Topic 11

Topic 11 on a page

Everything from this topic in one place, for the last read before an exam. If anything here is unfamiliar, follow the link back to the full page.

Constants and conversions

Quantity	Value
gas constant, R	8.31 J mol⁻¹ K⁻¹
temperature	K = °C + 273
activation energy	kJ mol⁻¹ × 1000 = J mol⁻¹ (use J with R)
time	s = min × 60
ln 2	0.693

The key relationships

To find	Use	Page
rate of loss of a reactant	−gradient of the tangent on a concentration–time graph	Rate and order
rate equation	$\text{rate} = k[\text{A}]^m[\text{B}]^n$ , with $m$ and $n$ = 0, 1 or 2, from experiment	Rate and order
units of k	rearrange for k, substitute units, cancel	The rate constant
order from two experiments	change one concentration, compare the factors	Orders from initial rates
rate from a clock reaction	rate ∝ 1 ÷ t	Initial rates and continuous monitoring
Arrhenius, as a line	$\ln k = -(E_\text{a}/R)(1/T) + \ln A$ ; gradient = $-E_\text{a}/R$	Activation energy from a graph
Arrhenius, two temperatures	$\ln(k_2 \div k_1) = (E_\text{a}/R)(1/T_1 - 1/T_2)$	Activation energy by calculation

Orders, graphs and half-lives

Order in A	Double [A]	Concentration–time graph	Successive half-lives	Rate–concentration graph	Units of k (overall order)
0	rate × 1	straight line down	get shorter	horizontal line	mol dm⁻³ s⁻¹ (0)
1	rate × 2	curve that flattens	constant	straight line through the origin	s⁻¹ (1)
2	rate × 4	steep curve, long tail	each double the last	upward curve	dm³ mol⁻¹ s⁻¹ (2)

The units of k for overall order 3 are dm⁶ mol⁻² s⁻¹. See Orders from concentration–time graphs and Orders from rate–concentration graphs.

Definitions to learn word for word

Rate of reaction: the change in concentration of a reactant or product per unit time.
Rate equation: $\text{rate} = k[\text{A}]^m[\text{B}]^n$ , where k is the rate constant and $m$ and $n$ are 0, 1 or 2.
Order with respect to a substance: the power to which its concentration is raised in the rate equation.
Overall order: the sum of the orders, $m + n$ .
Rate constant: the constant of proportionality in the rate equation, with a fixed value for a given reaction at a given temperature.
Half-life: the time taken for the concentration of a reactant to fall to half of its initial value.
Rate-determining step: the slowest step in the mechanism; it controls the overall rate.
Activation energy: the minimum energy that colliding particles must have for a reaction to occur.
Homogeneous catalyst: in the same phase as the reactants. Heterogeneous catalyst: in a different phase.

Methods

Technique: name the species or property that changes (gas, colour, acid, ions) and why it can be measured. Gas syringe for a light gas, mass change for a dense one, colorimetry for one coloured species, titration after quenching, pH or conductivity for acid or ions (page 5).
Initial-rate method: separate experiments, one concentration changed, constant temperature and total volume, tangent at t = 0. A clock reaction is an approximation, with rate ∝ 1/t.
Order from a table: compare two experiments where only one concentration changes.
Order from a graph: half-lives on a curve; rate against concentration for the shape.
Mechanism from a rate equation: species in the rate equation are in the slow step (or in fast steps before it); zero-order species react after it; steps must add up to the overall equation and intermediates cancel.
Activation energy: plot ln k (or ln(1/t)) against 1/T, gradient × (−R), ÷ 1000.
Solid catalyst: adsorption, weakened bonds, reaction on the surface, desorption; large surface area.

SN1 and SN2 as evidence

	Primary, SN2	Tertiary, SN1
Rate equation	rate = k[RBr][OH⁻]	rate = k[RBr]
Doubling [OH⁻]	rate × 2	no change
Steps	one	two, through a carbocation
Slow step	the only step	loss of Br⁻ from RBr

Core practicals 9a and 9b

9a: iodine, propanone and H⁺; samples quenched in excess NaHCO₃(aq); titrate I₂ with thiosulfate; titre against time is a straight line, so zero order in I₂; doubling propanone or H⁺ doubles the rate.
9b: iodine clock; H₂O₂ + 2I⁻ + 2H⁺ → I₂ + 2H₂O is slow; thiosulfate removes I₂ until it runs out; total volume kept constant; 1/t against volume of KI is a straight line through the origin, so first order in I⁻; ln(1/t) against 1/T gives $E_\text{a}$ .

The ten traps

Copying the coefficients of the equation into the rate equation.
Saying that changing a concentration changes k.
Wrong or missing units for k.
Measuring every half-life from t = 0, or measuring only one.
Calling any curve first order without checking half-lives.
Comparing experiments in which two concentrations change.
Confusing the straight-line shape on a concentration–time graph (zero order) with the one on a rate–concentration graph (first order, through the origin).
Putting an intermediate in a rate equation, or a zero-order species in the slow step.
Using °C, or kJ with R in J, in an Arrhenius calculation.
Saying a catalyst gives particles more energy, or shifts the equilibrium.

Mixed test

Try these with no notes, then check the answers.

For rate = k[NO]²[H₂], [NO] is tripled and [H₂] is halved. By what factor does the rate change?
Find the orders, the rate equation and k (with units) from these initial-rate data (illustrative): experiment 1, [A] = 0.10, [B] = 0.20, rate 5.0 × 10⁻⁴; experiment 2, [A] = 0.30, [B] = 0.20, rate 1.5 × 10⁻³; experiment 3, [A] = 0.10, [B] = 0.40, rate 2.0 × 10⁻³ (concentrations in mol dm⁻³, rate in mol dm⁻³ s⁻¹).
[A] is 0.200 mol dm⁻³ at 0 s, 0.100 at 50 s and 0.050 at 100 s. State the order and the time for [A] to fall to 0.0125 mol dm⁻³.
The reaction NO₂ + CO → NO + CO₂ has rate = k[NO₂]². Which slow step is consistent, (a) NO₂ + CO → NO + CO₂ or (b) 2NO₂ → NO + NO₃? Give the fast step for (b).
Zinc reacts with hydrochloric acid to give hydrogen. Suggest a technique to follow the rate, and say why mass change is a poor choice.
Doubling [OH⁻] has no effect on the rate of hydrolysis of 2-bromo-2-methylpropane. Name the mechanism and describe the two curly arrows.
A graph of ln k against 1/T has a gradient of −5.0 × 10³ K. Calculate $E_\text{a}$ in kJ mol⁻¹.
The rate constant of a reaction doubles from 300 K to 310 K. Calculate $E_\text{a}$ .
Explain how a solid iron catalyst speeds up the Haber process.
In Core Practical 9a, why is each sample run into sodium hydrogencarbonate, and why is the time noted at that moment?

Answers cover these until you have tried the questions

NO is second order: 3² = 9. H₂ is first order: ½. Overall 9 × ½ = × 4.5.
1 and 2: [A] ×3, rate ×3: first order in A. 1 and 3: [B] ×2, rate ×4: second order in B. rate = k[A][B]². k = 5.0 × 10⁻⁴ ÷ (0.10 × 0.20²) = 5.0 × 10⁻⁴ ÷ 4.0 × 10⁻³ = 0.125 dm⁶ mol⁻² s⁻¹.
The half-life is 50 s, then 50 s: constant, so first order. 0.200 → 0.0125 is four halvings (0.100, 0.050, 0.025, 0.0125), so 4 × 50 = 200 s.
(b). It has two NO₂ and no CO in the slow step, giving rate = k[NO₂]². (a) would give rate = k[NO₂][CO]. The fast step is NO₃ + CO → NO₂ + CO₂. Adding the steps and cancelling NO₃ and one NO₂ gives NO₂ + CO → NO + CO₂.
A gas syringe: hydrogen is the only gas made, and its volume is proportional to the amount of zinc that has reacted. Mass change is poor because hydrogen is so light: 100 cm³ at RTP has a mass of only (100 ÷ 24 000) × 2.0 = 0.0083 g, less than the smallest division of a 0.01 g balance.
SN1. The rate does not depend on [OH⁻], so OH⁻ is not in the rate-determining step. Step 1 (slow): a curly arrow from the C–Br bond to Br gives a carbocation and Br⁻. Step 2 (fast): a curly arrow from a lone pair on the oxygen of OH⁻ to the positive carbon gives the alcohol.
$E_\text{a}$ = 5.0 × 10³ × 8.31 = 41 550 J mol⁻¹ = 41.6 kJ mol⁻¹.
ln 2 = 0.6931. 1/300 − 1/310 = 1.0753 × 10⁻⁴ K⁻¹. $E_\text{a}$ = 8.31 × 0.6931 ÷ 1.0753 × 10⁻⁴ = 5.36 × 10⁴ J mol⁻¹ = 53.6 kJ mol⁻¹.
Nitrogen and hydrogen adsorb onto active sites on the surface, and their bonds are weakened. The atoms react on the surface to form ammonia, which desorbs. This gives an alternative route with a lower activation energy.
The sodium hydrogencarbonate neutralises the H⁺ catalyst, which stops the reaction, so the titre measures [I₂] at the moment of sampling. The time is noted then because that is when the reaction in that sample stopped.