Chapter 6: Equilibrium

Most reactions you have seen so far are irreversible - they go in one direction, run out of limiting reagent, and stop. But many reactions are reversible: the products can react to re-form the reactants. When a reversible reaction is placed in a closed system, something remarkable happens.

At first, the forward reaction dominates because reactants are abundant and products are scarce. As products accumulate, the reverse reaction speeds up. Eventually, the rate of the forward reaction equals the rate of the reverse reaction. Concentrations stop changing. The system has reached equilibrium.

Dynamic vs. Static Equilibrium

This is the single most important distinction in this chapter. Dynamic equilibrium means both reactions are still happening - molecules are still converting back and forth. It is not that everything stopped. A static equilibrium would mean nothing is happening at all, like a book sitting on a table.

At dynamic equilibrium:

• The forward reaction rate equals the reverse reaction rate
• Concentrations of all species remain constant (but not necessarily equal)
• The system is at its minimum Gibbs free energy and maximum entropy
• No net change is observable, even though molecular-level reactions continue

Recognizing Reversible Reactions

Reversible reactions are written with a double arrow (⇌) instead of a single arrow. For example:

N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g)

This tells you that nitrogen and hydrogen can combine to form ammonia (forward), AND ammonia can decompose back into nitrogen and hydrogen (reverse). In a closed container, both processes happen simultaneously.

What Determines the Position of Equilibrium?

The “position” of equilibrium refers to the relative amounts of products and reactants at equilibrium. Two factors determine this position:

1. The nature of the reaction itself - encoded in the equilibrium constant K
2. Temperature - the only external factor that changes the value of K

Changing concentration or pressure can shift the system temporarily, but these changes do not alter K. Only temperature does. We will return to this critical point in section 6.8.

Now that you understand what equilibrium looks like, let’s put a number on it. The equilibrium constant tells you exactly where the balance point lies - does this reaction barely make any products, or does it essentially go to completion?

The answer comes from the law of mass action. For a generic reversible reaction:

aA + bB ⇌ cC + dD

This ratio is constant at a given temperature. It does not matter how much of each substance you start with - once the system reaches equilibrium, this ratio always equals K.

Where Does Keq Come From?

The law of mass action connects equilibrium to kinetics. Consider a one-step reversible reaction:

2A ⇌ B + C

The forward rate is $\text{rate}_f = k_f[A]^2$ and the reverse rate is $\text{rate}_r = k_r[B][C]$. At equilibrium, these rates are equal:

$\displaystyle k_f[A]^2 = k_r[B][C]$

Rearranging gives:

$\displaystyle K_{eq} = \dfrac{k_f}{k_r} = \dfrac{[B][C]}{[A]^2}$

The equilibrium constant is simply the ratio of the forward and reverse rate constants. A large K_eq means $k_{f}$ >> $k_{r}$, so the forward reaction is much faster and products are heavily favored.

What Goes in the Expression?

Not every species appears in the equilibrium expression:

Interpreting the Size of K

The value of K tells you which side the equilibrium favors:

Properties of K to Memorize

1. Reverse the reaction - the new K is 1/K_original
2. Multiply the reaction by a factor n - the new K is (K_original)^n
3. Add two reactions - the new K is K₁ × K₂
4. K is temperature-dependent - changing T is the ONLY way to change K
5. K has no units on the MCAT - it is technically based on activities, which are dimensionless

There are two ways to express the equilibrium constant, and the MCAT expects you to know both. The difference is simply what you plug into the expression - concentrations or partial pressures.

K_c uses molar concentrations (mol/L) for all species. This is the version you have already seen.

K_p uses partial pressures (atm) for gaseous species. It is used when the problem gives you pressures instead of concentrations.

Both describe the same equilibrium. They just speak different “languages.”

When to Use Each

• Use K_c when the problem gives concentrations in mol/L
• Use K_p when the problem gives partial pressures in atm
• For dilute aqueous solutions, K_c = K_eq (they are interchangeable)

Example: Writing Kc and Kp

For the Haber process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

K_c = [NH₃]² / ([N₂][H₂]³)

K_p = ($P_{\text{NH}}$₃)² / ($P_{N}$₂ × ($P_{H}$₂)³)

Same structure, different units plugged in.

The Formula Connecting Kc and Kp

Key Insight: When Kp = Kc

When delta-n = 0 (the number of moles of gas is the same on both sides), then (RT)⁰ = 1, so Kp = Kc.

Example: H₂(g) + I₂(g) ⇌ 2HI(g)

delta-n = 2 - (1 + 1) = 0, so Kp = Kc for this reaction.

Worked Example

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), if Kc = 0.50 at 400°C, find Kp.

Step 1: Calculate delta-n = 2 - (1 + 3) = -2

Step 2: Convert temperature: T = 400 + 273 = 673 K

Step 3: Kp = Kc(RT)^delta-n = 0.50 × (0.0821 × 673)^(-2)

= 0.50 × (55.2)^(-2) = 0.50 × ($\frac{1}{3047}$) = 0.50 / 3047

$K_p \approx 1.6 \times 10^{-4}$

Because delta-n is negative (fewer moles of gas on the product side), Kp < Kc for this reaction.

You know K tells you where equilibrium IS. But what if you are not at equilibrium yet? How do you figure out which way the reaction needs to go? That is the job of Q, the reaction quotient.

Calculating Q

Q has the exact same mathematical form as K:

The critical difference: K uses equilibrium concentrations (which are constant). Q uses current concentrations (which may or may not be at equilibrium).

Comparing Q to K

The Connection to Gibbs Free Energy

The relationship between Q, K, and delta-G is one of the most important connections on the MCAT:

• When Q < K: delta-G < 0 (forward reaction is spontaneous)
• When Q = K: delta-G = 0 (system is at equilibrium)
• When Q > K: delta-G > 0 (reverse reaction is spontaneous)

Worked Example

Consider the reaction: CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g), with K = 1.0 at a certain temperature.

If the current concentrations are [CO] = 0.1 M, [H₂O] = 0.1 M, [CO₂] = 0.5 M, [H₂] = 0.5 M, which way will the reaction shift?

Step 1: Calculate Q = [CO₂][H₂] / [CO][H₂O] = (0.5)(0.5) / (0.1)(0.1) = 0.25 / 0.01 = 25

Step 2: Compare Q to K: Q (25) > K (1.0)

Step 3: Since Q > K, there are too many products. The reaction shifts to the left (reverse direction) until Q decreases to equal K.

You now know how to calculate K and Q, and how to predict which direction a reaction shifts. Le Chatelier’s principle is the conceptual framework that ties it all together - and it is tested on virtually every MCAT.

The Principle

Le Chatelier’s principle states: if a stress is applied to a system at equilibrium, the system will shift in the direction that partially relieves that stress.

The key word is “partially.” The system never completely undoes the change - it just moves toward a new equilibrium that partially offsets whatever you did.

Three Types of Stress

Le Chatelier’s principle applies to three types of changes:

1. Changes in concentration (adding or removing a reactant or product)
2. Changes in pressure or volume (for gas-phase reactions)
3. Changes in temperature (the only one that changes K)

Each of the next three sections covers one of these in depth. Here is the summary table:

How to Think About It Using Q

Le Chatelier’s principle is really just Q vs. K logic in disguise:

• Add reactant - the denominator of Q gets bigger, so Q decreases below K. Since Q < K, the reaction shifts forward.
• Remove product - the numerator of Q gets smaller, so Q decreases below K. Since Q < K, the reaction shifts forward.
• Add product - the numerator gets bigger, Q > K, so the reaction shifts backward.

Every Le Chatelier prediction can be confirmed by checking what happens to Q relative to K.

Concentration changes are the most intuitive application of Le Chatelier’s principle. The rule is simple: the system shifts away from whatever you added and toward whatever you removed.

Adding a Reactant or Removing a Product

Both of these push the reaction forward (to the right). Here is why:

Adding reactant increases the denominator of Q, so Q drops below K. The system responds by consuming the extra reactant - converting it into products - until Q climbs back up to equal K.

Removing product decreases the numerator of Q, so Q drops below K. The system responds by making more product to replace what was lost.

Adding a Product or Removing a Reactant

Both of these push the reaction backward (to the left).

Adding product increases the numerator of Q, so Q rises above K. The system responds by converting the excess product back into reactants.

Removing reactant decreases the denominator of Q, so Q rises above K. The system shifts left to replenish the lost reactant.

Important Exception: Pure Solids and Liquids

Adding or removing a pure solid or pure liquid does NOT shift equilibrium. Since these species do not appear in the equilibrium expression (their activity = 1), their amounts cannot change Q.

Example: CaCO₃(s) ⇌ CaO(s) + CO₂(g), K = [CO₂]

Adding more CaCO₃ does NOT shift the equilibrium because CaCO₃ is a solid and is not in the K expression. The equilibrium CO₂ pressure remains the same.

What Happens to Concentrations After the Shift?

A common misconception: if you add more reactant A and the equilibrium shifts right, does [A] end up higher, lower, or the same as before?

[A] ends up higher than the original equilibrium, but lower than right after you added it. The system partially consumes the added A, but it cannot completely remove it. The new equilibrium has higher [A] and higher [products] than the original equilibrium.

Pressure and volume changes only matter for reactions involving gases. Liquids and solids are essentially incompressible, so changing the pressure of the container has no effect on their concentrations.

The Core Rule

When pressure increases (or volume decreases), the system shifts toward the side with fewer moles of gas. When pressure decreases (or volume increases), the system shifts toward the side with more moles of gas.

Why? The system is trying to relieve the stress. If you squeeze the container (increase pressure), the system responds by reducing the total number of gas molecules - which reduces the pressure.

Example: The Haber Process

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

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

• Increase pressure: Shifts right (toward 2 moles, fewer gas molecules)
• Decrease pressure: Shifts left (toward 4 moles, more gas molecules)

This is why the Haber process uses high pressure (200 atm) - it pushes the equilibrium toward ammonia production.

When Moles of Gas Are Equal

If delta-n = 0 (same number of gas moles on both sides), pressure changes have no effect on the equilibrium.

Example: H₂(g) + I₂(g) ⇌ 2HI(g) has 2 moles of gas on each side. Changing pressure does not shift this equilibrium.

This distinction - inert gas at constant volume vs. constant pressure - is a classic MCAT trap question. Remember: partial pressures determine Q, not total pressure.

Why Does This Work? The Q Explanation

When you decrease the volume of a container, all gas concentrations increase (same moles, smaller volume). But they do not all increase equally in the Q expression - the side with more moles of gas gets “raised to higher powers” in the expression. This changes Q in a predictable direction:

• If more moles of gas are in the numerator (products), Q increases above K, so the reaction shifts left
• If more moles of gas are in the denominator (reactants), Q decreases below K, so the reaction shifts right

The net result: the shift always goes toward the side with fewer moles of gas.

Temperature changes are fundamentally different from concentration and pressure changes. Changing concentration or pressure shifts the equilibrium, but K stays the same. Changing temperature actually changes the value of K itself.

This is the single most important fact in this section: temperature is the ONLY factor that changes K.

The Heat-as-a-Species Trick

The easiest way to predict the effect of temperature is to treat heat as if it were a chemical species:

• Exothermic reaction (delta-H < 0): heat is a “product”
  • A + B ⇌ C + D + heat
• Endothermic reaction (delta-H > 0): heat is a “reactant”
  • heat + A + B ⇌ C + D

Now apply Le Chatelier’s principle as if heat were a concentration:

Summary Table

Why K Changes (Unlike With Concentration/Pressure)

When you change concentration, Q changes but K stays fixed - the system shifts to bring Q back to K. But when you change temperature, K itself changes to a new value. The system then shifts so that the current Q moves toward this NEW K.

This happens because K depends on the rate constants $k_{f}$ and $k_{r}$ (recall K = $k_{f}$/$k_{r}$). Temperature changes both rate constants according to the Arrhenius equation, but it changes them by different amounts depending on the activation energies. The net result is that K itself changes.

Example: The Haber Process

N₂(g) + 3H₂(g) ⇌ 2NH₃(g), delta-H = -92 kJ/mol (exothermic)

• Increasing temperature: Shifts left (fewer products), K decreases
• Decreasing temperature: Shifts right (more products), K increases

This creates a dilemma in industrial chemistry: low temperature gives a better K (more product), but the reaction is too slow. The compromise is moderate temperature (~450 degrees C) with a catalyst to speed things up.

Students often assume that a catalyst “pushes” the reaction toward products. This is wrong - and the MCAT knows you might think this. A catalyst makes the reaction reach equilibrium faster, but it does not change where that equilibrium lies.

What a Catalyst Does

• Lowers the activation energy for both the forward and reverse reactions by the same amount
• Speeds up both reactions equally, so the ratio $k_{f}$/$k_{r}$ stays the same
• Reaches equilibrium faster - the system gets to equilibrium in seconds instead of hours
• Does not change K, Q, or the equilibrium position
• Is not consumed in the reaction - it participates but is regenerated

What a Catalyst Does NOT Do

• Does NOT change the equilibrium constant K
• Does NOT shift the equilibrium position
• Does NOT change the concentrations of products or reactants at equilibrium
• Does NOT change delta-G or delta-H for the reaction
• Does NOT appear in the equilibrium expression

Two Types of Catalysts

Energy Diagram With and Without a Catalyst

On a reaction coordinate diagram, a catalyst lowers the peak (transition state) without changing the starting energy (reactants) or ending energy (products). The activation energy for both the forward and reverse reactions decreases by the same amount. Since delta-G = G(products) - G(reactants), and neither of those changes, delta-G is unchanged.

The solubility product (Ksp) is just the equilibrium constant for an ionic solid dissolving in water. Everything you learned about K applies here - Ksp is simply K applied to the specific case of dissolution.

Writing the Ksp Expression

For the dissolution of a generic ionic solid:

$A_{x}$ $B_{y}$(s) ⇌ xA^(y+)(aq) + yB^(x-)(aq)

Examples

Notice how the coefficient (2 for F-, 3 for OH-) becomes the exponent.

Ksp and Molar Solubility

Molar solubility (s) is the number of moles of solid that dissolve per liter of solution to form a saturated solution. You can calculate s from Ksp:

Example: Find the molar solubility of AgCl ($K_{sp} = 1.8 \times 10^{-10}$).

AgCl(s) ⇌ Ag+(aq) + Cl-(aq)

If s moles dissolve, then [Ag+] = s and [Cl-] = s.

$K_{sp} = (s)(s) = s^2$

$s = \sqrt{1.8 \times 10^{-10}} \approx 1.3 \times 10^{-5}$ M

Example: Find the molar solubility of CaF₂ ($K_{sp} = 3.9 \times 10^{-11}$).

CaF₂(s) ⇌ Ca²+(aq) + 2F-(aq)

If s moles dissolve, then [Ca²+] = s and [F-] = 2s.

$K_{sp} = (s)(2s)^2 = 4s^3$

$s = (K_{sp}/4)^{1/3} = (3.9 \times 10^{-11} / 4)^{1/3} \approx (9.75 \times 10^{-12})^{1/3} \approx 2.1 \times 10^{-4}$ M

Predicting Precipitation: Qsp vs. Ksp

Just as Q vs. K predicts reaction direction, Qsp vs. Ksp predicts whether a precipitate will form:

The common ion effect is Le Chatelier’s principle applied to solubility. It explains why a salt is less soluble in a solution that already contains one of its ions.

How It Works

Consider dissolving AgCl in a solution of 0.10 M NaCl:

AgCl(s) ⇌ Ag+(aq) + Cl-(aq), $K_{sp} = 1.8 \times 10^{-10}$

In pure water, $s \approx 1.3 \times 10^{-5}$ M (as we calculated in the previous section).

But in 0.10 M NaCl, there is already 0.10 M Cl- in solution. This is the “common ion” - the Cl- is shared between NaCl and AgCl.

Step 1: Set up the $K_{sp}$ expression with the common ion already present.

• [Ag+] = s (from dissolved AgCl)
• [Cl-] = 0.10 + s ≈ 0.10 (since s will be tiny compared to 0.10)

Step 2: Solve for s.
$K_{sp} = [\text{Ag}^+][\text{Cl}^-] = (s)(0.10) = 1.8 \times 10^{-10}$

$s = 1.8 \times 10^{-10} / 0.10 = 1.8 \times 10^{-9}$ M

Result: AgCl is about 7,000 times LESS soluble in 0.10 M NaCl than in pure water. The common ion (Cl-) dramatically suppressed the solubility.

Why Does This Happen? (Q vs. Ksp)

Adding a common ion increases the ion product Qsp above Ksp. Since Qsp > Ksp, the system is supersaturated and shifts left (toward solid formation). Precipitation continues until the ion concentrations drop enough that Qsp = Ksp again.

Applications

Laboratory separations: The common ion effect is used to selectively precipitate compounds. If you have a solution containing both Ag+ and Cu²+ and you add NaCl, the AgCl (very low Ksp) precipitates while CuCl₂ (high solubility) stays in solution. Adding excess Cl- drives the precipitation of AgCl even further.

Buffer solutions: Buffers contain a weak acid and its conjugate base (or a weak base and its conjugate acid). The common ion effect explains why a buffer resists pH changes - the common ion from the salt suppresses the dissociation of the weak acid or base.

Common Ion vs. Non-Common Ion

Adding NaCl to a saturated AgCl solution decreases solubility (common ion: Cl-).

Adding NaNO₃ to a saturated AgCl solution has essentially no effect - Na+ and NO₃- are not part of the AgCl equilibrium, so they do not change Qsp.

Every quantitative equilibrium problem on the MCAT follows the same framework. You set up a table, plug into the K expression, and solve for an unknown. That framework is the ICE table - and once you master it, every equilibrium calculation becomes a fill-in-the-blank exercise.

What ICE Stands For

Setting Up the Table

Step 1: Write the balanced equation and the K expression.

Step 2: Create the ICE table with one column per species.

Step 3: Fill in the Initial row from the problem.

Step 4: Fill in the Change row using stoichiometric ratios. The species that increases gets +x (scaled by its coefficient), and the species that decreases gets -x (scaled by its coefficient). To determine which direction the reaction proceeds, compare Q (the reaction quotient calculated from initial concentrations) to K: if Q < K, the reaction goes forward (products increase); if Q > K, it goes backward (reactants increase).

Step 5: Add I + C to get the Equilibrium row.

Step 6: Substitute the Equilibrium expressions into the K equation and solve for x.

Worked Example 1: Starting From Reactants Only

Consider: H₂(g) + I₂(g) ⇌ 2HI(g), K_c = 50.0 at 448°C.

A flask initially contains 0.100 M H₂ and 0.100 M I₂ with no HI. Find the equilibrium concentrations.

Step 1: K_c = [HI]² / ([H₂][I₂]) = 50.0

Step 2-5: Build the ICE table. Since Q = 0 and K = 50, Q < K, so the reaction goes forward (products increase, reactants decrease).

Step 6: Plug into K:

50.0 = (2x)² / ((0.100 - x)(0.100 - x))

50.0 = (2x)² / (0.100 - x)²

Take the square root of both sides (a shortcut when both sides are perfect squares):

√50.0 = 2x / (0.100 − x)

7.07 = 2x / (0.100 - x)

7.07(0.100 - x) = 2x

0.707 = 2x + 7.07x = 9.07x

x = 0.0780

Equilibrium concentrations:

• [H₂] = 0.100 - 0.078 = 0.022 M
• [I₂] = 0.100 - 0.078 = 0.022 M
• [HI] = 2(0.078) = 0.156 M

Check: K = (0.156)² / (0.022)(0.022) = 0.0243 / 0.000484 = 50.2 (close to 50.0 - rounding accounts for the difference).

Worked Example 2: The Small-x Approximation

Consider: N₂O₄(g) ⇌ 2NO₂(g), $K_c = 4.6 \times 10^{-3}$

A flask initially contains 0.500 M N₂O₄. Find equilibrium concentrations.

$K_c = (2x)^2 / (0.500 - x) = 4x^2 / (0.500 - x)$

This leads to a quadratic, but K is very small ($10^{-3}$) compared to the initial concentration (0.500). This tells us that very little N₂O₄ will decompose, so x will be very small compared to 0.500.

The approximation: Assume x << 0.500, so 0.500 - x ≈ 0.500.

$4x^2 / 0.500 = 4.6 \times 10^{-3}$

$4x^2 = 2.3 \times 10^{-3}$

$x^2 = 5.75 \times 10^{-4}$

x = 0.024

Check the approximation: x / 0.500 = 0.024 / 0.500 = 4.8%. Since this is less than 5%, the approximation is valid.

When Can You Use the Small-x Approximation?

The approximation works when K is much smaller than the initial concentration:

ICE Tables for Ksp

ICE tables work for solubility equilibria too. The only difference: the solid does not get a column (its activity is 1).

Molar solubility is the number of moles of a salt that dissolve per liter of solution. We label it “s” in ICE tables. If s moles of PbI₂ dissolve, then s moles of Pb²⁺ appear in solution - and 2s moles of I⁻ appear (because each formula unit releases 2 iodide ions).

Example: Find the molar solubility of PbI₂ ($K_{sp} = 9.8 \times 10^{-9}$).

PbI₂(s) ⇌ Pb²+(aq) + 2I-(aq)

$K_{sp} = [\text{Pb}^{2+}][\text{I}^-]^2 = (s)(2s)^2 = 4s^3$

$9.8 \times 10^{-9} = 4s^3$

$s^3 = 2.45 \times 10^{-9}$

$s = \sqrt[3]{2.45 \times 10^{-9}} \approx 1.35 \times 10^{-3}$ M

The molar solubility is $1.35 \times 10^{-3}$ M. This means $[\text{Pb}^{2+}] = 1.35 \times 10^{-3}$ M and $[\text{I}^-] = 2.70 \times 10^{-3}$ M at equilibrium.

ICE Tables With a Common Ion

When solving for solubility in a solution that already contains a common ion, the initial concentration of that ion is NOT zero.

Example: Find the molar solubility of PbI₂ in 0.10 M NaI solution.

$K_{sp} = (s)(0.10 + 2s)^2 = 9.8 \times 10^{-9}$

Since $K_{sp}$ is tiny and the common ion concentration is 0.10, s will be extremely small. Approximate: 0.10 + 2s ≈ 0.10.

$(s)(0.10)^2 = 9.8 \times 10^{-9}$

$s(0.01) = 9.8 \times 10^{-9}$

$s = 9.8 \times 10^{-7}$ M

Compare this to $1.35 \times 10^{-3}$ M in pure water - the common ion reduced solubility by over 1,000-fold.

Common ICE Table Mistakes

1. Forgetting stoichiometric ratios. If the coefficient is 2, the change is 2x, not x.
2. Wrong sign on the Change row. If the reaction goes forward, reactants decrease (-x) and products increase (+x).
3. Approximating when K is not small enough. Always check the 5% rule after solving.
4. Including a solid in the table. Solids do not appear in ICE tables or K expressions.
5. Forgetting the common ion. If a common ion is already present, its initial concentration is NOT zero.