🧪 Contents · Interactive Lab Chapter 13 · Fundamental Equilibrium Concepts
Chemistry · Chapter 13

Fundamental Equilibrium Concepts

Some reactions never finish. They slow down, stop changing, and sit there holding both reactants and products at once. Nothing has stalled or died. The forward reaction and the reverse reaction are both still running, at the same speed. That balance is called equilibrium, and it rules a huge slice of chemistry. It sets how much ammonia a factory can make. It sets how much oxygen your blood can carry. In this chapter you will write the equilibrium constant, compare it with the reaction quotient, and predict how a mixture answers back when you push on it.

By the end of this chapter you can

  • Explain why an equilibrium is busy, not dead, and what has to stay constant.
  • Write Kc and Kp from a balanced equation, and leave out what does not belong.
  • Convert between Kc and Kp using Δn.
  • Find Q for any mixture and say which way the reaction must go.
  • Predict the shift when you add, remove, squeeze, heat, or add a catalyst.
  • Set up an ICE table and solve for every equilibrium amount.
  • Use the small-x shortcut, then test whether it was allowed.

Section 1Dynamic equilibrium

Two reactions in one tube

Seal a little N2O4 gas in a glass tube. The gas starts out colorless. Within seconds the tube turns brown. The brown gas is NO2. Then the color stops changing. It holds that same shade for as long as you watch.

The tube did not run out of material. Crack it open later and you still find both gases. What stopped was the net change. This is a reversible reaction, one that can run in either direction. We write it with a double arrow.

N2O4(g) ⇌ 2 NO2(g)

At the very start there is no NO2 at all. Only the forward reaction can happen, and it runs fast. As NO2 piles up, the reverse reaction switches on. Its rate climbs while the forward rate falls. At some moment the two rates match. From then on, every N2O4 that splits apart is replaced by two NO2 joining back up.

Forward and reverse rates meeting at equilibrium time → rate forward rate reverse rate the rates are now equal amounts hold steady from here on only Nâ‚‚Oâ‚„ can react at first splitting apart joining back up
Equilibrium is where the two curves meet. The forward rate drops as reactant is used up. The reverse rate climbs as product builds. Once they are equal, the measured amounts stop moving.

Equal rates, not equal amounts

Dynamic equilibrium is that state. The two rates are equal, so the amounts stay put. Dynamic means moving. Molecules keep reacting both ways the whole time. Only the totals hold still. Many people picture a dead flask. Picture a busy escalator instead, with as many riders stepping on as stepping off.

Equal rates do not mean equal amounts. One side can hold far more material than the other. A rate depends on the amounts and on the rate constant. A sluggish reverse step needs a big pile of product to keep pace with a brisk forward step.

Equilibrium also needs a closed system. If the NO2 can leak away, the reverse rate never catches up, and the reaction keeps creeping forward. One more thing is worth knowing. You reach the same resting mixture from either side. Start with pure NO2 instead, hold the same temperature, and you land in the same place.

Worked example 1 · Rates that match

Hydrogen and iodine form hydrogen iodide: H2(g) + I2(g) ⇌ 2 HI(g). Treat each direction here as a single step, so rate = k × the concentrations. The forward constant is kf = 1.0 × 10-2 L/mol·s. The reverse constant is kr = 2.0 × 10-4 L/mol·s. At equilibrium [H2] = [I2] = 0.0100 M. Show that the two rates really are equal.

Step 1 · find the ratio of the constantsSetting the rates equal gives kf[H2][I2] = kr[HI]2. Rearranged, [HI]2 ÷ ([H2][I2]) = kf ÷ kr. That ratio is (1.0 × 10-2) ÷ (2.0 × 10-4) = 50. The units L/mol·s cancel, so the 50 is a bare number.

Step 2 · get the amount of product[HI]2 = 50 × (0.0100)(0.0100) = 5.0 × 10-3 M2. Take the square root: [HI] = 0.0707 M.

Step 3 · check both ratesForward rate = (1.0 × 10-2 L/mol·s)(0.0100 M)(0.0100 M) = 1.0 × 10-6 mol/L·s. Reverse rate = (2.0 × 10-4 L/mol·s)(0.0707 M)2 = (2.0 × 10-4)(5.0 × 10-3) = 1.0 × 10-6 mol/L·s.

AnswerThe rates match at 1.0 × 10-6 mol/L·s, even though HI is seven times more plentiful than either starting gas. That ratio of 50 is the equilibrium constant, and the next section gives it a name.

Key points
  • A reversible reaction runs both ways, so it settles instead of finishing.
  • At equilibrium the forward and reverse rates are equal, not zero.
  • Measured amounts stop changing, but molecules keep reacting.
  • Equal rates say nothing about equal amounts.
  • The system must be closed, and both starting points lead to the same mixture.
Interactive Watch forward and reverse rates meet and hold. Open full ↑

Section 2The equilibrium constant

Products over reactants

Equilibrium is more than a yes or no. It has a number attached. That number is the equilibrium constant, written K. For the general reaction

a A + b B ⇌ c C + d D

the constant is one fraction. Products go on top. Reactants go on the bottom. Each concentration is raised to its own coefficient from the balanced equation.

Kc = [C]c[D]d ÷ [A]a[B]b

This rule is the law of mass action. The small c in Kc means concentrations in mol/L. Square brackets always mean molarity. K values are reported with no units at all. They really come from activities, which are ratios against a standard state, so every unit cancels.

K depends on temperature and on nothing else. Change the amounts and the mixture shifts, but K holds. Heat the flask and K itself takes a new value.

What stays out of the expression

Pure solids and pure liquids are left out. Their concentration cannot change as the reaction runs. A chunk of solid has the same packing whether the chunk is large or small. So for the roasting of limestone,

CaCO3(s) ⇌ CaO(s) + CO2(g)   with   Kp = PCO2

only the gas appears. A reaction with more than one phase like this one is a heterogeneous equilibrium. When everything is in the same phase it is a homogeneous equilibrium. Water used as the solvent drops out for the same reason: it is there in such excess that its amount barely budges.

Kp, and the two tricks

Gases are easier to measure by pressure than by concentration. Kp is the same fraction built from partial pressures in atm. The two constants are tied together.

Kp = Kc(RT)Δn

Here R = 0.08206 L·atm/mol·K and T is in kelvin. The Δn is the moles of gas on the right minus the moles of gas on the left. When Δn = 0 the factor is 1, so the two constants are the same number.

Two small tricks come up again and again. Flip an equation and K becomes 1 ÷ K. Multiply an equation through by n and K becomes Kn. Add two equations together and you multiply their constants.

What the size of K tells you 10-10 10-5 10-3 1 103 105 1010 mostly reactants left barely any product forms both sides matter you can measure all of them mostly products made only traces of reactant left Nâ‚‚Oâ‚„ gives K = 0.0046 at 25 °C, so it sits just left of the middle. Hâ‚‚ plus Iâ‚‚ gives K = 50, a little right of the middle.
The size of K is the whole story in one glance. The scale runs by powers of ten. A K below about 10-3 leaves the reactants nearly untouched. A K above about 103 all but uses them up.

Worked example 2 · From Kc to Kp

For N2O4(g) ⇌ 2 NO2(g), Kc = 4.6 × 10-3 at 25 °C. Find Kp. Then do the same for N2(g) + 3 H2(g) ⇌ 2 NH3(g), where a run at 500 K gives Kc = 0.0100.

Step 1 · count the gas molesFor the first one, 2 moles of gas on the right and 1 on the left. So Δn = 2 - 1 = +1.

Step 2 · build RTT = 25 + 273.15 = 298.15 K. RT = (0.08206 L·atm/mol·K)(298.15 K) = 24.47 L·atm/mol.

Step 3 · raise and multiplyKp = (4.6 × 10-3)(24.47)1 = 0.113.

Step 4 · now the ammonia caseGas moles: 2 on the right, 1 + 3 = 4 on the left, so Δn = 2 - 4 = -2. RT = (0.08206)(500) = 41.03 L·atm/mol, and (41.03)2 = 1683. A negative exponent means divide: Kp = 0.0100 ÷ 1683 = 5.94 × 10-6.

AnswerKp = 0.113 for the first reaction and 5.94 × 10-6 for the second. A negative Δn pushes Kp far below Kc.

Watch the sign of Δn. This sign trips people up more than anything else in the chapter. Count the gas moles on each side first, write Δn down, and only then reach for RT.
Key points
  • Kc is products over reactants, each raised to its coefficient.
  • Pure solids, pure liquids, and solvent water are left out.
  • Kp = Kc(RT)Δn, with Δn counted for gases only.
  • A big K means the products win; a small K means the reactants stay.
  • Flipping the equation inverts K, and scaling it by n raises K to the n.
  • Only a change in temperature changes the value of K.
Interactive Write K from an equation and see what big and small K mean. Open full ↑

Section 3The reaction quotient Q

The same fraction, any moment you like

K only describes the resting state. Most mixtures you meet are not resting. The reaction quotient Q fixes that. It uses the exact same fraction as K, but you plug in the amounts you have right now.

Q = [C]c[D]d ÷ [A]a[B]b   at any instant

Then you compare the two numbers. The comparison is the whole point, and it has only three outcomes.

If Q < K, the fraction is too small. There is not enough product on top. The reaction runs forward, to the right, making more product until Q climbs up to K.

If Q > K, the fraction is too large. There is too much product. The reaction runs in reverse, to the left, rebuilding reactant until Q drops to K.

If Q = K, the mixture is already at equilibrium. Nothing shifts. The net change is zero.

Two edge cases are easy once you see them. With no product at all, Q = 0, which is below any K, so the reaction must go forward. With no reactant at all, the bottom is zero, Q blows up, and the reaction must go in reverse.

Three ways Q can compare with K K Q Q < K goes right too little product, so the forward reaction wins for a while K Q sits on K Q = K no shift already at equilibrium, so both rates are matched K Q Q > K goes left too much product, so the reverse reaction wins for a while
Q always moves toward K, never away from it. The gold dot is where your mixture sits. The teal arrow is the direction of net change. The dot slides until it lands on the tick mark for K.

Why this beats guessing

Without Q you would have to guess. A flask holding plenty of product might still need to make more. Q settles it with one calculation. That is why Q comes before every ICE table in the next sections. It tells you the sign of the change.

You will meet Q again in later chapters. The same test decides whether a solid will drop out of a solution. There you compare Q with Ksp. The logic never changes, only the names.

Worked example 3 · Which way does it go?

For H2(g) + I2(g) ⇌ 2 HI(g), K = 50.0 at 731 K. A sealed flask at 731 K holds [H2] = 0.200 M, [I2] = 0.200 M, and [HI] = 0.500 M. Which way does the reaction move?

Step 1 · write Q in the same shape as KQ = [HI]2 ÷ ([H2][I2]).

Step 2 · put the numbers inTop: (0.500)2 = 0.250. Bottom: (0.200)(0.200) = 0.0400.

Step 3 · divideQ = 0.250 ÷ 0.0400 = 6.25.

Step 4 · compareQ = 6.25 and K = 50.0, so Q < K.

AnswerThe reaction runs forward. H2 and I2 keep being used up, and [HI] keeps rising, until Q reaches 50.0.

A second lookNow try [HI] = 2.00 M with [H2] = [I2] = 0.100 M. Q = (2.00)2 ÷ (0.100)(0.100) = 4.00 ÷ 0.0100 = 400. Here Q > K, so this flask runs in reverse and breaks HI back apart.

Key points
  • Q is built exactly like K but from the amounts present right now.
  • Q < K sends the reaction to the right, toward more product.
  • Q > K sends the reaction to the left, toward more reactant.
  • Q = K means the mixture is already at rest.
  • Q drifts toward K and then stops there.
Interactive Compare Q with K and predict which way the reaction shifts. Open full ↑

Section 4Le Chatelier's principle

Push on it and it pushes back

A mixture at equilibrium is not fragile. Disturb it and it rearranges itself. Le Chatelier's principle says it this way. When you put a stress on a system at equilibrium, the system shifts in the direction that partly relieves that stress.

Partly is the honest word. It moves only far enough to make Q equal K again, and then it stops. Each kind of stress has its own rule, and all four rules come from the same Q against K test.

Adding and removing

Add more of any species and the system shifts away from it. Take some away and the system shifts toward it. Check it with Q. Add product and the top of the fraction grows, so Q > K, so the reaction runs in reverse.

Factories use this on purpose. In the ammonia process the NH3 is condensed out as fast as it forms. Removing product keeps Q below K forever, so the reaction keeps running forward.

Squeezing and stretching

Changing the volume only matters when the gas moles differ across the arrow. Squeeze a gas mixture into half the space and every concentration doubles at once. Whether Q rises or falls then depends on Δn.

The rule is short. Squeezing shifts the reaction toward the side with fewer moles of gas. Expanding shifts it toward the side with more. If the gas moles match on both sides, nothing shifts at all.

One classic trap. Pumping in an unreactive gas like argon at fixed volume changes the total pressure but changes no concentration. Q does not move, so neither does the equilibrium.

Heating and cooling

Temperature is the only stress that changes K itself. The trick is to treat heat as if it were a species in the equation. For an endothermic forward reaction, heat belongs on the left.

heat + N2O4(g) ⇌ 2 NO2(g)    ΔH = +57.2 kJ/mol

Now the rule for adding a species does the work. Heating adds heat to the left, so the reaction shifts right and K rises. Cooling removes heat from the left, so it shifts back and K falls. That is why a sealed tube of these gases darkens in hot water and pales in ice water. For an exothermic reaction heat sits on the right, and every arrow flips.

Catalysts do not count

A catalyst speeds the forward step and the reverse step by the same factor. The rates both rise, so they stay matched. K is untouched and no shift happens. What you gain is time, since a slow reaction may now settle in minutes.

Squeezing the N2O4 and NO2 equilibrium 1. at equilibrium 1.00 L, Q = K = 0.0046 Nâ‚‚Oâ‚„ = 0.0500 M NOâ‚‚ = 0.0152 M halve the volume 2. the instant after 0.500 L, Q = 0.0092 every amount doubled Q is twice K, so it goes left shifts 3. settled again 0.500 L, Q = K = 0.0046 Nâ‚‚Oâ‚„ = 0.104 M NOâ‚‚ = 0.0219 M, down from 0.0303 Nâ‚‚Oâ‚„ NOâ‚‚
Squeezing shifts this one to the left. Big teal circles stand for N2O4, small gold ones for NO2. Two NO2 become one N2O4, so the crowding eases. The brown is paler than in panel 2, though still deeper than in panel 1.

Worked example 4 · Squeeze it and track Q

A 1.00 L flask at 25 °C holds N2O4 and NO2 at equilibrium, with Kc = 4.6 × 10-3. The N2O4 is 0.0500 M. Find [NO2]. Then push the piston to 0.500 L and show which way the reaction shifts.

Step 1 · get the starting [NO2]Kc = [NO2]2 ÷ [N2O4], so [NO2]2 = (4.6 × 10-3)(0.0500) = 2.30 × 10-4 M2. The square root gives [NO2] = 0.0152 M.

Step 2 · halve the volumeSame moles in half the space, so both concentrations double. [N2O4] = 0.1000 M and [NO2] = 0.0303 M.

Step 3 · find Q right after the squeezeQ = (0.0303)2 ÷ 0.1000 = (9.20 × 10-4) ÷ 0.1000 = 9.2 × 10-3.

Step 4 · compare with KQ = 9.2 × 10-3 is exactly twice Kc = 4.6 × 10-3. So Q > K.

AnswerThe reaction shifts left, toward N2O4. That is the side with 1 mole of gas instead of 2, so the shift lowers the crowding. Notice why Q doubled: Δn = +1, so halving the volume multiplies Q by 21.

Le Chatelier is a shortcut, not a separate law. Every one of these rules is just the Q against K test in a hurry. When a case confuses you, compute Q and trust the number instead.
Key points
  • A stressed system shifts in the direction that partly relieves the stress.
  • Adding a species shifts the reaction away from it; removing one pulls it back.
  • Squeezing shifts toward the side with fewer moles of gas.
  • An unreactive gas added at fixed volume changes nothing.
  • Heating shifts an endothermic reaction forward and raises K.
  • A catalyst changes the speed, never the position, of equilibrium.
Interactive Add, remove, squeeze and heat a system at equilibrium. Open full ↑

Section 5ICE tables

Three rows that organize everything

Most equilibrium problems give you starting amounts and K, then ask for the final amounts. An ICE table keeps that bookkeeping straight. The letters stand for Initial, Change, and Equilibrium. You get one column per species and three rows.

The Initial row holds what you mixed, before any reaction. The Change row holds how much each one moves. The Equilibrium row is just the first two rows added together.

The Change row is where the thinking happens. You call the extent of reaction x. Every species changes by its coefficient times x. Reactants lose, so they get a minus sign. Products gain, so they get a plus sign. A coefficient of 2 means the change is 2x, not x.

How an ICE table is built Nâ‚‚Oâ‚„ (g) ⇌ 2 NOâ‚‚ (g) row Nâ‚‚Oâ‚„ NOâ‚‚ I nitial 0.0400 0 C hange − x + 2x E quil. 0.0400 − x 2x the 2 comes from the coefficient, and the sign from the side Then substitute the bottom row into K: K = (2x)² ÷ (0.0400 − x) = 0.0046, and you solve that for x.
The bottom row is what you feed into K. Add the Initial and Change rows to get the Equilibrium row, then put those expressions into the equilibrium constant and solve for x.

From the table to an answer

Once the Equilibrium row is written, the rest is algebra. Substitute those expressions into the K expression. You now have one equation in one unknown. Solve for x, then feed x back into the Equilibrium row to get every concentration.

Two sanity checks are worth the ten seconds. First, no concentration may come out negative. If it does, you probably picked the wrong root or the wrong direction. Second, put your answers back into K. The fraction should give the K you started with.

Sometimes both sides of the equation are perfect squares. Then you can take the square root of both sides and skip the quadratic formula. That shortcut shows up in the example below.

One warning about units. Put concentrations in the table, not moles, unless you convert at the end. If the problem gives moles and a volume, divide first. Doing it later is where errors sneak in.

Worked example 5 · A full ICE table

1.00 mol of H2 and 1.00 mol of I2 go into a 1.00 L flask at 731 K. For H2(g) + I2(g) ⇌ 2 HI(g), Kc = 50.0. Find all three concentrations at equilibrium.

Step 1 · starting concentrations1.00 mol ÷ 1.00 L = 1.00 M for each gas. There is no HI yet, so [HI] = 0.

Step 2 · the tableInitial: 1.00, 1.00, 0. Change: −x, −x, +2x. Equilibrium: 1.00 − x, 1.00 − x, 2x. Since Q starts at 0, the reaction must go forward, so x is positive.

Step 3 · substituteKc = (2x)2 ÷ [(1.00 − x)(1.00 − x)] = 50.0, which is (2x)2 ÷ (1.00 − x)2 = 50.0.

Step 4 · use the perfect squareBoth sides are squares, so take the square root. 2x ÷ (1.00 − x) = √50.0 = 7.0711.

Step 5 · solve for x2x = 7.0711 − 7.0711x, so 9.0711x = 7.0711 and x = 0.7795 M.

Step 6 · fill the bottom row[H2] = [I2] = 1.00 − 0.7795 = 0.220 M. [HI] = 2(0.7795) = 1.559 M, or 1.56 M.

Answer and check[H2] = [I2] = 0.220 M and [HI] = 1.56 M. Check: (1.559)2 ÷ (0.2205)2 = 2.431 ÷ 0.04862 = 50.0. It matches.

Key points
  • ICE stands for Initial, Change, Equilibrium, one column per species.
  • Each Change entry is the coefficient times x, minus for reactants and plus for products.
  • The Equilibrium row is the Initial row plus the Change row.
  • Feed the Equilibrium row into K to get one equation in x.
  • Reject any root that makes a concentration negative, then check your answer in K.
Interactive Fill an ICE table and solve for the equilibrium amounts. Open full ↑

Section 6Equilibrium calculations

Four kinds of problem

Almost every equilibrium question is one of four types. Knowing which one you are holding saves a lot of flailing.

Type one gives you all the equilibrium concentrations and asks for K. Just plug them in. Type two gives initial amounts plus one equilibrium amount. Use the known change to find x, then fill the rest of the table. Type three gives K and the initial amounts and asks for the equilibrium amounts. That is the ICE table from the last section. Type four is type three with a tiny K, and it has a shortcut.

The small-x shortcut

When K is very small, almost nothing reacts. The reactant you started with barely drops. So the term (c0 − x) is nearly c0, and you can drop the x from that one spot. This is the small-x approximation. It turns an awkward quadratic into a square root.

You do not drop every x. Only the one being subtracted from a much larger number. The x values sitting alone in the numerator are the answer, so they stay.

Two tests tell you whether the shortcut was allowed. Before you start, compare c0 with K. If c0 ÷ K is larger than about 400, the shortcut is safe. After you solve, use the five percent rule. If x is under five percent of c0, keep the answer. If not, throw it out and solve the full quadratic.

x ÷ c0 × 100% < 5%   →   the shortcut holds

That same fraction has a name: the percent dissociation. It says what share of your starting material fell apart.

Deciding whether the small-x shortcut is allowed Work out c ÷ K start over the constant Is it above 400? if yes, the x you subtract hardly moves the answer yes Drop that one x take a square root no Keep every x solve the quadratic Last step: is x under five percent of c? if it is not, the shortcut answer has to go Both tests ask the same thing.
Two gates guard the shortcut. The ratio test runs before you solve. The five percent rule runs after. Failing either one sends you to the quadratic formula.

Worked example 6 · A tiny K, so the shortcut works

Phosgene breaks up a little at 100 °C: COCl2(g) ⇌ CO(g) + Cl2(g), with Kc = 2.2 × 10-10. You start with 0.100 M COCl2. Find every equilibrium concentration.

Step 1 · run the ratio testc0 ÷ K = 0.100 ÷ (2.2 × 10-10) = 4.5 × 108. That is far above 400, so the shortcut is fine.

Step 2 · the tableInitial: 0.100, 0, 0. Change: −x, +x, +x. Equilibrium: 0.100 − x, x, x.

Step 3 · drop the one xKc = x2 ÷ (0.100 − x) becomes x2 ÷ 0.100 = 2.2 × 10-10.

Step 4 · solvex2 = (2.2 × 10-10)(0.100) = 2.2 × 10-11 M2, so x = 4.69 × 10-6 M.

Step 5 · check the five percent rule(4.69 × 10-6) ÷ 0.100 × 100% = 0.0047%. Nowhere near 5%.

Answer[CO] = [Cl2] = 4.7 × 10-6 M, and [COCl2] = 0.100 M to three figures. Hardly any gas broke up, which is what a K of 10-10 promised.

When the shortcut breaks

The shortcut fails when K is not small compared with the starting amount. Then x is a real chunk of c0, and dropping it skews the answer badly. You go back to the full quadratic, in the form ax2 + bx + c = 0, and use the formula. Take the root that keeps every concentration positive.

x = [ −b + √(b2 − 4ac) ] ÷ 2a

Worked example 7 · When the shortcut fails

Start with 0.0100 M N2O4 at 25 °C. For N2O4(g) ⇌ 2 NO2(g), Kc = 4.6 × 10-3. Try the shortcut, see it fail, and then do it properly.

Step 1 · the ratio test warns youc0 ÷ K = 0.0100 ÷ (4.6 × 10-3) = 2.2. That is tiny, far under 400, so expect trouble.

Step 2 · try it anywayEquilibrium row: 0.0100 − x and 2x. Dropping the x gives (2x)2 ÷ 0.0100 = 4.6 × 10-3, so 4x2 = 4.6 × 10-5 and x = 3.39 × 10-3 M.

Step 3 · the five percent rule rejects it(3.39 × 10-3) ÷ 0.0100 × 100% = 34%. Way over 5%, so this answer is out.

Step 4 · set up the real quadratic(2x)2 ÷ (0.0100 − x) = 4.6 × 10-3. Cross multiply: 4x2 = 4.6 × 10-5 − (4.6 × 10-3)x. Rearranged: 4x2 + (4.6 × 10-3)x − 4.6 × 10-5 = 0.

Step 5 · use the formulab2 − 4ac = (4.6 × 10-3)2 + 4(4)(4.6 × 10-5) = 2.116 × 10-5 + 7.36 × 10-4 = 7.572 × 10-4. Its square root is 2.7517 × 10-2. Then x = (−4.6 × 10-3 + 2.7517 × 10-2) ÷ 8 = (2.2917 × 10-2) ÷ 8 = 2.86 × 10-3 M.

Step 6 · fill in and check[N2O4] = 0.0100 − 0.00286 = 7.14 × 10-3 M. [NO2] = 2(2.86 × 10-3) = 5.73 × 10-3 M. Check: (5.73 × 10-3)2 ÷ (7.14 × 10-3) = (3.28 × 10-5) ÷ (7.14 × 10-3) = 4.6 × 10-3. It matches Kc.

Answerx = 2.86 × 10-3 M, not 3.39 × 10-3 M. The shortcut was 18% high. The percent dissociation here is 29%, which is why no x could be dropped.

Do the ratio test first, every time. It costs one division and tells you which path to take before you waste a page of algebra on the wrong one.
Key points
  • Sort the problem first: find K, finish a table, or solve for x.
  • A tiny K lets you replace (c0 − x) with c0.
  • Never drop the x values that are your answer.
  • c0 ÷ K above about 400 means the shortcut is safe.
  • Then confirm with the five percent rule, and solve the quadratic if it fails.
  • Always put your answers back into K as a final check.
Interactive Work through K problems, including the small-x shortcut. Open full ↑

The chapter in one breath

  • Equilibrium is busy, not finished. The forward and reverse rates are equal, so the measured amounts hold still while molecules keep reacting.
  • K is one number for the resting state. Products over reactants, each raised to its coefficient, with pure solids and pure liquids left out.
  • Kp = Kc(RT)Δn. Count only gas moles for Δn, and use R = 0.08206 L·atm/mol·K with T in kelvin.
  • The size of K says where the reaction sits. Big K means the products win, small K means the reactants mostly stay.
  • Q is K's formula with today's amounts. Q < K goes right, Q > K goes left, Q = K sits still.
  • Le Chatelier is that same test in a hurry. Add, remove, squeeze, or heat, and the system shifts the way that partly relieves the push.
  • Only temperature changes K. Catalysts and unreactive gases change speed or pressure, never the position.
  • ICE tables carry the bookkeeping. Initial plus Change gives Equilibrium, and the Change entry is the coefficient times x.
  • Check every shortcut. Drop x only when c0 ÷ K is above about 400, then confirm with the five percent rule.

ReferenceWords worth knowing

These are the words this chapter leans on. Learn them in pairs where you can, since K and Q, or forward and reverse, only make sense together.

Catalyst
A substance that speeds both the forward and the reverse reaction by the same factor. It shortens the wait for equilibrium and leaves K and the final amounts alone.
Closed system
A container that lets no reactant or product escape. Equilibrium can only be reached in one.
Dynamic equilibrium
The state in which the forward and reverse rates are equal. Amounts stay constant while molecules keep reacting both ways.
Equilibrium constant (K)
The value of the products-over-reactants fraction once a reaction has settled. It depends only on temperature and is written without units.
Forward reaction
The change that runs left to right as the equation is written, turning reactants into products.
Heterogeneous equilibrium
An equilibrium with species in more than one phase, such as a solid and a gas. Pure solids and pure liquids are left out of K.
Homogeneous equilibrium
An equilibrium in which every species is in the same phase, such as all gases or all dissolved.
ICE table
A three-row grid of Initial, Change, and Equilibrium amounts. It organizes a problem so K gives one equation in one unknown.
Kc
An equilibrium constant built from concentrations in mol/L.
Kp
An equilibrium constant built from partial pressures in atm. It equals Kc(RT)Δn.
Law of mass action
The rule that gives the shape of K: products over reactants, each concentration raised to its coefficient.
Le Chatelier's principle
A stressed equilibrium shifts in the direction that partly relieves the stress.
Percent dissociation
The share of the starting material that reacted, found as x divided by c0 times 100%.
Reaction quotient (Q)
The same fraction as K, but filled in with the amounts present at this moment. Comparing Q with K gives the direction of change.
Reverse reaction
The change that runs right to left, turning products back into reactants.
Reversible reaction
A reaction that can run in either direction, written with a double arrow.
Shift
The net movement of a disturbed equilibrium toward one side until Q equals K again.
Small-x approximation
Replacing (c0 − x) with c0 when K is tiny. It is valid when c0 ÷ K is above about 400.
Stress
Any change forced on a system at equilibrium, such as adding a species, changing the volume, or changing the temperature.

Check yourself

1. Write the Kc expression for 2 SO2(g) + O2(g) ⇌ 2 SO3(g).
Kc = [SO3]2 ÷ ([SO2]2[O2]) Products go on top. Each exponent is that species' coefficient, and O2 has an unwritten coefficient of 1.
2. At equilibrium that same mixture holds [SO2] = 0.400 M, [O2] = 0.200 M, and [SO3] = 0.600 M. Find Kc.
Kc = 11.3 Top: (0.600)2 = 0.360. Bottom: (0.400)2(0.200) = (0.160)(0.200) = 0.0320. Divide: 0.360 ÷ 0.0320 = 11.25, which rounds to 11.3.
3. Convert that Kc of 11.25 to Kp at 1000 K.
Kp = 0.137 Gas moles: 2 on the right, 3 on the left, so Δn = −1. RT = (0.08206)(1000) = 82.06 L·atm/mol. A Δn of −1 means divide: 11.25 ÷ 82.06 = 0.137.
4. For PCl5(g) ⇌ PCl3(g) + Cl2(g), Kc = 1.80 at 523 K. A flask holds [PCl5] = 0.100 M, [PCl3] = 0.200 M, and [Cl2] = 0.300 M. Which way does it shift?
Forward, to the right Q = (0.200)(0.300) ÷ 0.100 = 0.0600 ÷ 0.100 = 0.600. Since 0.600 is less than 1.80, Q < K, so more PCl5 must break apart.
5. Kc = 50.0 for H2 + I2 ⇌ 2 HI. What is Kc for 2 HI ⇌ H2 + I2 at the same temperature?
0.0200 Flipping an equation inverts its constant. 1 ÷ 50.0 = 0.0200. The same mixture is being described from the other side.
6. Start with 0.500 M HI at 731 K, where 2 HI ⇌ H2 + I2 has Kc = 0.0200. Find all three equilibrium concentrations.
[HI] = 0.390 M, [H2] = [I2] = 0.0551 M The table gives 0.500 − 2x, x, x. So x2 ÷ (0.500 − 2x)2 = 0.0200. Both sides are squares, so x ÷ (0.500 − 2x) = √0.0200 = 0.14142. Then x = 0.070711 − 0.28284x, so 1.28284x = 0.070711 and x = 0.05512 M. Finally [HI] = 0.500 − 2(0.05512) = 0.390 M.
7. A sealed piston holds N2(g) + 3 H2(g) ⇌ 2 NH3(g) at equilibrium. You push it to half its volume. Which way does the reaction shift?
To the right, toward NH3 The left side has 4 moles of gas and the right side has 2. Squeezing favors the side with fewer gas moles, so more ammonia forms.
8. You pump argon into that same rigid flask, raising the total pressure. What happens to the equilibrium?
Nothing shifts Argon does not react and the volume did not change, so no concentration or partial pressure of a reacting gas changed. Q is untouched, so there is no shift.
9. For N2O4(g) ⇌ 2 NO2(g), ΔH = +57.2 kJ/mol. You warm the tube. What happens to Kc and to the color?
Kc rises and the brown deepens A positive ΔH puts heat on the reactant side. Adding heat shifts the reaction right, making more brown NO2, and the constant itself takes a larger value.
10. Adding a catalyst to a reaction at equilibrium does what to K and to the time needed to get there?
K is unchanged, the wait gets shorter A catalyst lowers the barrier for both directions equally. Both rates climb by the same factor, so they stay matched at the same final amounts.
11. A weak acid HA breaks up as HA ⇌ H+ + A− with K = 1.0 × 10-5. You start with 0.250 M HA. Find x and test the shortcut.
x = 1.6 × 10-3 M, and the shortcut is fine Ratio test: 0.250 ÷ (1.0 × 10-5) = 2.5 × 104, well above 400. So x2 ÷ 0.250 = 1.0 × 10-5, giving x2 = 2.5 × 10-6 and x = 1.58 × 10-3 M. Check: 1.58 × 10-3 ÷ 0.250 = 0.63%, under 5%.
12. Why does a pure solid never appear in an equilibrium constant?
Its concentration cannot change A solid has a fixed density, so the amount packed into each liter of solid is the same whether the lump is large or small. A constant value folds into K instead of varying inside it.