_3(g)$

If we add more N₂:

• $[\text{N}_2]$ increases immediately
• Q decreases (denominator gets bigger) → $Q < K$
• System shifts right to consume the added N₂
• At the new equilibrium: $[\text{NH}_3]$ is higher, $[\text{H}_2]$ is lower

Adding Product → Shift Left

If we add more NH₃:

• $[\text{NH}_3]$ increases immediately
• Q increases (numerator gets bigger) → $Q > K$
• System shifts left to consume the added NH₃
• At the new equilibrium: $[\text{N}_2]$ and $[\text{H}_2]$ are higher

Removing a Species → Opposite Shift

Key Insight

The system shifts to partially counteract the change. It never fully restores the original concentrations — it finds a new equilibrium position.

Worked Example

$\text{CO}(g) + 2\,\text{H}_2(g) \rightleftharpoons \text{CH}_3\text{OH}(g)$, $K_c = 14.5$

Original equilibrium: $[\text{CO}] = 0.20$, $[\text{H}_2] = 0.30$, $[\text{CH}_3\text{OH}] = 0.26$ M

Verify: $Q = \frac{0.26}{(0.20)(0.30)^2} = \frac{0.26}{0.018} = 14.4 \approx K$ ✓

Stress: Add CO to make $[\text{CO}] = 0.40$ M

Immediate Q: $Q = \frac{0.26}{(0.40)(0.30)^2} = \frac{0.26}{0.036} = 7.2$

Since $Q = 7.2 < K = 14.5$: the system shifts right.

At the new equilibrium:

• $[\text{CO}]$ is higher than 0.20 but lower than 0.40 (some consumed)
• $[\text{H}_2]$ is lower than 0.30 (consumed)
• $[\text{CH}_3\text{OH}]$ is higher than 0.26 (produced)

Le Chatelier — Concentration 🎯

Predicting Concentration Changes 🧮

For: $\text{H}_2(g) + \text{Cl}_2(g) \rightleftharpoons 2\,\text{HCl}(g)$ at equilibrium

1. If $\text{Cl}_2$ is added, the reaction shifts which direction? (Enter "right" or "left")

2. If HCl is removed, the reaction shifts which direction? (Enter "right" or "left")

3. If $\text{H}_2$ is removed and $\text{Cl}_2$ is simultaneously added, the reaction shifts which direction? (Enter "right" or "left")

Concentration Stress Effects 🔍

Exit Quiz — Concentration Stresses ✅

🔄 Le Chatelier — Pressure/Volume and Temperature

Part 4 of 7 — How Gases and Heat Affect Equilibrium

Pressure/volume changes affect gaseous equilibria by changing concentrations. Temperature changes are unique because they actually change the value of K.

Pressure and Volume Changes

The Rule

For gaseous equilibria, when volume decreases (pressure increases):

• The system shifts toward the side with fewer moles of gas
• This reduces the total number of gas molecules, partially relieving the pressure

When volume increases (pressure decreases):

• The system shifts toward the side with more moles of gas

Why?

Decreasing volume concentrates all species equally. The side with more moles of gas is affected more. Shifting toward fewer moles relieves the pressure stress.

Example

$\text{N}_2(g) + 3\,\text{H}_2(g) \rightleftharpoons 2\,\text{NH}_3(g)$

• Reactant side: $1 + 3 = 4$ moles of gas
• Product side: $2$ moles of gas

Special Case: Equal Moles

$\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\,\text{HI}(g) \quad (\Delta n = 0)$

Both sides have 2 moles of gas. No shift occurs with pressure/volume changes.

Temperature Changes

Temperature is unique — it's the only stress that changes the value of $K$.

The "Heat as a Species" Trick

Treat heat as a reactant or product:

Exothermic ($\Delta H < 0$): Heat is a product $A \rightleftharpoons B + \text{heat}$

Endothermic ($\Delta H > 0$): Heat is a reactant $\text{heat} + A \rightleftharpoons B$

Effect of Temperature Changes

Example

$\text{N}_2(g) + 3\,\text{H}_2(g) \rightleftharpoons 2\,\text{NH}_3(g) \quad \Delta H = -92$ kJ (exothermic)

$\text{N}_2 + 3\,\text{H}_2 \rightleftharpoons 2\,\text{NH}_3 + \text{heat}$

• Increase T → adds heat → shifts left → K decreases
• Decrease T → removes heat → shifts right → K increases

Key Point

Pressure, volume, and concentration changes shift the equilibrium position but do NOT change K. Temperature changes BOTH the position AND the value of K.

Pressure, Volume, and Temperature 🎯

Predicting Shifts 🧮

For: $\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)$, $\Delta H = +87$ kJ (endothermic)

1. If the volume is decreased, the reaction shifts which direction? (Enter "right" or "left")

2. If the temperature is increased, the reaction shifts which direction? (Enter "right" or "left")

3. Does increasing temperature increase or decrease K for this reaction? (Enter "increase" or "decrease")

Pressure & Temperature Effects 🔍

Exit Quiz — Pressure and Temperature ✅

🔄 Catalysts and Inert Gas Addition

Part 5 of 7 — Stresses That Don't Shift Equilibrium

Not every change to a system causes an equilibrium shift. Two important cases: catalysts and inert gas addition (at constant volume).

Catalysts and Equilibrium

What Catalysts Do

A catalyst speeds up a reaction by providing an alternative pathway with a lower activation energy ($E_a$).

Key Facts About Catalysts at Equilibrium

Why No Shift?

A catalyst speeds up both the forward and reverse reactions by the same factor. Since both rates increase equally, the ratio of rates (and hence the equilibrium position) doesn't change.

What Catalysts ARE Useful For

• Reaching equilibrium faster
• Making a slow reaction practical (e.g., the Haber process uses an iron catalyst)
• Allowing equilibrium to be reached at lower temperatures (which may favor products for exothermic reactions)

Example

The Haber process: $\text{N}_2 + 3\,\text{H}_2 \rightleftharpoons 2\,\text{NH}_3$

Without a catalyst, this reaction is impractically slow at lower temperatures. The iron catalyst allows the reaction to reach equilibrium quickly at moderate temperatures (~450°C), which is a compromise between speed and yield.

Adding an Inert Gas

An inert (noble) gas does not react with any species in the equilibrium. Its effect depends on the conditions:

At Constant Volume

Adding inert gas at constant volume:

• Increases total pressure
• Does NOT change the partial pressures of any reacting species
• Does NOT change concentrations
• Q is unchanged → No shift
• K is unchanged

At Constant Pressure

Adding inert gas at constant pressure:

• The container must expand to maintain constant pressure
• This effectively increases the volume
• All partial pressures of reacting species decrease
• This is equivalent to a volume increase → shifts toward more moles of gas

Summary

AP Exam Note

On the AP exam, "adding an inert gas" typically means at constant volume unless stated otherwise. The answer is usually no effect.

Catalysts and Inert Gas 🎯

Complete Summary of All Stresses

Stress Identification 🔍

Quick Checks 🧮

1. Does a catalyst change the value of K? (Enter "no")

2. Does adding an inert gas at constant volume shift the equilibrium? (Enter "no")

3. If a catalyst is removed from a system at equilibrium, does the position shift? (Enter "no")

Exit Quiz — Catalysts and Inert Gas ✅

🧮 Problem-Solving Workshop

Part 6 of 7 — Q, K, and Le Chatelier Calculations

This workshop combines Q vs K comparisons with Le Chatelier's principle predictions. These multi-step problems mirror AP exam formats.

Problem-Solving Strategy

For Q vs K Problems

1. Write the $Q$ expression (same form as $K$)
2. Plug in the current concentrations or pressures
3. Compare $Q$ to $K$:
   • $Q < K$ → shift right
   • $Q > K$ → shift left
   • $Q = K$ → at equilibrium

For Le Chatelier Problems

1. Identify the stress (concentration, pressure/volume, temperature, catalyst, inert gas)
2. Predict the direction of shift:
   • Concentration: shift away from added species
   • Volume: shift toward side with more/fewer moles of gas
   • Temperature: treat heat as a species
   • Catalyst/inert gas at const V: no shift
3. Determine effect on each species' concentration
4. Determine effect on K (only temperature changes K)

Worked Example 1

$\text{CO}(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}_2(g) + \text{H}_2(g)$, $K_c = 5.10$ at 700 K

A flask contains: $[\text{CO}] = 0.200$, $[\text{H}_2\text{O}] = 0.300$, $[\text{CO}_2] = 0.400$, $[\text{H}_2] = 0.500$ M

$Q = \frac{(0.400)(0.500)}{(0.200)(0.300)} = \frac{0.200}{0.060} = 3.33$

$Q = 3.33 < K = 5.10$ → shift right

At the new equilibrium:

• $[\text{CO}]$ decreases, $[\text{H}_2\text{O}]$ decreases
• $[\text{CO}_2]$ increases, $[\text{H}_2]$ increases

Practice Problem 1 🧮

$\text{N}_2\text{O}_4(g) \rightleftharpoons 2\,\text{NO}_2(g)$, $K_c = 4.60 \times 10^{-3}$

Current: $[\text{N}_2\text{O}_4] = 0.500$ M, $[\text{NO}_2] = 0.100$ M

1. Calculate Q. (Enter as a decimal to 2 places)

2. Is $Q > K$, $Q < K$, or $Q = K$? (Enter exactly, e.g. "Q > K")

3. Which direction does the reaction shift? (Enter "right" or "left")

Round all answers to 3 significant figures.

Practice Problem 2 — Le Chatelier Multi-Stress 🎯

$2\,\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\,\text{SO}_3(g)$, $\Delta H = -198$ kJ (exothermic)

Practice Problem 3 🧮

For: $\text{A}(g) + 2\,\text{B}(g) \rightleftharpoons 3\,\text{C}(g)$, $\Delta H = +150$ kJ, $K = 0.25$ at 500 K

1. If the volume is halved, which direction does the reaction shift? (Enter "right" or "left")

2. If temperature is increased to 600 K, does K increase or decrease? (Enter "increase" or "decrease")

3. How many moles of gas are on each side? Enter as "left:N, right:M" (e.g., "left:3, right:2")

Quick Stress Review 🔍

Exit Quiz — Problem-Solving Workshop ✅

🎓 Synthesis & AP Review

Part 7 of 7 — Reaction Quotient & Le Chatelier's Principle

This final part reviews Q vs K comparisons, Le Chatelier predictions for all types of stress, and combines them in AP-style questions.

Complete Concept Summary

Reaction Quotient Q

• Same expression as K, using current (not equilibrium) concentrations
• $Q < K$ → shift right | $Q > K$ → shift left | $Q = K$ → at equilibrium

Le Chatelier's Principle

Key Reminders

• Only temperature changes K
• A catalyst speeds up the approach to equilibrium but doesn't change position or K
• The system partially counteracts a stress — never fully

AP-Style Multiple Choice — Set 1 🎯

AP-Style Multiple Choice — Set 2 🎯

AP Free-Response Style 🧮

$\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)$, $K_c = 0.040$ at 250°C, $\Delta H = +87$ kJ

Current concentrations: $[\text{PCl}_5] = 0.50$, $[\text{PCl}_3] = 0.10$, $[\text{Cl}_2] = 0.10$ M

1. Calculate Q. (Enter as a decimal)

2. Which direction does the reaction shift? (Enter "right" or "left")

3. If the temperature is raised to 300°C, does K increase or decrease? (Enter "increase" or "decrease")

Round all answers to 3 significant figures.

Final Concept Review 🔍

Final Exit Quiz ✅