Factors That Affect the Rate of Reactions (Senior 6 Chemistry): free lessons | Ganzaa

Factors That Affect the Rate of Reactions

Lessons

Senior 6 · Chemistry

Factors That Affect the Rate of Reactions

3 lessons

Lesson 1 of 3

Reaction Kinetics and the Rate of Reaction

25 min

Why do diamonds last?

At 25 °C and 1 bar, diamond is less stable than graphite. The standard Gibbs energy change for $\mathrm{C(diamond) \rightarrow C(graphite)}$ is about $-2.9\,\mathrm{kJ\,mol^{-1}}$ , so thermodynamics says the change is spontaneous. Yet a diamond ring does not turn into pencil lead in a lifetime, or in a million lifetimes. The reaction is feasible but immeasurably slow.

Other reactions sit at the opposite extreme. Silver nitrate solution added to sodium chloride solution gives a white precipitate of silver chloride the moment the drops meet. Between the two lie reactions you can watch: an iron gate rusting over months, banana juice fermenting over a few days, milk souring overnight on a warm shelf.

Thermodynamics (your earlier units on energy changes and free energy) answers whether a reaction can go and how far. It says nothing about how fast. That second question belongs to reaction kinetics.

Definition

Reaction kinetics (chemical kinetics): the study of the rates of chemical reactions, the factors that change them, and the step-by-step pathway (mechanism) by which reactants become products.

Rate of reaction: the change in concentration of a reactant or product per unit time, scaled so that it has one value for the whole reaction (see below). Unit: $\mathrm{mol\,dm^{-3}\,s^{-1}}$ (or per minute, per hour for slow reactions). Example: if $[\mathrm{HCl}]$ falls by $0.020\,\mathrm{mol\,dm^{-3}}$ in 10 s, HCl is used up at $2.0 \times 10^{-3}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ .

Why kinetics matters

Where	The kinetic question
Medicines	How fast does the active ingredient decompose? The expiry date on a packet of tablets comes from rate measurements.
Food	How much does cooling slow the bacterial and enzyme reactions that spoil milk, meat and fruit?
Industry	A reaction with a good equilibrium yield is useless if it takes a week. Plants such as ammonia works choose temperatures, pressures and catalysts as a compromise between yield and rate.
Environment	How quickly do pollutants break down in air, water and soil, and how fast do CFCs destroy ozone?

One rate for the whole reaction

In $\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)}$ , hydrogen is used up three times as fast as nitrogen and ammonia forms twice as fast as nitrogen is used. To give the reaction a single rate, divide each rate of change by the species' stoichiometric coefficient, and put a minus sign in front of reactants (whose concentrations fall), so the rate is always positive. For $a\mathrm{A} + b\mathrm{B} \rightarrow c\mathrm{C} + d\mathrm{D}$ :

\text{rate} = -\frac{1}{a}\frac{\Delta[\mathrm{A}]}{\Delta t} = -\frac{1}{b}\frac{\Delta[\mathrm{B}]}{\Delta t} = \frac{1}{c}\frac{\Delta[\mathrm{C}]}{\Delta t} = \frac{1}{d}\frac{\Delta[\mathrm{D}]}{\Delta t}

Worked example 1: relating the rates of different species

Problem. In a reactor, $\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)}$ , hydrogen is consumed at $0.090\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ . Find (a) the rate of consumption of nitrogen, (b) the rate of formation of ammonia, (c) the rate of reaction.

Solution. $-\frac{\Delta[\mathrm{H_2}]}{\Delta t} = 0.090\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ .

(c) $\text{rate} = -\frac{1}{3}\frac{\Delta[\mathrm{H_2}]}{\Delta t} = \frac{0.090}{3} = 0.030\,\mathrm{mol\,dm^{-3}\,s^{-1}}$

(a) $-\frac{\Delta[\mathrm{N_2}]}{\Delta t} = 1 \times \text{rate} = 0.030\,\mathrm{mol\,dm^{-3}\,s^{-1}}$

(b) $\frac{\Delta[\mathrm{NH_3}]}{\Delta t} = 2 \times \text{rate} = 0.060\,\mathrm{mol\,dm^{-3}\,s^{-1}}$

Check: the ratio $0.030 : 0.090 : 0.060 = 1 : 3 : 2$ , the same as the coefficients.

Average, instantaneous and initial rates

Dinitrogen pentoxide dissolved in tetrachloromethane decomposes when warmed: $\mathrm{2N_2O_5 \rightarrow 4NO_2 + O_2}$ . The table gives illustrative concentrations at 45 °C (calculated for this lesson; they follow the real pattern of this reaction).

$t$ / s	0	400	800	1200	1600	2000
$[\mathrm{N_2O_5}]$ / $\mathrm{mol\,dm^{-3}}$	2.000	1.561	1.218	0.950	0.742	0.579

The average rate over an interval is the change divided by the time taken: the gradient of a chord joining two points on the curve. The instantaneous rate at a moment is the gradient of the tangent to the curve at that moment, in calculus $-\frac{d[\mathrm{N_2O_5}]}{dt}$ . The initial rate is the instantaneous rate at $t = 0$ , before the reactant concentrations have changed; Unit 13 uses it to find rate laws.

Figure 1.1 plots the data. The $\mathrm{N_2O_5}$ curve is steepest at the start and flattens as the reactant is used up: the reaction slows down because fewer reactant molecules are left.

Figure 1.1: Concentration against time for the decomposition of $\mathrm{N_2O_5}$ (illustrative data), with the tangent drawn at 800 s.

Worked example 2: rates from a concentration–time graph

Problem. Using the table and Figure 1.1, find (a) the average rate of disappearance of $\mathrm{N_2O_5}$ between 0 and 400 s and between 400 and 800 s; (b) the instantaneous rate of disappearance of $\mathrm{N_2O_5}$ at 800 s; (c) the rate of reaction and the rate of formation of $\mathrm{NO_2}$ at 800 s.

Solution.

(a) $\frac{2.000 - 1.561}{400 - 0} = \frac{0.439}{400} = 1.10 \times 10^{-3}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ ; $\frac{1.561 - 1.218}{400} = \frac{0.343}{400} = 8.58 \times 10^{-4}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ . The rate falls as $[\mathrm{N_2O_5}]$ falls.

(b) The tangent at 800 s passes through $(0, 1.82)$ and $(1600, 0.61)$ :

-\frac{d[\mathrm{N_2O_5}]}{dt} = -\frac{0.61 - 1.82}{1600 - 0} = \frac{1.21}{1600} = 7.6 \times 10^{-4}\,\mathrm{mol\,dm^{-3}\,s^{-1}}

(c) $\text{rate} = \frac{1}{2} \times 7.6 \times 10^{-4} = 3.8 \times 10^{-4}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ , and $\frac{d[\mathrm{NO_2}]}{dt} = 4 \times 3.8 \times 10^{-4} = 1.5 \times 10^{-3}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ .

Notice that the instantaneous rate at 800 s ( $7.6 \times 10^{-4}$ ) lies between the average rates of the intervals on either side ( $8.58 \times 10^{-4}$ and $6.70 \times 10^{-4}$ for 800 to 1200 s), as it should.

Warning

Common mistakes

Quoting a negative rate. The rate of reaction is always positive; the minus sign in front of a reactant term cancels the negative $\Delta[\mathrm{A}]$ .
Forgetting to divide by the coefficient. "The rate of reaction" and "the rate of formation of $\mathrm{NO_2}$ " are different numbers unless the coefficient is 1.
Finding an instantaneous rate from a chord. A chord gives an average over the interval; only a tangent gives the rate at one instant.
Reading a "rate" with no unit. Concentration per time: $\mathrm{mol\,dm^{-3}\,s^{-1}}$ .

Tip

To draw a good tangent, place a ruler so that it touches the curve at the point and makes equal angles with the curve on both sides. Then make the "triangle" as large as the grid allows: the larger the $\Delta y$ and $\Delta t$ , the smaller the reading error.

CrashCourse, Chemistry #32, Kinetics: Chemistry's Demolition Derby. Watch the first half for what a reaction rate is and why rates fall as reactants are used up; the later part previews rate laws, which you meet in Unit 13.

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Practice

Define rate of reaction and state its usual unit.
Explain the difference between thermodynamic feasibility and rate, using the conversion of diamond to graphite.
For $\mathrm{4NH_3(g) + 5O_2(g) \rightarrow 4NO(g) + 6H_2O(g)}$ , ammonia is consumed at $0.24\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ . Find the rate of consumption of oxygen, the rate of formation of water and the rate of reaction.
Use the $\mathrm{N_2O_5}$ table to find the average rate of disappearance of $\mathrm{N_2O_5}$ between 1200 s and 2000 s, and the average rate of formation of $\mathrm{O_2}$ in the same interval.
Bromine water is decolourised by methanoic acid: $\mathrm{Br_2(aq) + HCOOH(aq) \rightarrow 2Br^{-}(aq) + 2H^{+}(aq) + CO_2(g)}$ . In the first 50 s, $[\mathrm{Br_2}]$ falls from 0.0120 to $0.0101\,\mathrm{mol\,dm^{-3}}$ . Calculate the average rate of reaction and the average rate of formation of bromide ions.
A tangent to a product's concentration–time curve at $t = 30$ s passes through $(0\,\mathrm{s}, 0.010\,\mathrm{mol\,dm^{-3}})$ and $(60\,\mathrm{s}, 0.046\,\mathrm{mol\,dm^{-3}})$ . Find the instantaneous rate of formation of the product at 30 s.
Explain, in terms of the particles present, why the gradient of the $\mathrm{N_2O_5}$ curve decreases with time, and predict the gradient after a very long time.

Show answersHide answers

The change in concentration of a reactant or product per unit time (divided by its coefficient so that it has one value for the reaction); $\mathrm{mol\,dm^{-3}\,s^{-1}}$ .
$\Delta G^{\circ} \approx -2.9\,\mathrm{kJ\,mol^{-1}}$ , so the change is feasible, but its rate at room temperature is immeasurably small; thermodynamics says whether, kinetics says how fast.
Rate $= \frac{0.24}{4} = 0.060\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ ; $\mathrm{O_2}$ : $5 \times 0.060 = 0.30$ ; $\mathrm{H_2O}$ : $6 \times 0.060 = 0.36\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ .
$\frac{0.950 - 0.579}{800} = 4.6 \times 10^{-4}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ ; $\mathrm{O_2}$ forms at half this rate, $2.3 \times 10^{-4}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ .
$\frac{0.0120 - 0.0101}{50} = 3.8 \times 10^{-5}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ (coefficient of $\mathrm{Br_2}$ is 1, so this is also the rate of reaction); $\mathrm{Br^{-}}$ forms at $7.6 \times 10^{-5}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ .
$\frac{0.046 - 0.010}{60} = 6.0 \times 10^{-4}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ .
As $\mathrm{N_2O_5}$ is used up there are fewer molecules per unit volume to decompose each second, so the rate (gradient) falls; when the reactant is effectively used up, the gradient tends to zero and the curve becomes horizontal.

Sources and credits

Image: Rust on iron.jpg by Laitr Keiows, via Wikimedia Commons, licensed under CC BY-SA 3.0.
Video: Kinetics: Chemistry's Demolition Derby, Crash Course Chemistry #32, CrashCourse, embedded from YouTube.
The text, table, illustrative data, graph, worked examples and Practice items were written and drawn by Ganzaa.
This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1Why do diamonds last?

2Why kinetics matters

3One rate for the whole reaction

4Average, instantaneous and initial rates

Why do diamonds last?

Why kinetics matters

One rate for the whole reaction

Average, instantaneous and initial rates