🧪 Contents · Interactive Lab Chapter 12 · Kinetics
Chemistry · Chapter 12

Kinetics

Some reactions are over before you can blink. Others creep along for years without seeming to move. Kinetics is the study of how fast a reaction runs, and why. A balanced equation tells you what you get. It says nothing about how long you wait. Here you will learn to measure a rate, write a rate law, and pull a rate constant out of real data. You will see why a small jump in temperature can double a rate. You will see how a catalyst opens a shortcut. By the end you can look at a reaction and predict what will speed it up.

By the end of this chapter you can

  • State the rate of a reaction from the change in any one species.
  • Build a rate law from initial-rate data and name the order.
  • Use the integrated rate laws to find a concentration at any time.
  • Pick the order by seeing which of three plots comes out straight.
  • Find a half-life for zero, first and second order, and say when it shifts.
  • Explain a rate with collisions, energy and the angle of the hit.
  • Get Ea from two rate constants, and judge a mechanism against the measured rate law.

Section 1Reaction rates

What a rate measures

A rate is a change divided by a time. For a car that is miles per hour. For a flask it is moles per liter each second. A reaction rate tracks how fast concentration shifts. Reactant concentration falls, and product concentration climbs at the same time. We measure concentration in molarity, so most rates here carry units of M/s.

Watch one species and you can state two things. The rate of disappearance of a reactant is how fast it gets used up. The rate of appearance of a product is how fast it builds. Both of them are quoted as positive numbers by habit. So a minus sign goes in front of the reactant change.

rate of loss of A = −Δ[A] / Δt

One rate for the whole reaction

Species in the same flask do not change at the same speed. Take 2 N2O5 → 4 NO2 + O2. Here NO2 shows up four times as fast as O2. That makes a single rate awkward to report. So we divide each change by its own coefficient. The result is one shared number, the rate of reaction.

rate = −(1/2) Δ[N2O5]/Δt = (1/4) Δ[NO2]/Δt = Δ[O2]/Δt

Now every species gives the same answer. Report that one number and nobody has to ask which species you watched.

Average, instantaneous, initial

Plot concentration against time and you get a curve, not a line. Its slope keeps changing from one moment to the next. An average rate uses two points far apart. It is the slope of the straight line joining them. An instantaneous rate is the slope of the tangent at one moment. The initial rate is the instantaneous rate right at the start.

Why does the curve bend? It bends because the reactants are running low. Fewer particles means fewer useful crashes each second. So nearly every reaction fades as it goes. The initial rate is the one we trust most. At that instant no product has built up yet.

Reading a rate off a concentration curve0.00.20.40.60.81.00246810concentration (M)time (s)tangent at t = 0initial rate = 0.250 M/sstraight line, 2 s to 8 saverage rate = 0.0785 M/sproductreactant
Reading a rate off the curve. A single reactant turns into a single product. The dashed tangent at time zero has a slope of 0.250 M/s, which is the initial rate. The dotted line joining the 2 s and 8 s points has a slope of 0.0785 M/s, the average rate over that stretch. The curve bends because the reactant is running out.

Getting the numbers in the lab

You need a signal that follows concentration. Color is a handy one for any species that soaks up visible light. A colored species soaks up light, so a spectrometer reads it right through the glass. A gas reaction can be watched by pressure instead. An acid or base reaction can be watched with a pH meter. For a slow reaction you can pull small samples, chill them to stop the change, then titrate later.

Worked example · one rate, three species

In a flask, [N2O5] falls from 0.0200 M to 0.0160 M in 100 s. The reaction is 2 N2O5 → 4 NO2 + O2. Find the rate of reaction. Then find how fast NO2 and O2 appear.

Step 1 · the change in N2O5

Δ[N2O5] = 0.0160 M − 0.0200 M = −0.0040 M

Step 2 · rate of loss

−(−0.0040 M) / 100 s = 4.0 × 10−5 M/s

Step 3 · divide by the coefficient

N2O5 has a coefficient of 2. So rate = (1/2)(4.0 × 10−5 M/s) = 2.0 × 10−5 M/s.

Step 4 · scale back up for the products

NO2 has a coefficient of 4, so it appears four times faster than the reaction rate. Δ[NO2]/Δt = 4(2.0 × 10−5) = 8.0 × 10−5 M/s. For O2 the coefficient is 1, so it appears at 2.0 × 10−5 M/s.

Check it a second way

The balanced equation makes four NO2 for every one O2. Our answers agree. 8.0 × 10−5 is exactly four times 2.0 × 10−5.

Watch the convention. A rate constant only means something once you say which rate it belongs to. The rate of loss of N2O5 and the rate of the reaction differ by the coefficient 2, so the two constants differ by 2 as well. Good data tables always say which one they mean, and so should you.
Key points
  • A rate is a concentration change divided by a time. The usual units are M/s.
  • Divide each change by its coefficient and every species gives the same rate.
  • An average rate spans an interval. An instantaneous rate is the tangent slope at one instant.
  • The initial rate is the cleanest number, because no product is in the way yet.
  • Rates nearly always drop as the reactants thin out.
Interactive Track concentration against time and read the rate off the curve. Open full ↑

Section 2Rate laws and reaction order

Writing a rate law

Concentration is the lever that drives the rate. A rate law is the equation that says exactly how. For a reaction between A and B it takes this shape.

rate = k[A]m[B]n

The letter k is the rate constant. It does not shift when you change a concentration. It does shift when you change the temperature. The exponents m and n are the reaction orders. Add them up and you get the overall order.

Here is the trap that catches almost everyone once. Those exponents are not the coefficients from the balanced equation. You cannot read them off the page. They come from experiment, every single time. Now and then they happen to match the coefficients. Often they do not match at all. An order can even be zero, a fraction, or below zero.

What each order feels like

Zero order in A means [A] has no grip on the rate at all. Double it and nothing happens. First order in A means the rate tracks [A] straight across. Double it and the rate doubles. Second order in A means the rate follows the square. Double it and the rate goes up four times. Triple it and the rate goes up nine times.

What each reaction order does to the rate13579relative rate11[A]12[A]13[A]zero order11[A]22[A]33[A]first order11[A]42[A]93[A]second orderconcentration of A, as a multiple of the first run
What the order does. Each group shows the rate when [A] is raised to one, two or three times its first value. Zero order never moves. First order follows the concentration. Second order follows its square, so three times the concentration gives nine times the rate.
A reaction order is not a count of molecules. An order of 1.5 is perfectly real, and so is an order of −1. A negative order means the species slows the reaction down, usually because it ties up a catalyst or pushes an early step backward. Fractional orders are a hint that free radicals or a surface are involved.

The units of k are a useful check. They depend on the overall order. A zero-order k comes out in M/s. A first-order k is in s−1. A second-order k is in M−1s−1. A third-order k is in M−2s−1. The pattern behind those units is simple. Each step up in order strips one M off the top.

The method of initial rates

This is the standard way to find the orders. You run the reaction several times at the same temperature. Start each run with a different mix. Measure the initial rate of each one. Then compare two runs that differ in just one reactant. Any rate change you see belongs to that reactant alone.

The math behind it is nothing more than a ratio. Say you multiply [A] by f and the rate goes up by a factor r. Then fm = r. Most of the time you can finish that in your head. If [A] doubles and the rate quadruples, m is 2.

Initial-rate runs for 2 NO + O2 → 2 NO2, all at the same temperature.
Run[NO]0 (M)[O2]0 (M)Initial rate (M/s)
10.01000.01002.00 × 10−3
20.02000.01008.00 × 10−3
30.01000.03006.00 × 10−3

Worked example · order from three runs

Use the three runs in the table. Find the rate law and the value of k.

Step 1 · find the order in NO

Runs 1 and 2 hold [O2] fixed. [NO] doubles. The rate climbs from 2.00 × 10−3 to 8.00 × 10−3 M/s. That is a factor of 4. So 2m = 4, which makes m = 2.

Step 2 · find the order in O2

Runs 1 and 3 hold [NO] fixed. [O2] triples. The rate climbs from 2.00 × 10−3 to 6.00 × 10−3 M/s. That is a factor of 3. So 3n = 3, which makes n = 1.

Step 3 · write the rate law

rate = k[NO]2[O2]. The overall order is 2 + 1 = 3.

Step 4 · solve for k with any run

Use run 1. k = rate / ([NO]2[O2]) = (2.00 × 10−3 M/s) / [(0.0100 M)2(0.0100 M)]. The bottom is 1.00 × 10−6 M3. So k = 2.00 × 103 M−2s−1.

Check it with a different run

Try run 3. k[NO]2[O2] = (2.00 × 103)(1.00 × 10−4)(0.0300) = 6.00 × 10−3 M/s. The table says 6.00 × 10−3 M/s. The units also land on M/s, so the order is right.

Key points
  • A rate law links rate to concentration through k and the orders.
  • Orders come from data, never from the balanced coefficients.
  • Change one reactant at a time and the rate change gives that order.
  • Solve fm = r for the order, then plug any run back in to get k.
  • The units of k tell you the overall order, so they make a fast check.
Interactive Run initial-rate experiments and solve for the order. Open full ↑

Section 3Integrated rate laws and half-life

From a rate to a clock

A rate law gives the speed at this instant. It does not tell you where you will be in ten minutes. For that you need the integrated rate law. Calculus turns each rate law into one equation linking concentration and time. You do not have to do the calculus. You only need the three results and the shape of each one.

The three forms you need. In each one [A]0 is the starting concentration.
OrderRate lawIntegrated formStraight plotSlopeHalf-life
Zerorate = k[A] = [A]0 − kt[A] vs t−k[A]0 / 2k
Firstrate = k[A]ln[A] = ln[A]0 − ktln[A] vs t−k0.693 / k
Secondrate = k[A]21/[A] = 1/[A]0 + kt1/[A] vs t+k1 / k[A]0

Three plots, one straight line

Look at the middle column of the table. Each form is already in the shape of a straight line. That shape is the whole trick behind finding an order. Take your data and draw three graphs. Plot [A] against time. Plot ln[A] against time. Plot 1/[A] against time. Only one of the three comes out straight. That plot names the order.

Three plots of one data set, and only one is straight[A] against tcurves downtime (s)ln[A] against tstill curvedtime (s)1/[A] against ta straight linetime (s)
One data set, three plots. These points come from a second-order run with k = 0.0500 M−1s−1 and [A]0 = 0.500 M. Only 1/[A] against time falls on a line. Its slope is k, and its intercept is 1/[A]0 = 2.00 M−1.

The data drawn above is second order. Watch what happens to each of the three plots. The plain concentration plot curves downward the whole way. The log plot still curves, just more gently. Only the plot of 1/[A] lines up against a ruler. A graph of real data will not be perfect, so you compare how straight the three lines are and pick the best.

Worked example · a first-order run

A reactant A decays by a first-order path with k = 2.50 × 10−3 s−1. You start at [A]0 = 0.800 M. What is [A] after 400 s? What is the half-life?

Step 1 · build the exponent

kt = (2.50 × 10−3 s−1)(400 s) = 1.00. The seconds cancel, so the exponent has no units. It should not have any.

Step 2 · use the first-order form

ln[A] = ln(0.800) − 1.00 = −0.223 − 1.00 = −1.223

Step 3 · undo the log

[A] = e−1.223 = 0.294 M

Step 4 · the half-life

t1/2 = 0.693 / (2.50 × 10−3 s−1) = 277 s

Check it a second way

400 s is 1.44 half-lives, since 400/277 = 1.44. Half of 0.800 is 0.400, and half again is 0.200. Our 0.294 M sits between those two, just as it should. You can also note that kt = 1 always leaves 1/e of the start, and 0.800/e = 0.294 M.

Half-life

The half-life is the time it takes for half of a reactant to vanish. We write it t1/2. First order is the special case worth memorizing. Its half-life does not care how much you start with. It is always 0.693/k. That is why radioactive decay has one fixed half-life you can quote on a chart.

Zero and second order behave differently from each other. For zero order, t1/2 = [A]0/2k. Start with more and the first half takes longer. For second order, t1/2 = 1/k[A]0. Start with more and the first half goes faster. In both cases each later half-life differs from the one before.

Worked example · a second-order half-life

NO2 breaks down by 2 NO2 → 2 NO + O2. The loss of NO2 is second order in NO2, with k = 0.540 M−1s−1 for that loss. Start at 0.0100 M. Find the first half-life. Then find the time to reach 0.00250 M.

Step 1 · the first half-life

t1/2 = 1 / k[A]0 = 1 / [(0.540 M−1s−1)(0.0100 M)] = 1 / (0.00540 s−1) = 185 s

Step 2 · set up the integrated form

1/[A] − 1/[A]0 = kt. Here 1/0.00250 = 400 M−1 and 1/0.0100 = 100 M−1.

Step 3 · solve for the time

t = (400 − 100) M−1 / (0.540 M−1s−1) = 300 / 0.540 = 556 s

Check it a second way

Walk down in halves instead. The drop to 0.00500 M takes 185 s. Now restart at 0.00500 M, so the next half-life is 1/[(0.540)(0.00500)] = 370 s. Add them: 185 + 370 = 555 s, which matches 556 s apart from rounding. Notice the second half-life is twice the first. That is the mark of second order.

Key points
  • Zero order: [A] = [A]0 − kt, and a plot of [A] against time is straight.
  • First order: ln[A] = ln[A]0 − kt, and a plot of ln[A] against time is straight.
  • Second order: 1/[A] = 1/[A]0 + kt, and a plot of 1/[A] against time is straight.
  • A first-order half-life is 0.693/k and never changes during the run.
  • Zero- and second-order half-lives depend on how much you started with.
Interactive Plot the three straight-line forms and find the order. Open full ↑

Section 4Collision theory and activation energy

Nothing happens without contact

Collision theory starts from one simple idea about contact. Particles must meet before they can react. Molecules in a gas or a solution never stop moving. They crash into each other billions of times a second. Almost every crash does nothing at all. Only a thin slice of them make new bonds.

Three things have to line up for a crash to work.

That is why rate rises with concentration. Pack more particles into the same flask and the hits come faster. It is also why a powder burns faster than a solid block.

The barrier

The activation energy, written Ea, is the least energy a crash needs to work. Picture a hill between reactants and products. At the top sits the transition state, also called the activated complex. Up there the old bonds are half broken and the new ones are half made. It survives for about one molecular wobble.

Ea is not the same thing as ΔH. The value of ΔH is the height gap from start to finish. The value of Ea is the height of the hill you must climb on the way. An exothermic reaction can still sit behind a tall hill. Paper and air release energy when they react, yet a book on a shelf does not burst into flame. It needs a match to get over the barrier.

The reverse reaction climbs the very same hill from the other side. So Ea for the reverse equals Ea forward minus ΔH.

Activation energy, the transition state and the heat of reactionpotential energyreaction progressEaforward = 50 kJ/molEareverse = 80 kJ/molΔH= −30reactantsproductstransition state
The hill between start and finish. The forward barrier here is 50 kJ/mol, and the products land 30 kJ/mol below the reactants. So the reverse barrier is 50 − (−30) = 80 kJ/mol. The dot at the peak is the transition state, which exists for about one bond vibration.

Only a few molecules make the cut

Particles do not all carry the same energy. The spread is wide, and it shifts with temperature. At any instant a small tail of the sample has more than Ea. That tail is the share that can react. Its size is given by a short expression.

fraction over the barrier = e−Ea/RT

How many molecules clear the barrier at two temperaturesactivation energynumber of moleculeskinetic energycoolerhottershaded tails: these can reactThe axis has no scale here. The shape and the size of the tail are the point.
Why heat helps so much. Both curves hold the same number of molecules. The hotter one is flatter and shifted right. The shaded area past the dashed line is the share with enough energy, and it grows far faster than the temperature does.

Look at the shaded tails in the figure. Heat the sample and the whole curve flattens and slides right. The peak drops, but the tail past the line grows a lot. A tiny change in that tail is a big change in rate.

Angle matters too

Even a hard, fast hit can fail on geometry alone. Imagine Cl striking the molecule NOCl. Hit the Cl end and the atoms swap cleanly. Hit the O end and the pieces bounce apart. Most molecules have only a small target area that works.

Worked example · how thin is the tail?

A reaction has Ea = 50.0 kJ/mol. Find the share of collisions with enough energy at 298 K. Then do it again at 308 K. Use R = 8.314 J/mol·K.

Step 1 · match the units

Ea = 50.0 kJ/mol = 5.00 × 104 J/mol, because R is in joules.

Step 2 · the exponent at 298 K

RT = (8.314 J/mol·K)(298 K) = 2478 J/mol. So Ea/RT = 50000/2478 = 20.18.

Step 3 · the fraction at 298 K

e−20.18 = 1.72 × 10−9. That is about 2 collisions in a billion.

Step 4 · now at 308 K

RT = (8.314)(308) = 2561 J/mol, so Ea/RT = 50000/2561 = 19.52. Then e−19.52 = 3.32 × 10−9.

Step 5 · compare them

(3.32 × 10−9) / (1.72 × 10−9) = 1.9. Ten degrees nearly doubled the useful fraction.

Check it a second way

The ratio should also equal e raised to the change in the exponent. That change is 20.18 − 19.52 = 0.66, and e0.66 = 1.9. The two routes agree.

Key points
  • A crash must bring enough energy and the right angle, or nothing happens.
  • Ea is the hill between reactants and products, not the energy released.
  • Ea for the reverse step equals Ea forward minus ΔH.
  • The share of molecules past the barrier is e−Ea/RT, which is often very small.
  • Raising the temperature fattens the energy tail, and that is what speeds things up.
Interactive Collide molecules with the right energy and the right angle. Open full ↑

Section 5Temperature and the Arrhenius equation

Putting temperature into the rate constant

Temperature does not change a concentration, so it must work somewhere else. It works on the rate constant k. Svante Arrhenius found the link in 1889, and it still carries his name. The Arrhenius equation ties the rate constant to Ea and to the temperature.

k = A e−Ea/RT

The piece on the right is the fraction of crashes with enough energy. The letter A is the frequency factor. It gathers two things: how often particles meet, and how often they meet facing the right way. So A is the ceiling and the exponent is the toll gate.

Two habits of this equation are worth holding on to. A bigger Ea makes k smaller, since the exponent grows more negative. A higher T makes k larger, since the exponent moves toward zero. Temperature sits in the bottom of a fraction, inside an exponent. That is why small changes in T do so much.

The straight-line version

Take the natural log of both sides and the equation straightens out.

ln k = ln A − (Ea/R)(1/T)

This is a line if you plot ln k on the vertical axis and 1/T on the horizontal. The slope is −Ea/R. The intercept is ln A. A graph like this is called an Arrhenius plot. Multiply the slope by −R and you have Ea straight from the data.

Often you only have two measurements. Write the line twice and subtract, and the intercept drops out.

ln(k2/k1) = (Ea/R)(1/T1 − 1/T2)

Worked example · Ea from two rate constants

A first-order reaction has k = 1.00 × 10−3 s−1 at 300 K. At 320 K it has k = 8.00 × 10−3 s−1. Find Ea and then A.

Step 1 · the log of the ratio

k2/k1 = 8.00, and ln(8.00) = 2.079.

Step 2 · the temperature term

1/300 − 1/320 = 20/96000 = 2.083 × 10−4 K−1

Step 3 · solve for Ea

Ea = R × 2.079 / (2.083 × 10−4) = (8.314)(9981) = 8.30 × 104 J/mol = 83.0 kJ/mol

Step 4 · now get A

ln A = ln k + Ea/RT. At 300 K, RT = 2494 J/mol and Ea/RT = 83000/2494 = 33.28. So ln A = −6.908 + 33.28 = 26.37, and A = 2.84 × 1011 s−1.

Check it a second way

Put A and Ea back in at 320 K. There RT = 2660 J/mol and Ea/RT = 31.20. Then k = (2.84 × 1011)e−31.20 = 8.0 × 10−3 s−1. That matches the second measurement. Also note A carries the same units as k, which is right for a first-order reaction.

Why ten degrees matters so much

You may have heard that a rate doubles for every 10 °C. It is a rough guide, not a law. It holds near room temperature for an Ea of about 50 kJ/mol. With a bigger barrier the same 10 °C does more than double the rate. With a small barrier it does less. This is also why a refrigerator works. Cooling food from 25 °C to 4 °C does not stop spoilage reactions. It just slows them by a large factor.

Key points
  • k = A e−Ea/RT, so k climbs with T and falls as Ea grows.
  • A holds the collision rate and the share of hits aimed the right way.
  • A plot of ln k against 1/T is a line of slope −Ea/R.
  • Two values of k at two temperatures are enough to get Ea.
  • Keep Ea in joules per mole when R is 8.314 J/mol·K.
Interactive Raise the temperature and watch k climb. Open full ↑

Section 6Mechanisms and catalysis

Reactions happen in steps

A balanced equation is a summary, not a story. Very few reactions happen in one shot. A reaction mechanism is the real list of events from start to finish. Each event in it is an elementary step, meaning one real collision or one real break.

We sort steps by how many particles take part. A unimolecular step has one particle falling apart. A bimolecular step has two particles meeting. A termolecular step needs three at once, which is so unlikely that most mechanisms avoid it.

Here is the payoff. For an elementary step, and only there, the rate law does come from the coefficients. A step A + B → C has rate = k[A][B]. A step 2 A → D has rate = k[A]2.

Species that appear in the middle and are gone by the end are reaction intermediates. They are made in one step, used in a later one, and absent from the overall equation.

The slow step sets the pace

Steps in a mechanism rarely run at the same speed. The slowest one is the rate-determining step. Think of a line of shoppers that must pass one slow register. The whole line moves at the pace of that bottleneck.

So the rate law of the slow step is usually the rate law of the reaction. That gives you a test. Derive the law the mechanism predicts and compare it with the measured one. If they disagree, the mechanism is wrong. If they agree, it is allowed but not proven.

Two steps, one slow, and the shortcut a catalyst openspotential energyreaction progresstallest hillstep 1 sets the paceintermediatereactantsproductscatalyzed path: lower hills, same products
Two steps, and a shortcut. The solid line is a mechanism with two elementary steps. The first hill is taller, so step 1 is the rate-determining step. The dip between the hills is the intermediate. The dashed line is the same reaction on a catalyst: lower hills, identical start and finish.

Worked example · test a mechanism, then add a catalyst

At moderate temperature, NO2 + CO → NO + CO2 is second order in NO2 and zero order in CO. Test this idea. Step 1, slow: 2 NO2 → NO3 + NO. Step 2, fast: NO3 + CO → NO2 + CO2.

Step 1 · do the steps add up?

Add and cancel. One NO2 and one NO3 drop out, leaving NO2 + CO → NO + CO2.

Step 2 · name the intermediate

NO3 is made in step 1 and used in step 2, so it is the intermediate.

Step 3 · derive the rate law

Step 1 sets the rate, and it is bimolecular in NO2. So rate = k1[NO2]2.

Step 4 · compare with the data

That law is second order in NO2 and has no CO in it, so it is zero order in CO. The mechanism passes.

Step 5 · what a catalyst would do

Say a catalyst drops Ea from 83.0 to 55.0 kJ/mol at 300 K. The speed-up is e raised to ΔEa/RT = 28000/2494 = 11.23. So e11.23 = 7.5 × 104, about 75,000 times faster.

Check it a second way

Near 300 K each 5.74 kJ/mol removed buys a factor of 10. We removed 28.0 kJ/mol, or 4.88 such chunks, so 104.88 = 7.6 × 104. That agrees.

Matching rate laws is evidence, not proof. Two different mechanisms can predict the same rate law, so kinetics alone can never settle the question. Chemists add other tests: trapping an intermediate, labeling an atom with a heavier isotope, or watching the reaction with a laser pulse short enough to catch the transition state.

Catalysts

A catalyst speeds a reaction up without being used up. It takes part in one step and comes back out in a later one. So it never appears in the overall equation, even though the reaction needs it.

A catalyst works by opening a path with a lower Ea. Look again at the dashed line in the figure above. Same start, same finish, shorter hills. But it leaves ΔH and the position of equilibrium exactly where they were, because it speeds both directions by the same factor.

In homogeneous catalysis the catalyst shares the phase of the reactants. Chlorine atoms wrecking ozone are a gas-phase case. First Cl + O3 → ClO + O2, then ClO + O → Cl + O2. The two add up to O3 + O → 2 O2, and the Cl atom is back to do it again.

In heterogeneous catalysis the catalyst is a different phase, usually a solid. Gases stick to its surface, react there, and leave. A car's catalytic converter is the everyday case. Platinum and rhodium turn CO and NO into CO2 and N2.

Enzymes are the catalysts your body runs on. Each one is a protein with a pocket called the active site. The reacting molecule settles in at the right angle, with the doomed bond already strained.

Key points
  • A mechanism is a list of elementary steps that add up to the overall equation.
  • For an elementary step alone, the orders match the coefficients.
  • The slowest step sets the pace, so its rate law is usually the measured one.
  • A catalyst lowers Ea, is remade, and does not shift ΔH or the equilibrium.
Interactive Step through a mechanism and lower the barrier with a catalyst. Open full ↑

The chapter in one breath

  • A rate is a change over a time. Divide each species change by its coefficient and the whole reaction gets one number.
  • Rate laws come from data. The orders in rate = k[A]m[B]n are measured, not copied from the coefficients.
  • Initial rates find the orders. Change one reactant, see how the rate answers, and solve fm = r.
  • One of three plots is straight. [A] means zero order, ln[A] means first order, 1/[A] means second order.
  • Half-life depends on order. First order gives 0.693/k every time, while the other two shift with [A]0.
  • Collisions need energy and aim. Only the share e−Ea/RT of them brings enough energy to clear the hill.
  • Temperature works on k. A plot of ln k against 1/T is a line, and its slope is −Ea/R.
  • The slow step rules. The rate-determining step usually hands you the rate law, which is how a mechanism gets tested.
  • Catalysts lower the hill. They are remade each cycle and they leave ΔH and the equilibrium untouched.

ReferenceWords worth knowing

These are the words this chapter leans on. Come back to the list when a term stops feeling solid.

activated complex
The arrangement of atoms at the top of the energy hill, where old bonds are partly broken and new ones partly formed. Also called the transition state.
activation energy (Ea)
The least energy a collision must carry for the reaction to happen. It is the height of the hill, measured up from the reactants.
Arrhenius equation
k = A e−Ea/RT. It links the rate constant to the barrier and the temperature.
bimolecular step
An elementary step in which two particles collide. Its rate is first order in each of them.
catalyst
A substance that speeds a reaction by offering a lower-barrier path, and is remade by the end. Enzymes are the catalysts in living cells.
collision theory
The picture of reaction rates built on particles meeting, with enough energy and in the right orientation.
elementary step
A single molecular event in a mechanism. Its rate law can be read straight from its coefficients.
frequency factor (A)
The front term in the Arrhenius equation. It bundles how often particles collide with how often they are aimed correctly.
half-life (t1/2)
The time for half of a reactant to be used up. It is constant only for a first-order reaction.
heterogeneous catalysis
Catalysis in which the catalyst is a different phase from the reactants, usually a solid surface holding gases or liquids.
homogeneous catalysis
Catalysis in which the catalyst is mixed into the same phase as the reactants.
initial rate
The instantaneous rate at the moment the reaction starts, before any product has built up.
instantaneous rate
The rate at one moment, equal to the slope of the tangent to the concentration curve. An average rate instead uses two points spread apart.
integrated rate law
The equation that gives concentration as a function of time. Each order has its own form and its own straight-line plot.
rate constant (k)
The number in front of a rate law. It does not depend on concentration, but it does depend on temperature.
rate-determining step
The slowest elementary step in a mechanism. It sets the pace, so its rate law is usually the overall rate law.
rate law
An equation giving rate in terms of concentrations raised to measured powers.
reaction intermediate
A species made in one step and used up in a later one. It is missing from the overall balanced equation.
reaction mechanism
The ordered set of elementary steps that add up to the overall reaction.
reaction order
The power on a concentration in the rate law. Add the powers and you get the overall order.

Check yourself

1. For 2 N2O5 → 4 NO2 + O2, O2 appears at 3.0 × 10−5 M/s. How fast does N2O5 disappear, and how fast does NO2 appear?
N2O5: 6.0 × 10−5 M/s. NO2: 1.2 × 10−4 M/s. O2 has a coefficient of 1, so the rate of reaction is 3.0 × 10−5 M/s. Multiply by 2 for N2O5: 2(3.0 × 10−5) = 6.0 × 10−5 M/s. Multiply by 4 for NO2: 4(3.0 × 10−5) = 1.2 × 10−4 M/s.
2. In two runs [B] is held fixed. When [A] doubles, the rate goes up eight times. What is the order in A?
Third order in A. Set 2m = 8. Since 23 = 8, m = 3. Nothing about B can be learned from this pair, because [B] never changed.
3. A first-order reaction has a half-life of 120 s. Find k, then find the fraction left after 360 s.
k = 5.78 × 10−3 s−1, and 12.5% is left. k = 0.693/120 s = 5.78 × 10−3 s−1. Then 360 s is exactly three half-lives, since 360/120 = 3. Half three times is (1/2)3 = 1/8, which is 0.125 or 12.5%.
4. A second-order reaction has k = 0.250 M−1s−1 and starts at 0.200 M. What is [A] after 100 s?
[A] = 0.0333 M. Use 1/[A] = 1/[A]0 + kt. Here 1/0.200 = 5.00 M−1 and kt = (0.250)(100) = 25.0 M−1. Adding gives 30.0 M−1, so [A] = 1/30.0 = 0.0333 M.
5. A rate constant doubles when the temperature rises from 300 K to 310 K. Find Ea.
About 53.6 kJ/mol. ln(2) = 0.693. Also 1/300 − 1/310 = 10/93000 = 1.075 × 10−4 K−1. So Ea = (8.314)(0.693)/(1.075 × 10−4) = 5.36 × 104 J/mol. This is the barrier size behind the old rule of thumb about 10 degrees.
6. A rate constant is reported as 0.45 M−2s−1. What is the overall order?
Third order overall. The units of k are M1−orders−1. Setting 1 − order = −2 gives order = 3. A quick check: rate = k[A]3 has units of (M−2s−1)(M3) = M/s, which is right.
7. Which plot comes out straight for data that follow rate = k[A], and what does its slope give you?
ln[A] against time, with slope −k. The first-order integrated form is ln[A] = ln[A]0 − kt. That is a line in ln[A] and t. A straight plot of [A] itself would mean zero order, and a straight plot of 1/[A] would mean second order.
8. A reaction has Ea = 40.0 kJ/mol. What fraction of collisions carries enough energy at 300 K?
About 1.1 × 10−7. RT = (8.314)(300) = 2494 J/mol. Then Ea/RT = 40000/2494 = 16.04, and e−16.04 = 1.1 × 10−7. So about one collision in ten million has the energy needed.
9. Forward Ea is 75 kJ/mol and ΔH is −40 kJ/mol. What is Ea for the reverse reaction?
115 kJ/mol. The reverse climbs the same peak, but from lower down. Ea(reverse) = Ea(forward) − ΔH = 75 − (−40) = 115 kJ/mol. The reverse barrier is taller because the products sit 40 kJ/mol below the reactants.
10. A catalyst lowers Ea by 20.0 kJ/mol at 298 K. By what factor does k grow?
About 3.2 × 103. The ratio is e raised to ΔEa/RT. Here RT = (8.314)(298) = 2478 J/mol, so the exponent is 20000/2478 = 8.07. Then e8.07 = 3.2 × 103, since A is taken as unchanged.
11. A mechanism is step 1 slow, 2 A → C, then step 2 fast, C + B → D. Predict the rate law and name the intermediate.
rate = k1[A]2, and C is the intermediate. The slow step is rate determining, and it is bimolecular in A, so its rate law is k1[A]2. B is absent from the rate law, so the reaction is zero order in B. C is made in step 1 and used in step 2, so it is the intermediate.
12. A catalyst doubles the forward rate of a reaction at equilibrium. What happens to the reverse rate and to K?
The reverse rate also doubles, and K does not change. A catalyst lowers the same barrier for both directions, so both rate constants grow by the same factor. Since K is the ratio of those constants, the ratio holds steady. A catalyst gets you to equilibrium sooner, not to a different equilibrium.