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Honors Chemistry Unit 9 Review: Chemical Kinetics and Equilibrium

Unit 9 asks two questions about any reaction: how fast does it go, and how far does it go before it settles? By the end you should be able to explain rate changes with collision theory, write and calculate equilibrium constants, and predict how an equilibrium shifts when you change concentration, pressure, or temperature.

Read the overview first, then work through each topic's notes and checkpoint question before testing yourself on the common mistakes and key terms.

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What is Honors Chemistry unit 9?

This unit links two ideas that are easy to mix up. Kinetics tells you how quickly a reaction happens, while equilibrium tells you where the reaction ends up. A reaction can be fast and barely make product, or slow and almost completely convert to product.

Honors Chemistry Unit 9 covers chemical kinetics and chemical equilibrium. You learn the factors that change reaction rate (concentration, temperature, surface area, pressure for gases, and catalysts), explained through collision theory and activation energy, along with simple rate laws like Rate=k[A]m[B]nRate = k[A]^m[B]^n. You then study reversible reactions that reach a dynamic equilibrium, write and calculate equilibrium constants (KcK_c and KpK_p), and use Le Chatelier's principle to predict shifts. The Haber process for making ammonia shows both halves of the unit working together.

Reactions happen only when particles collide with enough energy and the right orientation.

Collision theory says most collisions do nothing. A collision has to supply at least the activation energy, EaE_a, and line up the right atoms. Raising temperature, concentration, or surface area increases effective collisions, and a catalyst lowers EaE_a so more collisions succeed.

Equilibrium is a balance of two ongoing reactions, not a stopping point.

At equilibrium the forward and reverse reactions still run at equal rates, so concentrations stay constant. The equilibrium constant KK compares products to reactants at that point. A large KK means products are favored and a small KK means reactants are favored.

A disturbed equilibrium shifts to partially undo the change.

Le Chatelier's principle predicts which way the system moves after you add or remove a substance, change the volume of a gas mixture, or heat or cool it. Only a temperature change alters the value of KK. The other stresses move the position of equilibrium while KK stays the same.

Rate and extent are separate questions, and you need both to control a reaction.

Kinetics decides how fast you reach equilibrium, and equilibrium decides what the mixture looks like once you get there. A catalyst gets you there faster but doesn't change the final mixture. Changing temperature affects both, which is why chemists often settle on a compromise temperature.

Honors Chemistry unit 9 topics

9.3

Le Chatelier’s Principle

Le Chatelier's principle says that when you stress a system at equilibrium, it shifts in the direction that partially counteracts the stress.

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Unit 9 review notes

9.1

Factors that Affect the Rate of Reactions

Reaction rate is how fast reactants are used up or products form, usually in molarity per second (M/s). Collision theory explains every factor. Particles must collide with at least the activation energy, EaE_a, and the correct orientation.

Higher concentration packs more particles into the same space, so collisions happen more often. Higher temperature makes particles move faster, and more importantly a bigger fraction of collisions now have energy above EaE_a. A common rule of thumb is that a 10°C rise roughly doubles the rate of many reactions, but that is only an approximation. Grinding a solid into powder increases surface area, so more particles are exposed. Squeezing a gas into a smaller volume raises its concentration. A catalyst provides a different pathway with a lower EaE_a and is not used up.

A rate law shows how concentration affects rate, Rate=k[A]m[B]nRate = k[A]^m[B]^n. The orders mm and nn come from experiments, not from the coefficients in the balanced equation.

Worked example. Trial 1 has [A]=0.10[A] = 0.10 M, [B]=0.10[B] = 0.10 M, and a rate of 2.0×10−32.0 \times 10^{-3} M/s. Doubling [A][A] to 0.20 M gives 8.0×10−38.0 \times 10^{-3} M/s, which is 4 times faster, so A is second order (22=42^2 = 4). Doubling [B][B] instead gives 4.0×10−34.0 \times 10^{-3} M/s, so B is first order. The rate law is Rate=k[A]2[B]Rate = k[A]^2[B], and k=2.0×10−3(0.10)2(0.10)=2.0 M−2s−1k = \frac{2.0 \times 10^{-3}}{(0.10)^2(0.10)} = 2.0\ M^{-2}s^{-1}.

  • Activation energy ($E_a$): The minimum energy a collision needs in order to turn reactants into products.
  • Catalyst: A substance that speeds up a reaction by lowering the activation energy and comes out of the reaction unchanged.
  • Reaction order: The exponent on a reactant's concentration in the rate law, found by experiment.

Check yourself. In a reaction with Rate=k[A][B]2Rate = k[A][B]^2, what happens to the rate if you triple [B][B] and keep [A][A] the same? (Answer: The rate increases by a factor of 32=93^2 = 9.)

FactorWhat changesWhy the rate goes up
Higher concentrationMore particles per literCollisions happen more often
Higher temperatureParticles move fasterMore collisions per second, and a larger fraction have energy above $E_a$
More surface areaSolid broken into smaller piecesMore particles exposed to collisions
Higher pressure (gases only)Same gas squeezed into less volumeGas concentration rises, so collisions happen more often
CatalystNew reaction pathwayLower $E_a$, so more collisions succeed
9.2

Chemical Equilibrium

A reversible reaction can run both ways. If you start with reactants and no products, the reverse rate is initially zero, while the forward reaction makes products. As products build up, the reverse rate increases. The forward and reverse rates eventually become equal if the system reaches equilibrium. At that point reactions are still happening, but concentrations stay constant. Equilibrium needs a closed system, which keeps reactants and products from entering or leaving.

For aA+bB⇌cC+dDaA + bB \rightleftharpoons cC + dD, the equilibrium constant is Kc=[C]c[D]d[A]a[B]bK_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}. Coefficients become exponents. KpK_p uses partial pressures of gases instead of molarities. Leave out pure solids and pure liquids, because their concentrations don't change.

If K>1K > 1, products are favored at equilibrium. If K<1K < 1, reactants are favored. For a given reaction, KK depends only on temperature, not on how much you started with.

Worked example. For N2(g)+3H2(g)⇌2NH3(g)N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g), a mixture at equilibrium has [N2]=0.50[N_2] = 0.50 M, [H2]=0.20[H_2] = 0.20 M, and [NH3]=0.10[NH_3] = 0.10 M. Write Kc=[NH3]2[N2][H2]3K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}. Then Kc=(0.10)2(0.50)(0.20)3=0.010(0.50)(0.0080)=0.0100.0040=2.5K_c = \frac{(0.10)^2}{(0.50)(0.20)^3} = \frac{0.010}{(0.50)(0.0080)} = \frac{0.010}{0.0040} = 2.5. Since Kc>1K_c > 1, products are somewhat favored at this temperature.

  • Dynamic equilibrium: The state where forward and reverse reactions run at equal rates, so concentrations stay constant.
  • $K_c$: The equilibrium constant written with molar concentrations, products over reactants, each raised to its coefficient.
  • $K_p$: The equilibrium constant written with partial pressures of gases instead of concentrations.

Check yourself. Write the KcK_c expression for CaCO3(s)⇌CaO(s)+CO2(g)CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g). (Answer: Kc=[CO2]K_c = [CO_2], because the two solids are left out.)

9.3

Le Chatelier’s Principle

Le Chatelier's principle says that when you stress a system at equilibrium, it shifts in the direction that partially counteracts the stress. It never fully undoes the change, but it moves to a new equilibrium position.

For concentration, the system shifts away from what you add and toward what you remove. Adding N2N_2 to the ammonia equilibrium pushes it toward NH3NH_3.

For temperature, treat heat like a chemical. In an exothermic reaction, write heat as a product, so heating shifts toward reactants and cooling shifts toward products. In an endothermic reaction, heat is a reactant, so the pattern flips. Temperature is the only stress that changes the value of KK.

For pressure, only gases count. Decreasing volume (raising pressure) shifts toward the side with fewer moles of gas. If both sides have the same number of gas moles, nothing shifts. Adding an inert gas like argon at constant volume doesn't shift anything, because the partial pressures of the reacting gases don't change.

A catalyst never shifts equilibrium. It speeds up the forward and reverse reactions equally.

The Haber process, N2+3H2⇌2NH3N_2 + 3H_2 \rightleftharpoons 2NH_3 with ΔH=−92\Delta H = -92 kJ, has 4 moles of gas on the left and 2 on the right. Factories use high pressure (around 200 atm) to favor ammonia. Low temperature would also favor ammonia but would be too slow, so they compromise near 450°C and add an iron catalyst.

  • Stress: Any change in concentration, pressure or volume, or temperature that disturbs an equilibrium.
  • Shift: The net movement of a reaction toward products or reactants as it reaches a new equilibrium.
  • Exothermic reaction: A reaction that releases heat, so heat can be written on the product side.

Check yourself. For 2SO2(g)+O2(g)⇌2SO3(g)2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g), which way does the equilibrium shift if you decrease the volume? (Answer: Toward SO3SO_3, because the right side has 2 moles of gas and the left side has 3.)

StressDirection of shiftDoes K change?
Add a reactantToward productsNo
Remove a productToward productsNo
Decrease volume of a gas mixtureToward fewer moles of gasNo
Raise temperature, exothermic reactionToward reactantsYes, K decreases
Raise temperature, endothermic reactionToward productsYes, K increases
Add a catalystNo shiftNo

Key terms

TermDefinition
Reaction rateThe change in concentration of a reactant or product per unit time, often in M/s.
Collision theoryThe model that says reactions happen when particles collide with enough energy and the correct orientation.
Activation energyThe minimum collision energy needed to start a reaction.
CatalystA substance that lowers activation energy and speeds up a reaction without being used up.
Rate lawAn equation like $Rate = k[A]^m[B]^n$ that links reaction rate to reactant concentrations.
Reversible reactionA reaction where products can react to re-form the original reactants.
Chemical equilibriumThe state where forward and reverse reaction rates are equal and concentrations stay constant.
Equilibrium constant (K)The ratio of product to reactant concentrations or pressures at equilibrium, each raised to its coefficient.
Le Chatelier's principleThe rule that a system at equilibrium shifts to partially counteract any stress placed on it.
Haber processThe industrial reaction $N_2 + 3H_2 \rightleftharpoons 2NH_3$, run at high pressure with an iron catalyst to make ammonia.

Mistakes to avoid in Chemical Kinetics and Equilibrium

Thinking equilibrium means equal amounts of reactants and products

Equilibrium means equal rates, not equal concentrations. A reaction with K=10−5K = 10^{-5} strongly favors reactants, but you need the balanced reaction and the starting amounts to calculate the actual equilibrium mixture. Check whether the rates are equal and the concentrations are constant, not whether the amounts match.

Saying a catalyst shifts equilibrium toward products

A catalyst lowers the activation energy for both the forward and reverse reactions, so it speeds both up by the same factor. You reach equilibrium sooner, but the final mixture and the value of KK stay the same.

Reading reaction orders straight from the balanced equation

In Rate=k[A]m[B]nRate = k[A]^m[B]^n, the exponents come from experimental data, such as comparing initial rates when one concentration doubles. Coefficients do become exponents in the equilibrium expression KK, which is why the two get mixed up.

Do not include pure solids or pure liquids in K, and do not assume that extra pure solid shifts equilibrium

Pure solids and pure liquids have constant concentrations, so they stay out of KK expressions. Adding more of a pure solid, like extra CaCO3CaCO_3, to an existing equilibrium doesn't shift it as long as some of that solid is already present. Adding a liquid solvent such as water is a different case, because it dilutes the dissolved substances and can shift an aqueous equilibrium. Changing the concentrations or partial pressures of reacting gases and dissolved substances can shift equilibrium, and a temperature change shifts it and changes the value of KK.

Study unit 9 other ways

Ready to review Unit 9?Start with the notes, check the topic cards, and use the practice or resource links when they are available for this course.