Dr. Ray's Learning Center

Reaction Rates and the Factors That Affect Them

General Chemistry · Chemical Reaction Kinetics

Dr. Ray’s Learning Center · covers: Reaction Rates and the Factors That Affect Them

  • 21 to 29 min
  • 6 figures
  • 3 worked examples
  • 4 practice sets
  • 1 simulator

Before this lesson:

A match flares in a fraction of a second, an iron gate rusts over years, and milk sours in a few days in a warm kitchen but lasts for weeks in the refrigerator. All three are chemical reactions; what differs is how fast they go. The branch of chemistry that measures and explains the speed of reactions is called chemical kinetics.

In this lesson you will define the rate of a reaction from concentration data, link the rates of different species through the balanced equation, tell an average rate from an instantaneous one, and see the four factors that make a reaction faster or slower (Figure 1).

  1. Data[X] at two times
  2. Δ[X]Δt\dfrac{Δ[\mathrm{X}]}{Δt}change in [X] over change in t
  3. Rate of change of Xnegative for a reactant
  4. ×1a\times \frac{1}{a}a is its coefficient; minus sign for a reactant
  5. Rate of reactionalways positive, M/s
  6. ×\timesthe coefficient of any other species
  7. Answerrate of that species

For 2A→B+3C2\mathrm{A} \rightarrow \mathrm{B} + 3\mathrm{C}:rate=−12Δ[A]Δt=Δ[B]Δt=13Δ[C]Δt\text{rate} = -\frac{1}{2}\frac{Δ[\mathrm{A}]}{Δt} = \frac{Δ[\mathrm{B}]}{Δt} = \frac{1}{3}\frac{Δ[\mathrm{C}]}{Δt}

Figure 1. The reaction rate roadmap: a change in concentration over a change in time, divided by the coefficient (with a minus sign for a reactant), gives one positive rate of reaction; multiplying by another coefficient gives the rate of any other species.

Learning Objectives

After this lesson, you should be able to:

  1. Express average and instantaneous rates from concentration-time data and stoichiometry.
  2. Explain how concentration, temperature, surface area and catalysts change the rate.

Key Ideas

1. What a Reaction Rate Measures

As a reaction runs, reactants are used up and products form, so their concentrations change with time. The rate of a reaction tells how fast a concentration changes. Over a time interval ΔtΔt, a concentration changes by Δ[X]Δ[\mathrm{X}], final minus initial:

rate of change of [X]=Δ[X]Δt\text{rate of change of } [\mathrm{X}] = \dfrac{Δ[\mathrm{X}]}{Δt}

Concentrations are in M (mol/L) and times in s, so rates are in M/s. A reactant’s concentration falls, so its Δ[X]Δ[\mathrm{X}] is negative; a product’s concentration rises. Figure 2 shows illustrative data for the hypothetical reaction 2A→B+3C2\mathrm{A} \rightarrow \mathrm{B} + 3\mathrm{C}: [A] falls from 0.800 M while B and C build up. Every curve is steep at the start and flattens out as the reactant runs low.

Concentrations of A, B and C against timeA graph drawn to scale for the reaction 2 A goes to B plus 3 C, time from 0 to 180 seconds and concentration from 0 to 1.2 molar. The reactant A falls from 0.800 molar at time 0 to 0.400 at 25 seconds, 0.200 at 75 seconds and 0.100 at 175 seconds; its curve is steep at first and then flattens. The product C rises from 0 to 0.600, 0.900 and 1.05 molar at the same times, three times as far as B, which rises to 0.200, 0.300 and 0.350 molar.05010015000.40.81.2time (s)concentration (M)[C], product[B], product[A], reactant
Figure 2. Illustrative data for 2A→B+3C2\mathrm{A} \rightarrow \mathrm{B} + 3\mathrm{C}: [A] falls while [B] and [C] rise; [C] rises three times as fast as [B] and [A] falls twice as fast as [B] rises. Every curve is steep at first and then flattens, because the reaction slows down.

Key idea: a reaction rate is a change in concentration divided by the time it took, in M/s. Reactant concentrations fall and product concentrations rise.

2. Rates and the Balanced Equation

In 2A→B+3C2\mathrm{A} \rightarrow \mathrm{B} + 3\mathrm{C}, every time one B forms, two A are used up and three C form. So [A] falls twice as fast as [B] rises, and [C] rises three times as fast. To get one number for the whole reaction, divide each species’ rate of change by its coefficient, and put a minus sign on each reactant so the rate comes out positive:

rate=−12Δ[A]Δt=Δ[B]Δt=13Δ[C]Δt\text{rate} = -\dfrac{1}{2}\dfrac{Δ[\mathrm{A}]}{Δt} = \dfrac{Δ[\mathrm{B}]}{Δt} = \dfrac{1}{3}\dfrac{Δ[\mathrm{C}]}{Δt}

This single positive number is the rate of reaction (Whitten Sec. 16-1). Any one species’ data gives it, and from it you get the rate of every other species (Figure 3).

From A (reactant, coefficient 2): rate=−12Δ[A]Δt\text{rate} = -\dfrac{1}{2}\dfrac{Δ[\mathrm{A}]}{Δt}

From B (product, coefficient 1): rate=Δ[B]Δt\text{rate} = \dfrac{Δ[\mathrm{B}]}{Δt}

From C (product, coefficient 3): rate=13Δ[C]Δt\text{rate} = \dfrac{1}{3}\dfrac{Δ[\mathrm{C}]}{Δt}

Figure 3. One rate of reaction from any species: divide each rate of change by its coefficient, with a minus sign for the reactant.

Key idea: the rate of reaction is the rate of change of any species divided by its coefficient, with a minus sign for reactants. It is always positive.

3. Average and Instantaneous Rates

An average rate uses two points of the concentration-time data: it is the slope of the straight line, or secant, that joins them. Because the curve flattens, the average rate depends on the interval you choose. The instantaneous rate at one moment is the slope of the tangent line that touches the curve at that time (Figure 4). Chemists usually report instantaneous rates; the rate at t=0t = 0 is the initial rate, used in the next lesson.

An average rate and an instantaneous rate on the A curveThe concentration of A against time from 0 to 100 seconds, drawn to scale. A solid straight secant joins the points at 0 seconds, 0.800 molar, and 25 seconds, 0.400 molar; its slope is the average rate of change, minus 0.0160 molar per second. A dashed tangent touches the curve at 25 seconds and passes through 0 seconds, 0.600 molar and 75 seconds, 0 molar; its slope is the instantaneous rate of change at 25 seconds, minus 0.00800 molar per second. The tangent is less steep than the secant because the reaction is slowing down.025507510000.40.8time (s)[A] (M)secant 0 to 25 s:average ratetangent at 25 s:instantaneous rate
Figure 4. The slope of the secant (solid) gives the average rate of change of [A] from 0 to 25 s; the slope of the tangent at 25 s (dashed) gives the instantaneous rate of change at that moment. The tangent is less steep, because the reaction has slowed.

Shrinking the interval around a moment brings the average rate closer to the instantaneous rate. For the data in Figure 2, the average rate of reaction from 15 to 35 s is about 0.004167 M/s, from 20 to 30 s about 0.004040 M/s, and from 24 to 26 s about 0.004002 M/s, closing in on the tangent value at 25 s found in Worked Example 3.

Note: when the times are given in min, multiply the interval by 60 to get it in s before you divide, so the rate comes out in M/s.

Key idea: average rate: the slope of the secant between two points. Instantaneous rate: the slope of the tangent at one time. Both are converted to the rate of reaction with the coefficients.

Try it: measure a reaction rate

Pick a data set, drag the two time markers to see the secant and its average rate, switch to the tangent view for the instantaneous rate, and change the species to watch the coefficient do its work. The worked-example buttons replay each example on the graph.

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4. Four Factors That Change a Rate

Molecules must collide, with enough energy and the right orientation, to react. Anything that changes how often or how effectively they collide changes the rate (Whitten Sec. 16-2; Figure 5):

  • Nature of the reactants: some substances react faster than others, and the physical state matters. A solid reacts only at its surface, so a powder, with much more surface area, reacts faster than one lump of the same mass.
  • Concentration: more particles in the same volume collide more often, so the rate usually increases with concentration. The next lesson makes this exact with the rate law.
  • Temperature: at a higher temperature, molecules move faster and more of their collisions have enough energy to react. Many reactions go roughly twice as fast for a 10 °C rise near room temperature, which is why food keeps longer in the refrigerator.
  • Catalyst: a catalyst speeds up a reaction without being used up, by opening a lower-energy path. Enzymes are the catalysts of living cells.
Surface area and ratePanel (a), surface area: one large lump of a solid beside the same amount cut into 25 small pieces. The powder has much more surface, so it reacts faster.(a)Surface areaone lumppowderslowerfasterConcentration and ratePanel (b), concentration: a box with 4 reactant particles beside a box of the same size with 12. More particles in the same volume collide more often, so the reaction is faster.(b)Concentrationlower [X]higher [X]fewer collisionsmore collisionsTemperature and ratePanel (c), temperature: two boxes with 6 particles each; the particles in the hotter box have longer motion arrows. Faster molecules collide more often and harder, so more collisions lead to reaction.(c)Temperaturelower Thigher Tslower moleculesfaster, harder hitsA catalyst and ratePanel (d), catalyst: an energy profile from reactants to products. The path without a catalyst rises over a high barrier; the dashed path with a catalyst rises over a lower barrier between the same start and end, so the reaction is faster.(d)Catalystwithout catalystwith catalystprogress of reactionenergylower barrier, faster
Figure 5. Four ways to speed up a reaction. Panel (a): a powder has more surface than one lump. Panel (b): a higher concentration gives more collisions. Panel (c): at a higher temperature, molecules move faster and hit harder. Panel (d): a catalyst offers a path with a lower energy barrier.

Key idea: the nature and surface area of the reactants, their concentration, the temperature and a catalyst all change how fast a reaction goes.

5. A Strategy for Rate Problems

Follow the roadmap in Figure 1:

  1. Write the balanced equation and the rate expression with every coefficient.
  2. Find the change: Δ[X]Δ[\mathrm{X}] is final minus initial, and ΔtΔt is in s (multiply a time in min by 60). For an instantaneous rate, read the slope of the tangent instead.
  3. Divide by the coefficient, with a minus sign for a reactant, to get the rate of reaction.
  4. Convert to any other species: multiply the rate of reaction by that species’ coefficient. A reactant’s concentration changes with a minus sign; a rate is never negative.

Exam tip: check the sign. A rate of reaction or a rate of formation is positive; only Δ[X]Δt\frac{Δ[\mathrm{X}]}{Δt} of a reactant is negative. A negative rate of reaction means a minus sign is missing.

Worked Example 1: An Average Rate of Reaction

Problem: For the hypothetical reaction 2A→B+3C2\mathrm{A} \rightarrow \mathrm{B} + 3\mathrm{C} (data of Figure 2), [A] falls from 0.800 M at t=0st = 0\,\text{s} to 0.400 M at t=25st = 25\,\text{s}. What is the average rate of reaction over this interval, in M/s?

Step 1. Find the changes, final minus initial.

Δ[A]=0.400M−0.800M=−0.400MΔt=25s−0s=25s\begin{aligned} Δ[\mathrm{A}] &= 0.400\,\text{M}-0.800\,\text{M} = -0.400\,\text{M} \\ Δt &= 25\,\text{s}-0\,\text{s} = 25\,\text{s} \end{aligned}

Step 2. Divide to get the rate of change of [A].

Δ[A]Δt=−0.400M25s=−0.0160M/s\begin{aligned} \dfrac{Δ[\mathrm{A}]}{Δt} &= \dfrac{-0.400\,\text{M}}{25\,\text{s}} = -0.0160\,\text{M/s} \end{aligned}

Step 3. Divide by the coefficient of A (2) and change the sign, because A is a reactant.

rate=−12Δ[A]Δt=−12(−0.0160M/s)=0.00800M/s\begin{aligned} \text{rate} &= -\dfrac{1}{2}\dfrac{Δ[\mathrm{A}]}{Δt} = -\dfrac{1}{2}(-0.0160\,\text{M/s}) = 0.00800\,\text{M/s} \end{aligned}

Final Answer: 0.00800 M/s

Every value here is exact, so no rounding is needed. The rate is positive even though [A] fell, as a rate of reaction must be.

  1. ChangesΔ[A]=−0.400MΔ[\mathrm{A}] = -0.400\,\text{M}, Δt=25sΔt = 25\,\text{s}Step 1
  2. ×125s\times \dfrac{1}{25\,\text{s}}Step 2: divide
  3. Rate of change of [A]Δ[A]Δt=−0.0160M/s\dfrac{Δ[\mathrm{A}]}{Δt} = -0.0160\,\text{M/s}negative
  4. ×(−12)\times \left(-\dfrac{1}{2}\right)Step 3: coefficient 2, reactant
  5. Rate of reactionrate=0.00800M/s\text{rate} = 0.00800\,\text{M/s}positive
Figure 6. Worked Example 1: the change in [A] over the change in time, then divide by the coefficient of A and change the sign.

Your turn: an average rate of reaction

Same steps as Worked Example 1: find Δ[X]Δ[\mathrm{X}] and ΔtΔt in s, divide, then divide by the coefficient (with a minus sign for a reactant).

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Worked Example 2: The Rate of Formation of a Product

Problem: In the same reaction, 2A→B+3C2\mathrm{A} \rightarrow \mathrm{B} + 3\mathrm{C}, [A] falls from 0.400 M at 25 s to 0.200 M at 75 s. At what average rate does C form over this interval, in M/s?

Step 1. Find the rate of reaction from A, as in Worked Example 1.

rate=−12Δ[A]Δt=−12×−0.200M50s=0.00200M/s\begin{aligned} \text{rate} &= -\dfrac{1}{2}\dfrac{Δ[\mathrm{A}]}{Δt} = -\dfrac{1}{2} \times \dfrac{-0.200\,\text{M}}{50\,\text{s}} = 0.00200\,\text{M/s} \end{aligned}

Step 2. Multiply by the coefficient of C (3). The rate expression says rate=13Δ[C]Δt\text{rate} = \frac{1}{3}\frac{Δ[\mathrm{C}]}{Δt}, so:

Δ[C]Δt=3×rate=3×0.00200M/s=0.00600M/s\begin{aligned} \dfrac{Δ[\mathrm{C}]}{Δt} &= 3 \times \text{rate} = 3 \times 0.00200\,\text{M/s} = 0.00600\,\text{M/s} \end{aligned}

Final Answer: 0.00600 M/s

Check with the C data: [C] rises from 0.600 M to 0.900 M, a change of 0.300 M in 50 s, which is 0.00600 M/s. B forms at 0.00200 M/s, the rate of reaction itself, since its coefficient is 1.

Your turn: the rate of one species from another

Same steps as Worked Example 2: divide the given species’ rate by its coefficient to get the rate of reaction, then multiply by the coefficient of the species asked for.

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Worked Example 3: An Instantaneous Rate from a Tangent

Problem: In Figure 4 (the data of Figure 2), the tangent to the [A] curve at t=25st = 25\,\text{s} passes through the points (0 s, 0.600 M) and (75 s, 0 M). Find Δ[A]Δt\frac{Δ[\mathrm{A}]}{Δt} at 25 s, including its sign, and the instantaneous rate of reaction.

Step 1. Find the slope of the tangent from its two points.

Δ[A]Δt=0M−0.600M75s−0s=−0.00800M/s\begin{aligned} \dfrac{Δ[\mathrm{A}]}{Δt} &= \dfrac{0\,\text{M}-0.600\,\text{M}}{75\,\text{s}-0\,\text{s}} = -0.00800\,\text{M/s} \end{aligned}

Step 2. Divide by the coefficient of A and change the sign.

rate=−12(−0.00800M/s)=0.00400M/s\begin{aligned} \text{rate} &= -\dfrac{1}{2}(-0.00800\,\text{M/s}) = 0.00400\,\text{M/s} \end{aligned}

Final Answer: Δ[A]Δt=−0.00800M/s\frac{Δ[\mathrm{A}]}{Δt} = -0.00800\,\text{M/s} at 25 s; rate of reaction 0.00400 M/s

C forms at 3 times the rate of reaction, 0.0120 M/s, at this moment. The average rate from 0 to 25 s (Worked Example 1, 0.00800 M/s) is twice as large, because the curve is steeper early on.

Your turn: the signed rate of change of a reactant

Same idea as Worked Example 3: a reactant’s concentration falls, so its Δ[X]Δt\frac{Δ[\mathrm{X}]}{Δt} is negative and equals minus its coefficient times the rate of reaction.

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Common Mistakes

Common mistake: Forgetting the coefficient. In Worked Example 1, stopping at Δ[A]Δt\frac{Δ[\mathrm{A}]}{Δt} and dropping the sign gives 0.0160 M/s, twice the rate of reaction. Divide each species’ rate by its coefficient.

Common mistake: Using the coefficient the wrong way. In Worked Example 2, dividing the rate of reaction by 3 instead of multiplying gives about 0.000667 M/s for C, nine times too small. The species with the larger coefficient changes faster.

Common mistake: Getting the sign wrong. The rate of reaction is always positive: in Worked Example 1, Δ[A]Δt=−0.0160M/s\frac{Δ[\mathrm{A}]}{Δt} = -0.0160\,\text{M/s}, but the rate of reaction is 0.00800 M/s, because of the minus sign in front of each reactant term. A reactant’s own rate of change stays negative: in Worked Example 3, Δ[A]Δt\frac{Δ[\mathrm{A}]}{Δt} is -0.00800 M/s, not +0.00800 M/s, because its concentration falls.

Common mistake: Using a concentration instead of a change. In Worked Example 2, dividing [C] = 0.900 M by 50 s gives 0.0180 M/s, three times the true rate. Use Δ[C]Δ[\mathrm{C}], final minus initial.

Common mistake: Leaving the time in min. A rate in M/s needs ΔtΔt in s. If the times are in min, multiply the interval by 60 first; otherwise the answer is 60 times too large.

Common mistake: Using an average rate for an instantaneous one. The average rate from 0 to 25 s, 0.00800 M/s, is not the rate at 25 s (0.00400 M/s). An instantaneous rate needs the tangent at that time.

Key Takeaway

A reaction rate is a change in concentration over time, in M/s. Dividing each species’ rate of change by its coefficient, with a minus sign for reactants, gives one positive rate of reaction. An average rate is the slope of a secant; an instantaneous rate is the slope of the tangent. Concentration, temperature, surface area and catalysts all change how fast a reaction goes.

Practice: randomized problems

Each problem below uses fresh numbers every time and mixes every kind of problem in this lesson: work it out, check your answer, then open the worked solution to compare your steps.

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Your results

This summary collects every practice problem you try on this page and suggests what to review next.

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Figures and simulator: original work by Dr. Ray’s Learning Center, licensed CC BY 4.0. Method: Whitten, Davis, Peck and Stanley, Chemistry, 9th ed. (Brooks/Cole, Cengage Learning, 2010), Section 16-1 (pp. 608 to 614) for the rate of a reaction, the rate expression and average and instantaneous rates; Sections 16-2 (p. 614), 16-3 (pp. 614 to 622), 16-8 (pp. 638 to 642) and 16-9 (from p. 642) for the factors that affect rates. Data: illustrative concentrations generated from a stated rate law and checked with exact fractions in our own Python; they are not measured values. The simulator and the practice problems are audited; see the lesson audit log.