D. Energy Changes and Rates of Reaction · Energy Changes and Rates of ReactionReaction Rates and How to Measure Them
D3.5 D2.1 A1.2 A1.6 A1.12 A1.13About 70 minutes
Learning goals- I can define reaction rate and calculate average rates from data, with correct units.
- I can find an instantaneous rate from the slope of a tangent to a concentration-time graph.
- I can relate the rates of different reactants and products using the balanced equation.
- I can choose a suitable method to measure the rate of a given reaction.
Words to know
chemical kinetics
The study of the rates of chemical reactions and of the steps by which they occur.
rate of reaction
The change in concentration (or amount) of a reactant or product per unit time, often in mol/(L·s).
average rate
The rate over a time interval; the slope of a secant on a concentration-time graph.
instantaneous rate
The rate at one moment; the slope of the tangent to a concentration-time graph.
initial rate
The instantaneous rate at the start of a reaction (t = 0).
Fast, slow and everything in between
A precipitate forms the instant two solutions mix. A match flares in a fraction of a second. Bread dough rises over an hour, an iron railing rusts over years, and the limestone and dolostone of the Niagara Escarpment slowly dissolve in rainwater over thousands of years. Thermochemistry tells us how much energy a reaction releases or absorbs; it says nothing about how fast the reaction happens. The second half of this unit is about chemical kinetics: the study of reaction rates and of the steps by which reactions happen.
What is a reaction rate?
The rate of reaction is the change in the amount or concentration of a reactant or product per unit time. For a substance in solution or a gas, we usually use molar concentration, written in square brackets: [A] means "the concentration of A in mol/L".
average rate = Δ[product] ÷ Δt or average rate = −Δ[reactant] ÷ Δt
The minus sign for a reactant makes the rate positive, because the reactant's concentration decreases. Units are usually mol/(L·s).
Rates can also be expressed in other measurable units: millilitres of gas per second, grams lost per minute, or the change in pressure per second.
Rates change as a reaction proceeds
For most reactions the rate is fastest at the start and slows down as reactants are used up. On a graph of concentration versus time:
- The average rate over an interval is the slope of the secant (the straight line joining two points).
- The instantaneous rate at one moment is the slope of the tangent to the curve at that moment. (In Calculus and Vectors, MCV4U, this is exactly the derivative.)
- The initial rate is the instantaneous rate at t = 0.
Sample data for the decomposition of a reactant A. The curve is steepest at the start. The dashed secant gives the average rate from 0 s to 200 s (0.380 ÷ 200 = 1.90 × 10⁻³ mol/(L·s)); the solid tangent at 100 s gives the instantaneous rate there.| t (s) | 0 | 100 | 200 | 300 | 400 |
|---|
| [A] (mol/L) | 0.800 | 0.580 | 0.420 | 0.305 | 0.221 |
Sample data for the graph above.Worked exampleExample 1: Average rates from data
Use the table to find the average rate of consumption of A (a) from 0 s to 100 s and (b) from 300 s to 400 s.
- (a) rate = −Δ[A] ÷ Δt = −(0.580 − 0.800) mol/L ÷ (100 − 0) s = 0.220 ÷ 100 = 2.20 × 10⁻³ mol/(L·s).
- (b) rate = −(0.221 − 0.305) ÷ (400 − 300) = 0.084 ÷ 100 = 8.4 × 10⁻⁴ mol/(L·s).
(a) 2.20 × 10⁻³ mol/(L·s); (b) 8.4 × 10⁻⁴ mol/(L·s). The rate in the last interval is less than half the rate at the start, because less A remains to react.
Worked exampleExample 2: Instantaneous rate from a tangent
The tangent drawn at t = 100 s passes through the points (20 s, 0.729 mol/L) and (200 s, 0.394 mol/L). Find the instantaneous rate at 100 s.
- Slope of the tangent = Δ[A] ÷ Δt = (0.394 − 0.729) ÷ (200 − 20) = −0.335 ÷ 180 = −1.86 × 10⁻³ mol/(L·s).
- Rate of consumption = −slope.
1.86 × 10⁻³ mol/(L·s) at t = 100 s.
Rates and coefficients
In a balanced equation, substances are used up and produced in the mole ratio of their coefficients, so their rates are related the same way. For
2N₂O₅(g) → 4NO₂(g) + O₂(g)
whenever 2 mol of N₂O₅ react, 4 mol of NO₂ and 1 mol of O₂ form. So NO₂ forms twice as fast as N₂O₅ is used up, and O₂ forms half as fast:
rate of NO₂ formation = 2 × rate of N₂O₅ consumption; rate of O₂ formation = ½ × rate of N₂O₅ consumption.
Worked exampleExample 3: Using the mole ratio
In 2N₂O₅(g) → 4NO₂(g) + O₂(g), N₂O₅ is consumed at 3.6 × 10⁻⁴ mol/(L·s). Find the rates of formation of NO₂ and O₂.
- NO₂: (4/2) × 3.6 × 10⁻⁴ = 7.2 × 10⁻⁴ mol/(L·s).
- O₂: (1/2) × 3.6 × 10⁻⁴ = 1.8 × 10⁻⁴ mol/(L·s).
NO₂ forms at 7.2 × 10⁻⁴ mol/(L·s); O₂ forms at 1.8 × 10⁻⁴ mol/(L·s).
How rates are measured
Choose a property that changes as the reaction goes and that you can measure without stopping the reaction:
- Volume of gas produced (gas syringe, or a graduated cylinder of water turned upside down): Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g).
- Mass lost as a gas escapes from an open flask on a balance: CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g).
- Colour intensity, measured with a colorimeter or spectrophotometer (for example, purple permanganate being used up).
- pH (when H⁺ or OH⁻ is used up or produced) or electrical conductivity (when the number of ions changes).
- Pressure of a gas mixture in a closed container (when the number of moles of gas changes).
- Time for a fixed change, such as the time for a precipitate to hide a mark under the flask. Then rate is proportional to 1 ÷ time.
Two common ways to follow a reaction. A: a gas syringe collects the gas, so the volume can be read at regular times. B: an open flask on a balance loses mass as carbon dioxide escapes (the cotton plug stops liquid spraying out but lets the gas through).Worked exampleExample 4: Rate from mass loss
Marble chips (CaCO₃) react with excess hydrochloric acid in an open flask on a balance. The mass falls from 152.38 g to 151.72 g in the first 60 s. Find the average rate of CO₂ production in g/s and in mol/s (M(CO₂) = 44.01 g/mol).
- Mass lost = mass of CO₂ escaped = 152.38 − 151.72 = 0.66 g.
- Rate = 0.66 g ÷ 60 s = 0.011 g/s.
- In moles: 0.011 g/s ÷ 44.01 g/mol = 2.5 × 10⁻⁴ mol/s.
0.011 g/s, or 2.5 × 10⁻⁴ mol of CO₂ per second. (Two significant digits, because the mass lost, 0.66 g, has two.)
Hands-on activityFollowing a reaction on a kitchen scale
You need: a digital kitchen scale (1 g resolution is fine), a tall jug or large measuring cup, 250 mL of vinegar, 2 tablespoons of baking soda on a piece of paper, a timer, graph paper, and a grown-up nearby. Do this in a sink or a large bowl in case it foams over.
- Put the jug with the vinegar on the scale. Put the paper with the baking soda on the scale beside the jug. Record the total mass.
- Start the timer as you tip the baking soda into the vinegar (leave the paper on the scale). Record the mass every 15 s for 3 minutes.
- Make a table of time and mass lost (= starting mass − mass). Graph mass lost (y) against time (x).
- Find the average rate (g/s) for the first 30 s and for the last 30 s. Draw a tangent at 60 s and find the instantaneous rate there.
What to look for: the mass drops quickly at first, then more slowly, and levels off when the acetic acid (or the baking soda) is used up. The total loss is the mass of CO₂ that escaped, often a few grams. Questions: Why does the rate decrease? Why might the measured total be a little smaller than the theoretical mass of CO₂? (Some CO₂ stays dissolved in the liquid.)
Just for funChemists sometimes describe very slow reactions as "energetically downhill but in no hurry". An iron railing rusting over many winters is in no hurry. Sodium metal dropped into water is in a very big hurry.
Your turn
1Define the rate of a reaction and give its usual units.
The change in concentration (or amount) of a reactant or product per unit time; usually mol/(L·s).
2[B] increases from 0.000 mol/L to 0.150 mol/L in 25 s. Find the average rate of formation of B.
0.150 ÷ 25 = 6.0 × 10⁻³ mol/(L·s).
3Why is a minus sign used when the rate is calculated from a reactant's concentration?
The reactant's concentration decreases, so Δ[reactant] is negative; the minus sign makes the rate positive.
4For N₂(g) + 3H₂(g) → 2NH₃(g), H₂ is consumed at 0.030 mol/(L·s). Find the rates for N₂ and NH₃.
N₂: (1/3)(0.030) = 0.010 mol/(L·s) consumed. NH₃: (2/3)(0.030) = 0.020 mol/(L·s) formed.
5What is the difference between an average rate and an instantaneous rate?
An average rate is over a time interval (slope of a secant); an instantaneous rate is at a single moment (slope of a tangent).
6Suggest a way to measure the rate of Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g).
Collect and measure the volume of hydrogen gas over time (gas syringe or inverted cylinder), or follow the mass loss as H₂ escapes (small, so a precise balance is needed), or track the pH rising as H⁺ is used.
7Why does the rate of most reactions decrease with time?
The reactants are being used up, so their concentrations decrease and there are fewer reacting collisions per second.
More practice: warm-up
The molar concentration of A, in mol/L.
2Which line on a concentration-time graph gives the instantaneous rate?
The tangent at that time.
3When during a reaction is the rate usually greatest?
4If 40 mL of gas forms in 20 s, what is the average rate?
5If it takes 50 s for a precipitate to hide a mark, what is the rate in s⁻¹ (using rate ∝ 1/t)?
More practice: on your own
1[X] falls from 1.20 mol/L to 0.90 mol/L in the first 2.0 min, and from 0.50 to 0.40 mol/L between 6.0 and 8.0 min. Find both average rates in mol/(L·min) and explain the difference.
First: 0.30 ÷ 2.0 = 0.15 mol/(L·min). Later: 0.10 ÷ 2.0 = 0.050 mol/(L·min). The rate is lower later because the concentration of X is lower.
2For 2H₂O₂(aq) → 2H₂O(l) + O₂(g), O₂ forms at 1.5 × 10⁻³ mol/(L·s). Find the rate of consumption of H₂O₂.
2 mol H₂O₂ per 1 mol O₂: 2 × 1.5 × 10⁻³ = 3.0 × 10⁻³ mol/(L·s).
3The volume of H₂ collected from Mg + HCl is 0, 22, 38, 48, 53, 55 mL at 0, 10, 20, 30, 40, 50 s. Find the average rate for 0 to 10 s and for 40 to 50 s, and explain what is happening at 50 s.
0 to 10 s: 22 ÷ 10 = 2.2 mL/s. 40 to 50 s: 2 ÷ 10 = 0.2 mL/s. The reaction is nearly finished: the limiting reactant (magnesium or the acid) is almost used up.
4A tangent to a [P]-versus-time curve at 30 s passes through (0 s, 0.10 mol/L) and (60 s, 0.46 mol/L). What is the instantaneous rate of formation of P at 30 s?
(0.46 − 0.10) ÷ 60 = 6.0 × 10⁻³ mol/(L·s).
5Why is the mass-loss method poor for the reaction of magnesium with acid, but good for marble chips with acid?
Hydrogen has a tiny molar mass (2.02 g/mol), so the mass lost is very small and hard to measure; CO₂ (44.01 g/mol) gives a much larger, easily measured mass loss.
Challenge
1In 4PH₃(g) → P₄(g) + 6H₂(g), the concentration of H₂ rises by 0.048 mol/L in 12 s. Find the average rate of consumption of PH₃ and of formation of P₄.
Rate of H₂ = 0.048 ÷ 12 = 4.0 × 10⁻³ mol/(L·s). PH₃: (4/6)(4.0 × 10⁻³) = 2.7 × 10⁻³ mol/(L·s). P₄: (1/6)(4.0 × 10⁻³) = 6.7 × 10⁻⁴ mol/(L·s).
2For the sample data in this lesson, the instantaneous rate is always 0.0032 s⁻¹ × [A]. Find the instantaneous rate at 0 s and at 300 s, and compare with the average rates found in Example 1.
At 0 s: 0.0032 × 0.800 = 2.56 × 10⁻³ mol/(L·s), larger than the 0 to 100 s average (2.20 × 10⁻³) because the rate falls during that interval. At 300 s: 0.0032 × 0.305 = 9.8 × 10⁻⁴, larger than the 300 to 400 s average (8.4 × 10⁻⁴) for the same reason. Each interval's average lies between the instantaneous rates at its ends.
3Design a way to measure the rate of the reaction between purple potassium permanganate and oxalic acid in acidic solution, which ends colourless. What would you measure, and how would you turn it into a rate?
Measure the colour intensity (absorbance) with a colorimeter or spectrophotometer at regular times. Absorbance is proportional to the permanganate concentration, so convert absorbance to [MnO₄⁻] with a calibration line, then calculate −Δ[MnO₄⁻] ÷ Δt. A simpler version: time how long the colour takes to disappear and use rate ∝ 1/t.
Check your understandingQuiz
1. [A] decreases from 0.50 mol/L to 0.30 mol/L in 40 s. The average rate of consumption of A is
2. The slope of a tangent to a concentration-time curve gives
3. For 2SO₂(g) + O₂(g) → 2SO₃(g), SO₃ forms at 0.060 mol/(L·s). O₂ is consumed at
4. Which method is best for following CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g) in an open flask?
5. Why does a reaction's rate usually decrease as it proceeds?
6. A reaction makes a mark disappear in 40 s at one temperature and in 20 s at another. Using rate ∝ 1/time, the second rate is
For parents and teachersBig ideas (D3.5, D2.1): the rate of a reaction is the change in concentration (or another measurable quantity) per unit time. Average rates come from secants and instantaneous rates from tangents on a concentration-time graph. Rates of different substances are related by the coefficients of the balanced equation. Rates can be measured by gas volume, mass loss, colour, pH, conductivity, pressure or the time for a fixed change.
How to help: Do the kitchen-scale activity together and help your teen draw a smooth curve (not dot-to-dot) and a tangent. Ask: "Why is the curve levelling off?"
Common difficulties:
- Negative rates: rates are reported as positive numbers.
- Using the coefficients upside down when converting between substances (check: does the substance with the larger coefficient change faster?).
- Drawing a tangent that cuts through the curve rather than touching it at one point.
- Confusing "fast" with "releases a lot of energy": rate and ΔH are separate ideas.
Activity safety: vinegar and baking soda foam; work over a sink. Keep the scale dry.