Module 6: Chemical Equilibrium

All MCAT topics related to Chemical Equilibrium are accounted for in the planned module sections:

Planned Section	Covers Topics ✔️
Section 1: Introduction	Conceptual equilibrium, dynamic vs. static
Section 2: K Expressions	Kc, Kp, solid/liquid rules, stoichiometry
Section 3: Reaction Quotient (Q)	Q vs. K logic and application
Section 4: Le Châtelier’s Principle	All stress types + MCAT strategies
Section 5: ICE Tables	Quantitative solving, assumptions, math tricks
Section 6: Thermo vs. Kinetic Control	Energy diagram interpretation, product types
Section 7: Strategy & Summary	Tips, ΔG vs. K, wrap-up and study strategies

Dynamic Nature of Chemical Equilibrium

What Is Chemical Equilibrium?

Chemical equilibrium occurs in reversible reactions when the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentrations of reactants and products over time. However, the system remains dynamic — both reactions are still occurring at the molecular level.

This balance is a key feature of many biological and chemical systems and is central to understanding biochemical pathways, physiological regulation, and laboratory reaction optimization.

Important: Equilibrium does not mean equal concentrations of reactants and products — only that their levels remain constant.

Dynamic, Not Static

Even though macroscopic changes stop at equilibrium, the reaction remains dynamic at the molecular level. Molecules continuously form and break down, but their net amounts stay unchanged.

Analogy: Imagine two equally fast conveyor belts exchanging boxes between two stations. Each belt moves constantly, but the number of boxes at each station doesn’t change.

Key Characteristics of Equilibrium:

• The system must be closed — no matter or energy exchange with the surroundings.
• Equilibrium can be established from either direction — starting with reactants or products.
• Concentrations of reactants/products remain constant (but not necessarily equal).
• Forward and reverse reaction rates are equal, not necessarily the concentrations.
• Forward reaction: Reactants → Products
• Reverse reaction: Products → Reactants
• Reversible reaction: A reaction that can proceed in both directions
• Dynamic equilibrium: A balance where the forward and reverse reactions occur at the same rate

MCAT Analogy: A Balanced Tug of War

Envision two teams playing tug of war with equal force. The rope doesn’t move, even though both sides are actively pulling. That’s chemical equilibrium — forces (reactions) in balance.

Examples of Reversible Reactions at Equilibrium

Example 1:

$$\mathrm{N_2 (g) + 3H_2 (g) \leftrightarrow 2NH_3 (g)}$$

Haber process for ammonia synthesis (heavily tested on the MCAT!)

Example 2:

$$\mathrm{CO_2 (g) + H_2O (l) \leftrightarrow H_2CO_3 (aq)}$$

Carbon dioxide in blood equilibrating with carbonic acid (a key MCAT biology crossover topic)

Not All Reactions Reach Equilibrium

Some reactions go to completion (irreversible). Others are reversible and reach equilibrium under proper conditions.

• Irreversible Reactions: Combustion, strong acid/base neutralizations
• Reversible Reactions: Weak acid dissociation, gas-phase synthesis, biochemical pathways

Why Equilibrium Matters for the MCAT

Equilibrium concepts appear throughout general chemistry, biochemistry, and physiology. Some specific MCAT applications include:

• Hemoglobin binding oxygen (oxy- vs. deoxy-hemoglobin)
• Buffer systems (e.g., bicarbonate buffer)
• Metabolic reaction directionality
• Enzyme kinetics (reversible enzyme reactions)

MCAT Strategy Tips

• Don’t confuse equal concentrations with equilibrium — they are often not equal.
• Watch for “open” system” traps on the MCAT. Open systems cannot achieve true chemical equilibrium.
• Know that equilibrium applies only to reversible reactions. Irreversible reactions go to completion and are not in equilibrium.
• A single arrow (→) means the reaction goes to completion.
• A double arrow (⇌) indicates a reversible reaction at equilibrium.

Common Pitfalls

Misconception	Correction
Reactant and product concentrations must be equal at equilibrium	False – they remain constant, but not necessarily equal
Forward and reverse rates stop at equilibrium	False – they continue equally
Only forward reaction contributes to equilibrium	False – both forward and reverse matter
Open systems can reach equilibrium	False – only closed systems can

The Equilibrium Constant (K)

What is the Equilibrium Constant?

At equilibrium, the forward and reverse reaction rates are equal, and the concentrations of reactants and products remain constant. The equilibrium constant, denoted K, quantitatively expresses the ratio of product concentrations to reactant concentrations at equilibrium.

$$ aA + bB \leftrightarrow cC + dD $$

The equilibrium expression is:

$$ K = \frac{[C]^c[D]^d}{[A]^a[B]^b} $$

Where:

$$ [ \, ] \text{ means molar concentration (mol/L).} $$ $$ K_c \text{ refers to equilibrium in terms of concentration.} $$ $$ \text{For gas-phase reactions, we may use } K_p, \text{ defined using partial pressures.} $$

In other words:

$$ K = \frac{\text{

}^{\text{coefficients}}}{\text{[reactants]}^{\text{coefficients}}} $$

Examples of K Expressions

Example 1:

$$ \text{H}_2 (g) + \text{I}_2 (g) \leftrightarrow 2\text{HI} (g) $$ $$ K = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} $$

Example 2:

$$ \text{N}_2 (g) + 3\text{H}_2 (g) \leftrightarrow 2\text{NH}_3 (g) $$ $$ K = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} $$

Understanding the Value of K

• K > 1 → Products are favored (forward reaction is favored).
• K < 1 → Reactants are favored (reverse reaction is favored).
• K = 1 → Neither side is strongly favored.

MCAT Insight: You don’t need to memorize specific K values but should interpret what a relative size means.

Changing pressure, volume, or concentration does not alter the value of K.

Only temperature affects K.

When Not to Include Species in K Expressions

Key Rule: Pure solids and pure liquids are never included in equilibrium expressions because their concentrations are considered constant and irrelevant to the position of equilibrium.

Include in K expressions:

• Aqueous species
• Gaseous species

Exclude from K expressions:

• Pure solids
• Pure liquids

MCAT Trap Alert: If you see H2O(l) or a solid in the reaction, do not include them in your K expression!

Example:
For the reaction:

$$
\ce{CaCO3(s) <=> Ca^{2+}(aq) + CO3^{2-}(aq)}
$$

The correct equilibrium expression is:

$$
K_{sp} = [\ce{Ca^{2+}}][\ce{CO3^{2-}}]
$$

Interpreting the Magnitude of Kc

The magnitude of K reveals the extent to which a reaction proceeds.

K Value	What It Means	Equilibrium Position
K much greater than 1	Products dominate	Reaction lies to the right
K much less than 1	Reactants dominate	Reaction lies to the left
K approximately equal to 1	Mix of products and reactants	Near balanced

MCAT Strategy Tip:
The larger the K, the more “complete” the reaction appears — meaning more product is present at equilibrium. But remember: K tells you how far a reaction goes, not how fast. The rate is governed by kinetics, not equilibrium constants.

Equilibrium Constants: Overview

At equilibrium, the forward and reverse reaction rates are equal, and the concentrations of products and reactants no longer change. This balance point is described mathematically using equilibrium constants, which express the ratio of product to reactant concentrations (or partial pressures) at equilibrium.

MCAT-relevant types include:

$$
K_c: \text{Equilibrium constant in terms of concentration (mol/L)}
$$

$$
K_p: \text{Equilibrium constant in terms of partial pressure (atm)}
$$

$$
K_a: \text{Acid dissociation constant}
$$

$$
K_b: \text{Base dissociation constant}
$$

$$
K_w: \text{Autoionization constant of water}
$$

$$
K_{sp}: \text{Solubility product constant}
$$

Detailed Definitions and Use Cases

$$
K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}
\quad \text{Used for reactions in aqueous solution; calculated with molar concentrations at equilibrium}
$$

$$
K_p = \frac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b}
\quad \text{Used for gas-phase reactions; calculated using partial pressures in atm}
$$

$$
K_a = \frac{[H^+][A^-]}{[HA]}
\quad \text{Measures the strength of an acid by quantifying its dissociation in water}
$$

$$
K_b = \frac{[OH^-][BH^+]}{[B]}
\quad \text{Measures the strength of a base by quantifying its dissociation in water}
$$

$$
K_w = [H^+][OH^-] = 1.0 \times 10^{-14} \quad \text{at 25^\circ \text{C}}
\quad \text{Equilibrium constant for the self-ionization of water}
$$

$$
K_{sp} = [A^+]^m[B^-]^n
\quad \text{Solubility product for salts that partially dissolve in solution}
$$

Quick Reference Table

Constant	Stands For	What It Measures	Example
Kc	Concentration constant	[Products]/[Reactants] (mol/L)	$$ K_c = \frac{[C][D]}{[A][B]} $$
Kp	Pressure constant	Partial pressures in atm	$$ K_p = \frac{(P_C)(P_D)}{(P_A)(P_B)} $$
Ka	Acid dissociation	Strength of weak acids	$$ K_a = \frac{[H^+][A^-]}{[HA]} $$
Kb	Base dissociation	Strength of weak bases	$$ K_b = \frac{[B][OH^-]}{[BH^+]} $$
Kw	Water autoionization	Water → H⁺ + OH⁻	$$ K_w = [H^+][OH^-] = 1.0 \times 10^{-14} $$
Ksp	Solubility product	Salt solubility in water	$$ K_{sp} = [A^+]^m [B^-]^n $$

MCAT Strategy Tip

• Know how to write the K expression.
• Do not include pure solids or liquids in the expression.
• If a question gives K and asks which direction the reaction will shift, compare to Q (reaction quotient) — coming soon.

Worked Example: Writing the Equilibrium Expression

Question:
For the following reaction at equilibrium:

$$
N_2 (g) + 3H_2 (g) \rightleftharpoons 2NH_3 (g)
$$

Write the equilibrium expression Kc for this reaction and explain each step in detail.

Step-by-Step Solution:

Step 1: Identify the balanced chemical equation.
We’re given:

$$
N_2 (g) + 3H_2 (g) \rightleftharpoons 2NH_3 (g)
$$

This equation is already balanced, with the coefficients:

• 1 for N₂
• 3 for H₂
• 2 for NH₃

These stoichiometric coefficients will be used as exponents in the equilibrium expression.

Step 2: Write the general form of the equilibrium expression.
Use this format:

$$
K_c = \frac{[\text{products}]^{\text{coefficients}}}{[\text{reactants}]^{\text{coefficients}}}
$$

Or more precisely, in generalized form:

$$
K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}
$$

So for our reaction:

$$
K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}
$$

Step 3: Label the meaning of each component.

• [NH₃]²: The concentration of ammonia, squared because the coefficient is 2
• [N₂]: The concentration of nitrogen gas
• [H₂]³: The concentration of hydrogen gas, raised to the power of 3 due to its coefficient

This expression reflects how the equilibrium constant relates the relative concentrations of reactants and products at equilibrium.

Key Points:

• Only aqueous and gaseous species appear in the equilibrium expression.
• Solids and liquids are omitted (they do not appear in K expressions).
• Coefficients from the balanced equation become exponents in the expression.

Final Answer:

$$
K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}
$$

Reaction Quotient (Q) and Direction of Shift

What is the Reaction Quotient (Q)?

The reaction quotient, denoted Q, is a numerical value calculated using the same expression as the equilibrium constant KKK, but with a crucial difference: Q reflects the current concentrations (or partial pressures) of reactants and products at any given point in time, regardless of whether the system is at equilibrium. In contrast, KKK specifically refers to the ratio of these quantities only at equilibrium. This distinction makes Q an invaluable diagnostic tool in chemistry — by calculating Q and comparing it to the known value of KKK, you can predict the direction in which a reaction must proceed to reach equilibrium.

The utility of Q lies in its ability to capture a reaction “in progress.” For example, if you mix the reactants in a flask and measure the concentrations after a short time, the system is likely not at equilibrium yet. Calculating Q with these initial or intermediate concentrations allows you to assess whether the system needs to shift forward (to the right) to produce more products, or shift in reverse (to the left) to consume some of the excess products. If Q<K, the reaction proceeds forward; if Q>K, the reaction shifts in reverse; and if Q=K, the system is at equilibrium.

In laboratory settings, this approach allows chemists to monitor the progress of a reaction and make predictions about how it will respond to changes in concentration, pressure, or other disturbances. It also plays a role in chemical engineering, pharmacology, and biochemistry, where reactions often occur in dynamic, open systems where equilibrium is not guaranteed. On the MCAT, understanding Q is critical not just for calculation questions, but also for interpreting passages involving homeostatic feedback, buffering systems, and enzyme reversibility. It’s a flexible concept that bridges the gap between theory and real-world application — and mastering it gives you powerful predictive control over chemical behavior.

Equation Form:

For a general reaction:

$$
aA + bB \leftrightarrow cC + dD
$$

The reaction quotient is:

$$
Q = \frac{[C]^c [D]^d}{[A]^a [B]^b}
$$

It is identical in form to K — same formula, same structure. The difference is when it’s measured.

Q vs. K: Predicting Direction of the Reaction

Comparing Q to K lets you predict whether the reaction will shift forward (toward products) or reverse (toward reactants) to reach equilibrium.

Comparison	Interpretation	Reaction Shift
Q < K	Too many reactants, not enough products	Shift right (→) to form products
Q > K	Too many products, not enough reactants	Shift left (←) to form reactants
Q = K	Reaction is at equilibrium	No shift (equilibrium)

MCAT Strategy Tip: Think of Q as a snapshot of a reaction at any time — it answers: “Are we at equilibrium yet?”

Analogy: Chemical “See-Saw”

Imagine a see-saw where:

• Reactants are on one side, Products on the other.
• At equilibrium (Q = K), the see-saw is perfectly balanced.
• If one side is heavier, the reaction shifts to rebalance the see-saw.

Q < K = not enough products → tip right to make more
Q > K = too many products → tip left to get rid of excess

MCAT Traps & Tips

• Don’t confuse Q with K! They use the same math, but only K reflects equilibrium.
• On test day, expect Q to be calculated from given concentrations, while K is typically a known constant.
• If you’re given initial concentrations, use them to calculate Q, then compare to K.

Worked Example: Q vs. K Problem

Question:

At 450°C, the equilibrium constant for the reaction:

$$
\mathrm{N_2 (g) + 3H_2 (g) \leftrightarrow 2NH_3 (g)}
$$

is Kc = 0.040. A reaction mixture is found to have the following concentrations:

$$
[\text{N}_2] = 1.0 \, \text{mol/L} \
[\text{H}_2] = 1.0 \, \text{mol/L} \
[\text{NH}_3] = 0.10 \, \text{mol/L}
$$

Will the reaction shift forward, backward, or is it at equilibrium?

Step 1: Write the expression for Qc

Same as Kc:

$$
Q_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}
$$

Step 2: Plug in the values

$$
Q_c = \frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(0.10)^2}{(1.0)(1.0)^3} = \frac{0.01}{1.0} = 0.01
$$

Step 3: Compare Qc to Kc

$$
Q_c = 0.01 \
K_c = 0.040
$$

Since $Q_c < K_c$, the system has too many reactants and not enough products.

Answer: The reaction will shift right, toward products, to reach equilibrium.

Summary Points

• Q is calculated like K, but using current concentrations.
• Compare Q to K to predict the direction of the reaction.
• Q < K: shift right
• Q > K: shift left
• Q = K: at equilibrium

This is a very testable MCAT concept, especially when paired with ICE tables or Le Châtelier’s Principle.

Le Châtelier’s Principle

Conceptual Overview: What Is Le Châtelier’s Principle?

Le Châtelier’s Principle is one of the most important tools in chemical equilibrium. It predicts how a system at equilibrium will respond to a disturbance or stress — that is, a change in conditions such as concentration, pressure, or temperature.

Core Idea:

When a system at equilibrium is disturbed, it will shift in the direction that counteracts the disturbance, in order to re-establish equilibrium.

Think of equilibrium like a balanced scale. If you place extra weight (a stressor) on one side, the system shifts to rebalance. In chemistry, this means the forward or reverse reaction temporarily increases to offset the change and return the system to equilibrium.

MCAT Analogy: Chemical Tug of War

Imagine a game of tug-of-war between the forward and reverse reactions. At equilibrium, both sides pull with equal force, and the rope doesn’t move. But if you suddenly add more people to one side (e.g., increase reactant concentration), the rope shifts in the opposite direction – the reaction compensates by favoring the forward reaction to use up that excess.

General Response Rules:

Type of Stress	System Response
Add reactants	Shift right to make more products
Remove reactants	Shift left to make more reactants
Add products	Shift left to consume products
Remove products	Shift right to make more products
Increase pressure	Shift to side with fewer gas molecules
Decrease pressure	Shift to side with more gas molecules
Increase volume	Same as decreasing pressure
Decrease volume	Same as increasing pressure
Increase temperature	Shift away from heat (depends on endo/exothermic)
Decrease temperature	Shift toward heat

1. Changes in Concentration

One of the most intuitive applications of Le Châtelier’s Principle involves changes in the concentration of reactants or products. If a system at equilibrium experiences a concentration change, the system will shift in the direction that consumes the added species or replenishes the removed one in order to restore balance.

Core Principle

If you add a reactant or product, the reaction shifts away from it.
If you remove a reactant or product, the reaction shifts toward it.

This response occurs because the equilibrium ratio (Kc​) is disrupted. The system compensates by adjusting concentrations through shifts in the forward or reverse reaction until the equilibrium expression is satisfied again.

Worked Example

Consider the equilibrium reaction:

$$
\ce{N_2(g) + 3H_2(g) <=> 2NH_3(g)}
$$

At equilibrium, concentrations remain stable. Now, suppose more H2​ is added.

What happens?

• The system now has an excess of reactants.
• To counteract this, the reaction shifts to the right, converting more N2 and H2​ into NH3​, thus producing more ammonia.

This shift reduces the added H2 and restores the balance dictated by the equilibrium constant Kc​.

General Shifts Summary

Change	Direction of Shift	Why?
Add reactant	Toward products (right)	To consume the added reactant
Remove reactant	Toward reactants (left)	To replace the removed species
Add product	Toward reactants (left)	To reduce product buildup
Remove product	Toward products (right)	To produce more product

MCAT Trap Alert

• Don’t confuse concentration changes with changes in Kc​.
• K stays the same unless temperature is changed.
• The system adjusts concentrations, not the constant itself.

Pressure & Volume Changes

In gas-phase equilibrium systems, pressure and volume are inversely related (Boyle’s Law: PV=constant). So any change in one causes the opposite change in the other. Le Châtelier’s Principle tells us that the system will shift to counteract this stress.

When Pressure Increases (Volume Decreases)

• System shifts toward the side with fewer gas molecules.
• This decreases total pressure, helping relieve the stress.

Example:

$$
\ce{N2(g) + 3H2(g) <=> 2NH3(g)}
$$

• Left side: 1 + 3 = 4 moles of gas
• Right side: 2 moles of gas
• If you increase pressure (or decrease volume), the system shifts right to form more ammonia and reduce the number of gas molecules.

When Pressure Decreases (Volume Increases)

• System shifts toward the side with more gas molecules to raise the pressure again.

Important Exception:

• If both sides have equal moles of gas, a change in pressure has no effect on equilibrium position.

Example:

$$
\ce{H2(g) + I2(g) <=> 2HI(g)} \quad (1\ \text{mol vs.}\ 1\ \text{mol on each side})
$$

Pressure changes do not affect this equilibrium.

Solids & Liquids Don’t Count

Pressure/volume changes only matter for gases. Solids and liquids are incompressible – their volume doesn’t change with pressure, so they don’t influence gas pressure-based shifts.

Temperature Changes

Unlike concentration or pressure, changing temperature actually alters the value of K – meaning it changes the equilibrium position permanently.

Think of Heat as a Reactant or Product

Treat heat (ΔH) as if it were part of the equation:

• Exothermic reaction:

$$
\ce{A + B <=> C + D + heat}
$$

• Increasing temperature adds more “product” (heat), so the system shifts left.
• Decreasing temperature removes heat, system shifts right.

• Endothermic reaction:

$$
\ce{A + B + heat <=> C + D}
$$

• Increasing temperature adds heat to the reactants, so the system shifts right.
• Decreasing temperature shifts left.

MCAT Insight:

• If you know the ΔH value for a reaction, you can predict the direction of shift with temperature changes.
• If ΔH is not given, look for clues or context – it’s a common MCAT trick to ask about a temperature shift without explicitly saying “exo” or “endo.”

Summary Table: Volume, Pressure, and Temperature

Change	System Response	Direction of Shift
↑ Pressure (↓ Volume)	Reduce total gas moles	Toward fewer gas particles
↓ Pressure (↑ Volume)	Increase total gas moles	Toward more gas particles
↑ Temperature (Endo rxn)	Use added heat	Toward products
↑ Temperature (Exo rxn)	Counter excess heat	Toward reactants
↓ Temperature (Endo rxn)	Remove heat from reactants	Toward reactants
↓ Temperature (Exo rxn)	Remove heat from products	Toward products

ICE Tables – Solving Equilibrium Problems

What Are ICE Tables?

ICE Tables are a systematic way of organizing equilibrium problems where you’re asked to calculate unknown concentrations or equilibrium constants. The acronym stands for:

• I – Initial concentrations (before any reaction occurs)
• C – Change in concentrations (how each species changes as the reaction progresses)
• E – Equilibrium concentrations (what’s left at equilibrium)

This method is especially useful when you’re given partial information — like initial concentrations and a Kc value — and need to find unknowns like the concentration of one species at equilibrium.

Step-by-Step Algorithm for Solving ICE Table Equilibrium Problems (MCAT)

Use this procedure whenever you’re asked to solve for concentrations at equilibrium, especially when you’re given initial conditions and the equilibrium constant K (or vice versa).

Step 1: Write the Balanced Chemical Equation

• Ensure all stoichiometric coefficients are correct.
• Identify which species are aqueous or gaseous – only those will go in the K expression.

Step 2: Set Up an ICE Table

Species	Initial (I)	Change (C)	Equilibrium (E)
Reactant A	Given	-ax (use stoichiometry)	[A]0 – ax
Reactant B	Given	-bx	[B]0 – bx
Product C	Usually 0	+cx	cx
Product D	Usually 0	+dx	dx

• Use x to represent the change in concentration.
• Negative signs for reactants, positive for products.
• Coefficients from the balanced equation become multipliers of x.

Step 3: Write the Expression for K

• Use the general form:

$$
K = \frac{[\text{Products}]^{\text{coefficients}}}{[\text{Reactants}]^{\text{coefficients}}}
$$

• Plug in your expressions from the E row of the ICE table.

Step 4: Plug In K and Solve for x

• Use the given value of K (or Kc or Kp) and solve for x.
• This may involve:
  • Expanding a quadratic equation
  • Using a square root
  • Applying approximations (e.g. if K << 1, assume change is negligible)

Step 5: Plug x Back In to Find Equilibrium Concentrations

• Use your value of x to compute all final concentrations from the E row.

Step 6: (Optional) Check Your Answer by Recalculating Q

• Plug equilibrium concentrations back into the K expression to confirm that it equals the given K value.
• This validates your solution and is good exam-day strategy if time permits.

---

How ICE Tables Work

Let’s walk through the general process using the example reaction:

$$
\mathrm{A + B \leftrightarrow C}
$$

We assume all species are aqueous or gases (so they count in the equilibrium expression). The table is set up as follows:

Species	Initial (mol/L)	Change (mol/L)	Equilibrium (mol/L)
A	[A]₀	–x	[A]₀ – x
B	[B]₀	–x	[B]₀ – x
C	0	+x	x

• The coefficients from the balanced equation determine the relative changes (for instance, if you had 2A + B <=> C, the change would be –2x for A, –x for B, +x for C).
• Plug the equilibrium values into the Kc expression, solve for x, and then find the desired quantities.

Sample Problem: ICE Table With Quadratic

Given:

$$
\mathrm{H_2(g) + I_2(g) \leftrightarrow 2HI(g)} \quad K_c = 49.0
$$

Initial concentrations:

$$
[\ce{H_2}] = 1.00\ \text{mol/L} \
[\ce{I_2}] = 1.00\ \text{mol/L} \
[\ce{HI}] = 0
$$

Find the equilibrium concentration of all species.

Step 1: Set Up the ICE Table

Species	I (mol/L)	C (mol/L)	E (mol/L)
H₂	1.00	–x	1.00 – x
I₂	1.00	–x	1.00 – x
HI	0	+2x	2x

Step 2: Plug into the Equilibrium Expression

$$
K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(2x)^2}{(1.00 – x)(1.00 – x)} = \frac{4x^2}{(1.00 – x)^2}
$$

Set this equal to the known Kc = 49.0:

$$
\frac{4x^2}{(1.00 – x)^2} = 49.0
$$

Step 3: Solve for x

Take the square root of both sides:

$$
\frac{2x}{1.00 – x} = 7
$$

Multiply both sides:

$$
2x = 7(1.00 – x) \Rightarrow 2x = 7 – 7x
$$

Solve for x:

$$
2x = 7(1.00 – x) \Rightarrow 2x = 7 – 7x \Rightarrow 9x = 7 \Rightarrow x = \frac{7}{9} \approx 0.778
$$

Step 4: Solve for Equilibrium Concentrations

$$
[\ce{H_2}] = 1.00 – x = 0.222\ \text{mol/L}
$$

$$
[\ce{I_2}] = 1.00 – x = 0.222\ \text{mol/L}
$$

$$
[\ce{HI}] = 2x = 1.556\ \text{mol/L}
$$

Common Pitfalls

Mistake	Correction
Forgetting to square terms with coefficients	Coefficients become exponents in the K expression
Plugging in initial concentrations into the equilibrium expression	Use E row only
Dropping the ± sign too early	Keep track of direction of change carefully
Skipping the quadratic if approximation is invalid	Only assume x is small when K is very small or large

MCAT Strategy Tip

• When K is very small (e.g. < 10^-4), assume x is small and approximate.
• Always check your work by plugging back into the expression.
• Pay attention to units – Kc uses mol/L, while Kp uses atm.
• MCAT questions often ask you qualitatively about ICE logic even without numbers.

Thermodynamic vs. Kinetic Control

Overview

Some reactions can produce more than one possible product. Under certain conditions, one product is favored over the other. This is where the concept of kinetic vs. thermodynamic control comes in.

The MCAT expects you to distinguish between:

• A kinetically favored product (forms faster but may not be most stable)
• A thermodynamically favored product (forms more slowly but is more stable)

Basic Definitions

Term	Definition	Key Feature	Typical Conditions
Kinetic Control	Product forms fastest	Lowest activation energy (Ea)	Low temperature, short reaction time
Thermodynamic Control	Product is most stable	Lowest final energy (most negative ΔG)	High temperature, long reaction time

Energy Diagram

We’ll describe the general idea (since diagrams can’t be shown here):

• Imagine two pathways from reactants to products.
• One has a lower activation energy (Ea) → this is the kinetic product.
• One ends at a lower energy final state → this is the thermodynamic product.

The shape of the curve determines which pathway is favored under certain conditions.

Kinetic Product

• Formed more quickly because the activation energy (Ea) is lower.
• Typically the major product at low temperatures.
• May be less stable (higher final energy).
• Example: Forms via less-substituted carbocation or less stable intermediate.

MCAT Clue: “Fastest product formed” → Kinetic control

Thermodynamic Product

• Formed more slowly because the pathway has a higher activation energy.
• However, the product is more stable, with lower energy overall.
• Favored at high temperatures where there’s time and energy to overcome Ea.
• Example: Forms via more substituted alkene or resonance-stabilized intermediate.

MCAT Clue: “Most stable product” → Thermodynamic control

Real-World Analogy:

Think of two routes to a destination:

• One is short and easy but takes you to a noisy, less comfortable place.
• The other is longer and harder, but ends at a luxury resort.

Low energy (Ea) = easy/fast (kinetic),
Low final energy (ΔG) = better destination (thermodynamic).

Common Pitfalls

Misconception	Correction
Kinetic = most stable	No — it’s the fastest, not necessarily the most stable
Thermodynamic = fastest	No — it’s the most stable, not the fastest
Products always reach equilibrium	Not if you’re under kinetic control — the reaction may get “trapped” in the fast but less stable state

MCAT Example

Two isomeric products form from an acid-catalyzed alkene addition reaction. At low temperature, the less substituted alkene predominates. At high temperature, the more substituted alkene predominates.

Which is under kinetic control? Which is under thermodynamic control?

• Answer:
  • The less substituted alkene = kinetic product (formed faster, less stable)
  • The more substituted alkene = thermodynamic product (formed slower, more stable)

Recap

Feature	Kinetic Control	Thermodynamic Control
Speed	Fastest product	Slow, forms over time
Activation Energy (Ea)	Lower	Higher
Stability	Less stable	More stable
Temperature	Low	High
Reversibility	Irreversible	Reversible or at equilibrium