Chemistry · Study Guide

Exam 4 Study Guide

This guide covers chapters 10 through 12. First you will look at the forces between molecules, and how they set melting points, boiling points and the shape of a heating curve. Then you will mix things together and watch a solute change the solvent's freezing and boiling points. The last chapter asks a new question: not whether a reaction happens, but how fast. Read each "what to know" list first. Copy out the equations by hand. Work the examples with the answers covered, then run the ten self-test items. The cram sheet at the end holds every equation and constant.

Chapter 10 Chapter 11 Chapter 12 Cram sheet

What this guide covers

Chapter 10 · intermolecular forces, liquid properties, phase changes and heating curves, vapor pressure, phase diagrams and crystal lattices.
Chapter 11 · the dissolving process, solution energetics, concentration units, solubility and Henry's law, colligative properties and colloids.
Chapter 12 · reaction rates, rate laws and order, integrated rate laws and half-life, collision theory, the Arrhenius equation, mechanisms and catalysis.

Chapter 10Liquids and Solids

What to know

Bonds hold a molecule together. Intermolecular forces hold separate molecules near each other. Those forces are much weaker than bonds, but they decide whether a substance is a solid, a liquid or a gas at room temperature.

There are four forces to rank. Dispersion forces act in every substance, because electron clouds flicker out of balance for an instant. They grow stronger as a molecule gains electrons. Dipole-dipole forces need a lopsided molecule with a permanent dipole. A hydrogen bond is a strong special case, and it needs hydrogen bonded to nitrogen, oxygen or fluorine. Ion-dipole forces are the strongest of the four, and they are what pulls a salt into water.

Stronger forces raise the boiling point, the viscosity and the surface tension. Viscosity is a liquid's thickness. Surface tension is the skin that lets an insect stand on water. Capillary action is a liquid climbing a narrow tube, and it happens when adhesion to the tube beats cohesion within the liquid.

Six phase changes come in three pairs. Melting and freezing. Vaporizing and condensing. Subliming and depositing. During any phase change the temperature holds steady, even while heat keeps pouring in. That heat is breaking attractions, not speeding particles up.

A heating curve shows this directly. The sloped parts are q = mcΔT, where temperature climbs. The flat parts are q = mΔH, where the phase changes. Each part needs its own specific heat, because ice, water and steam all resist heating differently.

Vapor pressure is the push of the vapor sitting above a liquid. It rises steeply with temperature. Boiling starts when the vapor pressure matches the outside pressure. On a mountain the outside pressure is lower, so water boils below 100 °C. The Clausius-Clapeyron equation links two pressure and temperature pairs, and a plot of ln P against 1/T gives a line whose slope reveals the heat of vaporization.

A phase diagram maps pressure against temperature. The three lines are the boundaries. The triple point is where all three states coexist. The critical point is where the liquid and gas lines stop telling apart. Most substances have a melting line that leans right. Water's leans left, because ice is less dense than liquid water.

Solids come in four families: ionic, metallic, covalent network and molecular. A crystal's repeating box is the unit cell. A simple cubic cell holds 1 atom, a body-centered cubic cell holds 2, and a face-centered cubic cell holds 4. Corner atoms count as one eighth each, face atoms as one half, and a body-center atom as a whole.

Lock these in

Hydrogen bonding needs H bonded to N, O or F.
Dispersion forces exist in everything and grow with electron count.
Temperature holds flat through a phase change.
Water needs 334 J/g to melt and 2,260 J/g to boil.
Boiling happens when vapor pressure equals the outside pressure.
Water's melting line leans left, because ice floats.

Equations, units and symbols

Equation	What each symbol means	Units
q = mcΔT	the sloped part of a heating curve; c belongs to that one phase	J = g × (J/g·°C) × °C
q = mΔH_fus	the flat melting step; ΔH_fus is the heat of fusion	J = g × (J/g)
q = mΔH_vap	the flat boiling step; ΔH_vap is the heat of vaporization	J = g × (J/g)
q_total = Σ (each step)	add every sloped and flat piece of the path	J, then divide by 1,000 for kJ
ln(P₂/P₁) = −(ΔH_vap/R)(1/T₂ − 1/T₁)	Clausius-Clapeyron, linking two vapor pressures	ΔH_vap in J/mol; R = 8.314; T in K; P in any one unit
ρ = ZM ÷ (N_Aa³)	Z is atoms per cell, M is molar mass, a is the edge length	ρ in g/cm³; a in cm; M in g/mol
h = 2T cosθ ÷ (rρg)	capillary rise; T is surface tension, θ the contact angle, r the radius	h in m; all SI units
atoms per cell	corner × 1/8, face × 1/2, edge × 1/4, body center × 1	a plain count: 1, 2 or 4

Worked examples

Worked example 1 · a three-step heating curve

How much heat takes 50.0 g of ice at −10.0 °C up to liquid water at 25.0 °C?

Step 1 · warm the iceIce has c = 2.09 J/g·°C. So q₁ = 50.0 × 2.09 × 10.0 = 1,045 J.

Step 2 · melt the iceThis step is flat at 0 °C. q₂ = 50.0 g × 334 J/g = 16,700 J.

Step 3 · warm the waterWater has c = 4.184 J/g·°C. So q₃ = 50.0 × 4.184 × 25.0 = 5,230 J.

Step 4 · add them up1,045 + 16,700 + 5,230 = 22,975 J, or 23.0 kJ.

What it tells youMelting alone took 16,700 J, which is nearly three quarters of the total. Breaking the attractions in ice costs far more than warming the liquid afterward.

Worked example 2 · heat of vaporization from two vapor pressures

A liquid has a vapor pressure of 100.0 torr at 300.0 K and 400.0 torr at 330.0 K. Find its heat of vaporization.

Step 1 · the pressure ratioP₂/P₁ = 400.0 ÷ 100.0 = 4.000. So ln(4.000) = 1.3863.

Step 2 · the temperature term1/T₂ − 1/T₁ = 1/330.0 − 1/300.0 = 0.0030303 − 0.0033333 = −0.00030303 K⁻¹.

Step 3 · rearrangeΔH_vap = −R × ln(P₂/P₁) ÷ (1/T₂ − 1/T₁).

Step 4 · substituteΔH_vap = −(8.314)(1.3863) ÷ (−0.00030303) = 11.526 ÷ 0.00030303 = 38,035 J/mol.

Step 5 · report itThat is 38.0 kJ/mol.

Check the signThe two minus signs cancel, so ΔH_vap comes out positive. It has to be, since boiling always takes heat in.

Worked example 3 · density from a unit cell

Copper is face-centered cubic with an edge length of 361.5 pm. Its molar mass is 63.55 g/mol. Find its density.

Step 1 · count atoms per cellFace-centered cubic has 8 corners at 1/8 each, giving 1, plus 6 faces at 1/2 each, giving 3. So Z = 4 atoms.

Step 2 · convert the edge361.5 pm = 3.615 × 10⁻⁸ cm, since 1 pm = 10⁻¹⁰ cm.

Step 3 · cube ita³ = (3.615 × 10⁻⁸)³ = 4.724 × 10⁻²³ cm³.

Step 4 · build the top and bottomTop: ZM = 4 × 63.55 = 254.2 g/mol. Bottom: N_Aa³ = (6.022 × 10²³)(4.724 × 10⁻²³) = 28.45 cm³/mol.

Step 5 · divideρ = 254.2 ÷ 28.45 = 8.94 g/cm³.

Check it a second wayCopper's measured density is 8.96 g/cm³, so the cell model lands within 0.3%. Watch the unit on a: using meters instead of centimeters throws the answer off by a factor of a million.

Traps people fall into

Using q = mcΔT during a phase change. The flat parts use q = mΔH instead.
Carrying water's c through the whole curve. Ice is 2.09, water is 4.184 and steam is 1.84 J/g·°C.
Swapping ΔH_fus and ΔH_vap. Boiling water costs 2,260 J/g, nearly seven times melting's 334 J/g.
Claiming any H makes a hydrogen bond. It needs H on N, O or F. Methane has none.
Using Celsius in Clausius-Clapeyron. That equation needs kelvin and R = 8.314 J/mol·K.
Counting whole atoms in a unit cell. Corners are shared eight ways and faces two ways.
Expecting water's phase diagram to look normal. Its melting line leans left, which is why ice melts under pressure.

Chapter 10 self-test · ten items

1. What is the strongest intermolecular force in liquid NH₃?

Hydrogen bonding. Ammonia has hydrogen bonded straight to nitrogen, which is one of the three qualifying atoms. Dispersion and dipole-dipole forces are present too, but the hydrogen bonds dominate.

2. What is the strongest intermolecular force between CO₂ molecules?

Dispersion forces. CO₂ is linear and symmetric, so its two bond dipoles cancel and the molecule is nonpolar. With no permanent dipole, only dispersion is left. That is why CO₂ is a gas at room temperature.

3. How much heat melts 25.0 g of ice already sitting at 0 °C?

8.35 kJ. Use q = mΔH_fus = 25.0 g × 334 J/g = 8,350 J, which is 8.35 kJ. No ΔT appears, because the temperature never changes during melting.

4. How much heat boils 10.0 g of water already at 100 °C?

22.6 kJ. Use q = mΔH_vap = 10.0 g × 2,260 J/g = 22,600 J, or 22.6 kJ. Note that boiling 10.0 g costs more than melting 25.0 g, because vaporizing breaks every attraction rather than just loosening them.

5. Why does water boil below 100 °C on a mountain?

The outside pressure is lower. Boiling starts when the vapor pressure matches the pressure pushing down on the liquid. Higher up there is less air above you, so that match happens at a lower temperature. Food takes longer to cook, because the water is cooler.

6. How many atoms are in a body-centered cubic unit cell?

Two. The eight corner atoms each contribute one eighth, giving 8 × 1/8 = 1. The atom at the center belongs entirely to this cell, so it adds 1. The total is 2.

7. How many atoms are in a face-centered cubic unit cell?

Four. Corners give 8 × 1/8 = 1. Each of the six faces is split between two cells, so faces give 6 × 1/2 = 3. Adding them gives 4, which is why Z = 4 in the density equation for copper.

8. Which has the higher viscosity, water or glycerol, and why?

Glycerol. Glycerol carries three O–H groups per molecule, so it forms many more hydrogen bonds than water's two. Molecules that grip each other harder slide past each other less easily, which is what viscosity measures.

9. What is the name for a solid turning straight into a gas, and what everyday example shows it?

Sublimation, and dry ice is the example. Solid CO₂ sublimes at −78 °C under normal pressure. Its triple point sits above 1 atm, so no liquid phase is reachable at room pressure.

10. How much heat turns 100.0 g of water at 20.0 °C into steam at 100.0 °C?

259 kJ. Two steps. Warming: q = 100.0 × 4.184 × 80.0 = 33,472 J. Boiling: q = 100.0 × 2,260 = 226,000 J. Total: 33,472 + 226,000 = 259,472 J, or 259 kJ. The boiling step is about seven times the warming step.

Chapter 11Solutions and Colloids

What to know

A solution is one even phase. The solute is what dissolves, and the solvent is what it dissolves into. Dissolving trades old attractions for new ones. So like dissolves like. Polar solvents take polar solutes, and nonpolar solvents take nonpolar ones.

Water pulls ions apart with ion-dipole forces. It turns its oxygen end toward a cation and its hydrogen ends toward an anion. That wrapping is called hydration.

Picture dissolving as three energy steps. Pull the solute apart, which costs energy. Push the solvent apart to make room, which also costs energy. Then let solute and solvent attract each other, which pays energy back. Add the three and you get ΔH_soln. For salts it is a small leftover from two huge numbers, so its sign is hard to guess. A positive ΔH_soln can still dissolve, because spreading out raises entropy.

Four concentration units matter here. Molarity is moles of solute per liter of solution. Molality is moles of solute per kilogram of solvent. Mole fraction is one component's moles over the total moles. Percent by mass and parts per million both use mass of solute over mass of solution. Note that molarity uses solution volume while molality uses solvent mass, so density is what bridges them.

A saturated solution holds all it can at that temperature. Unsaturated holds less. Supersaturated holds more than it should, and it crashes out when disturbed. Solids follow their own solubility curve as temperature changes, and most dissolve better warm. Gases always dissolve less when warm. Gas solubility also follows Henry's law, so it rises with the partial pressure above the liquid.

Colligative properties count heads, not names. Only the number of dissolved particles matters. Solute lowers the solvent's vapor pressure, which is Raoult's law. That lowering pushes the boiling point up and the freezing point down. Osmotic pressure is the push needed to stop water flowing across a membrane into the solution.

Ionic solutes count more than once. The van 't Hoff factor, i, is the number of particles one formula unit releases. NaCl gives 2 and CaCl₂ gives 3. A molecular solute like sugar gives 1.

Three mixtures sort by particle size. A true solution has particles under 1 nm and is clear. A colloid has particles from 1 to 1,000 nm, never settles, and scatters a light beam. That scattering is the Tyndall effect. A suspension has larger particles still, and it settles out if you leave it alone.

Lock these in

Molarity is per liter of solution. Molality is per kilogram of solvent.
Colligative properties count particles, so ionic solutes need i.
Freezing point goes down and boiling point goes up.
Water's K_f is 1.86 °C/m and its K_b is 0.512 °C/m.
Gases dissolve less as temperature rises, and more as pressure rises.
A colloid scatters light and never settles. A suspension settles.

Equations, units and symbols

Equation	What each symbol means	Units
ΔH_soln = ΔH₁ + ΔH₂ + ΔH₃	the three steps: split the solute, open the solvent, then mix	kJ/mol
M = mol solute ÷ L solution	M is molarity, the crowding per unit volume	mol/L, written M
m = mol solute ÷ kg solvent	m is molality; it uses solvent mass, not solution volume	mol/kg, written m
X_A = n_A ÷ n_total	X_A is the mole fraction of component A	no unit; all fractions add to 1
percent by mass = (mass solute ÷ mass solution) × 100	mass solution is solute plus solvent	percent
ppm = (mass solute ÷ mass solution) × 10⁶	parts per million, for very dilute mixtures	ppm; 1 ppm is about 1 mg per kg
C = k_HP	Henry's law; C is dissolved gas, k_H is its constant, P is the partial pressure	C in mol/L; P in atm; k_H in mol/L·atm
P_solution = X_solvent × P°_solvent	Raoult's law; P° is the pure solvent's vapor pressure	torr or atm; note it uses the solvent's fraction
ΔT_b = iK_bm	boiling point elevation; K_b is the solvent's constant	°C; K_b in °C/m; m in mol/kg
ΔT_f = iK_fm	freezing point depression; subtract this from the normal freezing point	°C; K_f in °C/m
Π = iMRT	osmotic pressure; M is molarity, not molality	atm; R = 0.08206 L·atm/mol·K; T in K

Worked examples

Worked example 1 · molality, then a freezing point

You stir 15.0 g of NaCl into 250.0 g of water. Find the molality and the new freezing point.

Step 1 · grams to molesNaCl is 58.44 g/mol. So 15.0 ÷ 58.44 = 0.2567 mol.

Step 2 · solvent mass in kilograms250.0 g = 0.2500 kg.

Step 3 · dividem = 0.2567 ÷ 0.2500 = 1.03 mol/kg.

Step 4 · pick iNaCl splits into Na⁺ and Cl⁻, so i = 2.

Step 5 · find the dropΔT_f = iK_fm = 2 × 1.86 × 1.027 = 3.82 °C.

Step 6 · subtract itWater normally freezes at 0.00 °C. So the solution freezes at −3.82 °C.

Why i mattersWithout the factor of 2, you would predict only −1.91 °C. That is why salt works so well on an icy road: every formula unit delivers two particles.

Worked example 2 · boiling point and osmotic pressure

First find the boiling point of 0.500 m glucose in water. Then find the osmotic pressure of 0.0100 M NaCl at 25 °C.

Step 1 · glucose's iGlucose is molecular and does not split, so i = 1.

Step 2 · find the riseΔT_b = iK_bm = 1 × 0.512 × 0.500 = 0.256 °C.

Step 3 · add it onThe solution boils at 100.000 + 0.256 = 100.256 °C. It is a small shift, which is normal for colligative effects.

Step 4 · set up the osmosis problemΠ = iMRT, with i = 2 for NaCl, M = 0.0100 mol/L, R = 0.08206 and T = 25 + 273.15 = 298 K.

Step 5 · multiply0.08206 × 298 = 24.45. Then 24.45 × 0.0100 = 0.2445. Then 0.2445 × 2 = 0.489 atm.

Check the scaleA very dilute solution still generates nearly half an atmosphere of osmotic pressure. That is why osmotic pressure is the most sensitive of the colligative properties.

Worked example 3 · Henry's law, ppm and Raoult's law

Oxygen dissolves at 1.3 × 10⁻³ M under 1.0 atm of pure O₂. How much dissolves from air, where its partial pressure is 0.21 atm? Then express 0.0025 g of lead in 1.0 kg of water in ppm. Last, find the vapor pressure above a solution whose water mole fraction is 0.950 at 25 °C.

Step 1 · find k_Hk_H = C ÷ P = (1.3 × 10⁻³ M) ÷ (1.0 atm) = 1.3 × 10⁻³ mol/L·atm.

Step 2 · apply it to airC = k_HP = (1.3 × 10⁻³)(0.21) = 2.7 × 10⁻⁴ M. Lower pressure means less dissolved gas, which is why fish need moving water.

Step 3 · the ppm calculationMass of solution is about 1,000 g. So ppm = (0.0025 ÷ 1,000) × 10⁶ = 2.5 ppm.

Step 4 · Raoult's lawPure water's vapor pressure at 25 °C is 23.8 torr. So P = X_water × P° = 0.950 × 23.8 = 22.6 torr.

Check the logicRaoult's law uses the solvent's mole fraction, never the solute's. The solute takes up 5.0% of the surface, so the vapor pressure drops by about 5.0%.

Traps people fall into

Swapping molarity and molality. Molarity divides by liters of solution. Molality divides by kilograms of solvent.
Leaving out i. NaCl gives 2 and CaCl₂ gives 3, so the shift is that many times bigger.
Adding ΔT_f instead of subtracting it. The freezing point always goes down.
Using solvent mass as solution mass. Percent by mass and ppm both need solute plus solvent on the bottom.
Using Celsius in Π = iMRT. Convert to kelvin first.
Putting the solute's mole fraction into Raoult's law. It takes the solvent's fraction.
Assuming all solubility rises with heat. Gases always dissolve less when warm.

Chapter 11 self-test · ten items

1. Find the molality of 0.250 mol of solute dissolved in 500.0 g of water.

0.500 m. Convert the solvent mass: 500.0 g = 0.5000 kg. Then m = 0.250 mol ÷ 0.5000 kg = 0.500 mol/kg.

2. A mixture holds 2.00 mol of ethanol and 8.00 mol of water. Find the mole fraction of ethanol.

0.200. Total moles are 2.00 + 8.00 = 10.00. Then X = 2.00 ÷ 10.00 = 0.200. Water's fraction is 0.800, and the two add to 1.000.

3. You dissolve 25.0 g of NaCl in 175.0 g of water. What is the percent by mass?

12.5%. The solution's mass is 25.0 + 175.0 = 200.0 g. Then (25.0 ÷ 200.0) × 100 = 12.5%. Using 175.0 g on the bottom would wrongly give 14.3%.

4. Find the freezing point of 0.100 m CaCl₂ in water.

−0.558 °C. CaCl₂ releases one Ca²⁺ and two Cl⁻, so i = 3. Then ΔT_f = 3 × 1.86 × 0.100 = 0.558 °C. Subtract from 0.00 °C to get −0.558 °C.

5. Find the boiling point of 1.50 m sucrose in water.

100.77 °C. Sucrose is molecular, so i = 1. Then ΔT_b = 1 × 0.512 × 1.50 = 0.768 °C. Add it on: 100.000 + 0.768 = 100.768 °C, so about 100.77 °C.

6. Find the osmotic pressure of 0.0500 M glucose at 300. K.

1.23 atm. Glucose gives i = 1. Then Π = (1)(0.0500 mol/L)(0.08206 L·atm/mol·K)(300. K) = 1.2309 atm, so 1.23 atm.

7. At the same molality, which lowers the freezing point more, NaCl or glucose?

NaCl, by twice as much. Colligative properties count particles. NaCl splits into two ions, so i = 2. Glucose stays whole, so i = 1. The ratio of the two drops is 2 to 1.

8. Why does a warm soda go flat faster than a cold one?

Gas solubility falls as temperature rises. Warmer CO₂ molecules carry more kinetic energy, so more of them escape the liquid. Opening the can also drops the CO₂ partial pressure above the drink, and Henry's law says less gas can stay dissolved.

9. A solution has a water mole fraction of 0.900. Pure water's vapor pressure at 25 °C is 23.8 torr. Find the solution's vapor pressure.

21.4 torr. Use Raoult's law: P = 0.900 × 23.8 = 21.42 torr, so 21.4 torr. The drop of 2.4 torr is 10% of 23.8, matching the 10% of the surface the solute now occupies.

10. A beam of light shining through a mixture leaves a visible path. What does that tell you?

It is a colloid, not a true solution. That is the Tyndall effect. Colloid particles run from 1 to 1,000 nm, which is large enough to scatter visible light. A true solution's particles are under 1 nm, so the beam passes straight through unseen. Milk and fog both show this.

Chapter 12Kinetics

What to know

A rate is a change over a time. For a reactant the concentration falls, so the raw change is negative. Put a minus sign in front to report a positive rate. Divide each species' change by its coefficient, and the whole reaction gets one number.

Three kinds of rate come up. An average rate covers a span of time. An instantaneous rate is the slope of the curve at one moment. An initial rate is the instantaneous rate right at the start, which is the cleanest one to measure.

A rate law looks like rate = k[A]^m[B]ⁿ. The exponents are the orders, and they come from experiment. They are not the coefficients from the balanced equation. Only for a single elementary step do the two happen to match.

Each order behaves differently. Zero order means the rate ignores that reactant. First order means doubling it doubles the rate. Second order means doubling it quadruples the rate. The method of initial rates finds each one. Change one reactant, hold the rest, and see how the rate answers.

The units of k change with the overall order. Rate is always in M/s, so k has to carry whatever units make that work. Zero order gives M/s. First order gives s⁻¹. Second order gives M⁻¹s⁻¹.

Integrated rate laws turn a rate into a clock. Each order has one plot that comes out straight. Plot [A] against t for zero order. Plot ln[A] against t for first order. Plot 1/[A] against t for second order. Whichever plot is straight tells you the order, and its slope gives k.

Half-life is the time for half the reactant to disappear. Only first order has a half-life that never changes, and it equals 0.693/k. For zero order it is [A]₀/2k, and for second order it is 1/(k[A]₀), so both shift as the starting amount shifts.

Collision theory explains why reactions need help. Molecules have to touch, bring enough energy, and hit at the right angle. The energy threshold is the activation energy, E_a. The share of collisions that clears it is e^−E_a/RT, which is why a small temperature rise can change the rate so much.

The Arrhenius equation puts that into the rate constant: k = Ae^−E_a/RT. Taking logs gives a straight line, so a plot of ln k against 1/T has a slope of −E_a/R. With just two rate constants at two temperatures, you can solve for E_a directly.

Most reactions run in steps, and the set of steps is the mechanism. The slowest step sets the pace, so it is the rate-determining step. Its rate law is usually the overall rate law, which is how a proposed mechanism gets tested. An intermediate is made in one step and used up in a later one. A catalyst lowers E_a by offering an easier path. It gets remade each cycle, and it leaves ΔH and the equilibrium position untouched.

Lock these in

Orders come from data, never from coefficients.
The units of k depend on the overall order.
Only first order has a constant half-life, and it is 0.693/k.
Whichever of the three plots is straight names the order.
A plot of ln k against 1/T has slope −E_a/R.
A catalyst changes E_a only. It does not change ΔH or K.

Equations, units and symbols

Equation	What each symbol means	Units
rate = −(1/a)Δ[A]/Δt	a is A's coefficient; the minus sign makes a reactant's rate positive	M/s, which is mol/L·s
rate = +(1/b)Δ[B]/Δt	for a product, with no minus sign needed	M/s
rate = k[A]^m[B]ⁿ	k is the rate constant; m and n are measured orders	rate in M/s; k's unit depends on m + n
[A] = [A]₀ − kt	zero order; a plot of [A] against t is straight with slope −k	[A] in M; k in M/s
ln[A] = ln[A]₀ − kt	first order; a plot of ln[A] against t is straight with slope −k	k in s⁻¹
1/[A] = 1/[A]₀ + kt	second order; a plot of 1/[A] against t is straight with slope +k	k in M⁻¹s⁻¹
t_1/2 = [A]₀ ÷ 2k	zero order half-life; it shrinks as the reaction runs	s
t_1/2 = 0.693 ÷ k	first order half-life; it never changes	s
t_1/2 = 1 ÷ (k[A]₀)	second order half-life; it grows as the reaction runs	s
fraction over the barrier = e^−E_a/RT	the share of collisions with enough energy	no unit; E_a in J/mol with R = 8.314
k = Ae^−E_a/RT	the Arrhenius equation; A is the frequency factor	A carries k's units; T in K
ln k = ln A − (E_a/R)(1/T)	the straight-line form; slope is −E_a/R	slope in K; E_a = −slope × R
ln(k₂/k₁) = (E_a/R)(1/T₁ − 1/T₂)	the two-point form, for one E_a from two temperatures	E_a in J/mol; T in K

Worked examples

Worked example 1 · a rate law from initial rates

For A + B → C, three runs give these initial rates. Run 1: [A] = 0.100 M, [B] = 0.100 M, rate = 2.0 × 10⁻³ M/s. Run 2: [A] = 0.200 M, [B] = 0.100 M, rate = 8.0 × 10⁻³ M/s. Run 3: [A] = 0.100 M, [B] = 0.200 M, rate = 4.0 × 10⁻³ M/s. Find the rate law and k.

Step 1 · compare runs 1 and 2Only [A] changed, and it doubled. The rate went from 2.0 to 8.0, so it rose 4 times.

Step 2 · solve for mYou need 2^m = 4, so m = 2. The reaction is second order in A.

Step 3 · compare runs 1 and 3Only [B] changed, and it doubled. The rate went from 2.0 to 4.0, so it rose 2 times. You need 2ⁿ = 2, so n = 1.

Step 4 · write the rate lawrate = k[A]²[B]. The overall order is 2 + 1 = 3.

Step 5 · solve for kUse run 1. k = rate ÷ ([A]²[B]) = (2.0 × 10⁻³) ÷ [(0.100)²(0.100)] = (2.0 × 10⁻³) ÷ (1.0 × 10⁻³) = 2.0 M⁻²s⁻¹.

Check it a second wayTest run 2: k[A]²[B] = 2.0 × (0.200)² × 0.100 = 2.0 × 0.0400 × 0.100 = 8.0 × 10⁻³ M/s. That matches the measured rate.

Worked example 2 · a first-order half-life

A first-order reaction has k = 0.0231 s⁻¹. Find the half-life. Then find what fraction is left after 90.0 s.

Step 1 · the half-lifet_1/2 = 0.693 ÷ k = 0.693 ÷ 0.0231 s⁻¹ = 30.0 s.

Step 2 · count the half-lives90.0 s ÷ 30.0 s = 3.00 half-lives.

Step 3 · halve three times1 goes to 1/2, then 1/4, then 1/8. So 12.5% remains.

Check it a second wayUse the integrated form. ln[A] = ln[A]₀ − kt, so ln([A]/[A]₀) = −(0.0231)(90.0) = −2.079. Then [A]/[A]₀ = e^−2.079 = 0.125. Both routes give 12.5%.

Worked example 3 · activation energy from two temperatures

A reaction has k = 1.0 × 10⁻³ at 300. K and k = 8.0 × 10⁻³ at 330. K. Find E_a.

Step 1 · the log of the ratiok₂/k₁ = 8.0, so ln(8.0) = 2.0794.

Step 2 · the temperature term1/T₁ − 1/T₂ = 1/300. − 1/330. = 0.0033333 − 0.0030303 = 0.00030303 K⁻¹.

Step 3 · rearrangeE_a = R × ln(k₂/k₁) ÷ (1/T₁ − 1/T₂).

Step 4 · substituteE_a = (8.314)(2.0794) ÷ (0.00030303) = 17.288 ÷ 0.00030303 = 57,051 J/mol.

Step 5 · report itThat is 57.1 kJ/mol, a very typical value.

What it showsA 30 K rise multiplied the rate constant by 8. Nothing about the molecules changed. Only the share of collisions clearing the barrier changed.

Worked example 4 · reading a mechanism

A proposed mechanism has a slow first step: NO₂ + NO₂ → NO₃ + NO. Then a fast second step: NO₃ + CO → NO₂ + CO₂. Find the overall equation and the predicted rate law.

Step 1 · add the stepsLeft side: 2 NO₂ + NO₃ + CO. Right side: NO₃ + NO + NO₂ + CO₂.

Step 2 · cancel what appears on both sidesNO₃ cancels, and one NO₂ cancels. The overall equation is NO₂ + CO → NO + CO₂.

Step 3 · name the intermediateNO₃ is made in step 1 and used in step 2, so it is the intermediate. It never appears in the overall equation.

Step 4 · read the slow stepThe slow step is elementary, so its coefficients are its orders. Two NO₂ collide, so rate = k[NO₂]².

Step 5 · notice what is missingCO does not appear in the rate law, even though it is a reactant. Adding more CO will not speed the reaction up, because the bottleneck happens before CO is needed.

How a mechanism gets testedMeasure the real rate law. If it comes out as k[NO₂]², the mechanism survives. If CO shows up in it, the mechanism is wrong.

Traps people fall into

Reading orders off the coefficients. Orders are measured. Only an elementary step lets them match.
Giving k one fixed unit. It changes with order, from M/s to s⁻¹ to M⁻¹s⁻¹.
Treating every half-life as constant. Only first order has that property.
Using Celsius in Arrhenius. It needs kelvin, with R = 8.314 J/mol·K.
Forgetting the minus sign for a reactant. Its concentration falls, so Δ[A] is negative.
Skipping the coefficient when relating rates. Divide each species' change by its own coefficient.
Thinking a catalyst shifts the equilibrium. It speeds both directions equally, so K stays put.
Mixing up an intermediate and a catalyst. An intermediate is made then used. A catalyst is used then remade.

Chapter 12 self-test · ten items

1. For 2 N₂O₅ → 4 NO₂ + O₂, oxygen forms at 0.0050 M/s. How fast does N₂O₅ disappear?

0.010 M/s. The relation is −(1/2)Δ[N₂O₅]/Δt = Δ[O₂]/Δt. So Δ[N₂O₅]/Δt = −2 × 0.0050 = −0.010 M/s, which is a disappearance rate of 0.010 M/s. Two molecules are consumed for every O₂ made.

2. A reaction has rate = k[A][B]². What is the overall order?

Third order. Add the exponents: 1 + 2 = 3. It is first order in A and second order in B. Doubling B alone would raise the rate four times.

3. What are the units of k for a first-order reaction?

s⁻¹. Rate is in M/s and the rate law is k[A], with [A] in M. So k = (M/s) ÷ M = s⁻¹. The molarity cancels completely.

4. What are the units of k for a second-order reaction?

M⁻¹s⁻¹. Here rate = k[A]², so k = (M/s) ÷ M² = M⁻¹s⁻¹. You can also write it as L/mol·s.

5. A first-order reaction has k = 0.0500 min⁻¹. Find the half-life.

13.9 min. Use t_1/2 = 0.693 ÷ k = 0.693 ÷ 0.0500 min⁻¹ = 13.86 min. The unit of k sets the unit of the answer, so it comes out in minutes.

6. For a first-order reaction, what fraction remains after four half-lives?

1/16, which is 6.25%. Halve four times: 1/2, 1/4, 1/8, 1/16. In general the fraction left is (1/2)ⁿ for n half-lives.

7. Which plot comes out straight for a second-order reaction?

1/[A] against t. The integrated form is 1/[A] = 1/[A]₀ + kt, which is already in straight-line shape. Its slope is +k, and its intercept is 1/[A]₀. Zero order gives a straight [A] plot, and first order gives a straight ln[A] plot.

8. Doubling [A] leaves the rate unchanged. What is the order in A?

Zero. You need 2^m = 1, and that only works at m = 0. Since anything to the zero power is 1, the term drops out of the rate law. This often happens when a surface or an enzyme is already saturated.

9. A plot of ln k against 1/T has a slope of −5.00 × 10³ K. Find E_a.

41.6 kJ/mol. The slope equals −E_a/R, so E_a = −slope × R = (5.00 × 10³ K)(8.314 J/mol·K) = 41,570 J/mol, or 41.6 kJ/mol. The kelvin units cancel, leaving J/mol.

10. A catalyst is added. Which of these change: E_a, ΔH, or the equilibrium constant K?

Only E_a. A catalyst offers a lower-energy path, so it cuts E_a and raises k. The reactants and products are unchanged, so ΔH is the same. It speeds the forward and reverse directions by the same factor, so K holds. A catalyst gets you to equilibrium faster, not to a different one.

One pageCram sheet

Everything from chapters 10 to 12 in one place. Cover the right column and rebuild each row from memory.

Equations

Name	Equation	Units
Sloped part of a heating curve	q = mcΔT	J
Melting step	q = mΔH_fus	J
Boiling step	q = mΔH_vap	J
Clausius-Clapeyron	ln(P₂/P₁) = −(ΔH_vap/R)(1/T₂ − 1/T₁)	ΔH_vap in J/mol
Density from a unit cell	ρ = ZM ÷ (N_Aa³)	g/cm³
Capillary rise	h = 2T cosθ ÷ (rρg)	m
Energy of solution	ΔH_soln = ΔH₁ + ΔH₂ + ΔH₃	kJ/mol
Molarity	M = mol solute ÷ L solution	mol/L
Molality	m = mol solute ÷ kg solvent	mol/kg
Mole fraction	X_A = n_A ÷ n_total	no unit
Percent by mass	(mass solute ÷ mass solution) × 100	percent
Parts per million	(mass solute ÷ mass solution) × 10⁶	ppm
Henry's law	C = k_HP	mol/L
Raoult's law	P_solution = X_solvent × P°_solvent	torr or atm
Boiling point elevation	ΔT_b = iK_bm	°C
Freezing point depression	ΔT_f = iK_fm	°C
Osmotic pressure	Π = iMRT	atm
Reaction rate	rate = −(1/a)Δ[A]/Δt	M/s
Rate law	rate = k[A]^m[B]ⁿ	M/s
Zero order	[A] = [A]₀ − kt, t_1/2 = [A]₀/2k	k in M/s
First order	ln[A] = ln[A]₀ − kt, t_1/2 = 0.693/k	k in s⁻¹
Second order	1/[A] = 1/[A]₀ + kt, t_1/2 = 1/(k[A]₀)	k in M⁻¹s⁻¹
Arrhenius	k = Ae^−E_a/RT	E_a in J/mol
Arrhenius, line form	ln k = ln A − (E_a/R)(1/T)	slope = −E_a/R
Arrhenius, two points	ln(k₂/k₁) = (E_a/R)(1/T₁ − 1/T₂)	T in K

Constants and values

Quantity	Value	Note
Specific heat of ice	2.09 J/g·°C	for the sloped part below 0 °C
Specific heat of water	4.184 J/g·°C	for the sloped part from 0 to 100 °C
Specific heat of steam	1.84 J/g·°C	for the sloped part above 100 °C
Heat of fusion of water	334 J/g, or 6.01 kJ/mol	the flat step at 0 °C
Heat of vaporization of water	2,260 J/g, or 40.7 kJ/mol	the flat step at 100 °C
Water's vapor pressure	23.8 torr at 25 °C	rises steeply with temperature
Water's K_f	1.86 °C/m	for freezing point depression
Water's K_b	0.512 °C/m	for boiling point elevation
Van 't Hoff factors	glucose and sucrose 1, NaCl and KCl 2, CaCl₂ and Na₂SO₄ 3	count the ions one formula unit releases
Gas constant	R = 0.08206 L·atm/mol·K	use in Π = iMRT
Gas constant, energy form	R = 8.314 J/mol·K	use in Arrhenius and Clausius-Clapeyron
Avogadro's number	N_A = 6.022 × 10²³ /mol	use in the unit cell density
Half-life shortcut	ln 2 = 0.693	the numerator of the first-order half-life
Length conversion	1 pm = 10⁻¹⁰ cm = 10⁻¹² m	unit cell edges come in pm

Rules that are not equations

Topic	The rule
Force ranking	Ion-dipole beats hydrogen bonding, which beats dipole-dipole, which beats dispersion at equal size.
Hydrogen bonding	It needs hydrogen bonded directly to N, O or F.
Phase changes	Temperature holds flat while a phase change runs, so use q = mΔH there.
Boiling	It begins when vapor pressure equals the outside pressure, so altitude lowers it.
Phase diagrams	Triple point: all three states together. Critical point: liquid and gas stop differing. Water's melting line leans left.
Unit cells	Simple cubic 1 atom, body-centered 2, face-centered 4. Corners count 1/8 and faces 1/2.
Like dissolves like	Polar solvents take polar solutes. Nonpolar takes nonpolar.
Solubility and temperature	Most solids dissolve better warm. Gases always dissolve less warm.
Colligative properties	They count particles, so always apply i. Freezing point falls and boiling point rises.
Mixture sizes	Solution under 1 nm and clear. Colloid 1 to 1,000 nm, scatters light, never settles. Suspension larger, and it settles.
Finding the order	Change one reactant at a time and solve f^m = r, where f is the concentration factor and r the rate factor.
Reading a mechanism	Steps must add to the overall equation. The slow step gives the rate law, and its coefficients are its orders.
Intermediate or catalyst	An intermediate is produced then consumed. A catalyst is consumed then regenerated.
What a catalyst does	It lowers E_a only. ΔH and K stay exactly the same.