Unit 14 · Physical Chemistry

Rate Laws & Measurements

Rate equations, reaction orders, integrated rate laws, half-life expressions, the Arrhenius equation, and reaction mechanisms.

14.1

Rate Equations and Reaction Orders

Rate Equation (Rate Law)For a reaction aA + bB → products, the rate equation is: rate = k[A]m[B]n. The powers m and n are the orders with respect to A and B respectively; (m+n) is the overall order. Orders must be determined experimentally — they cannot be predicted from the stoichiometry of the overall equation.
OrderRate vs [A]Units of kExample
Zero (0)Rate = k (constant)mol L−1 s−1Some enzyme reactions at saturation; NH3 decomposition on W
First (1)Rate = k[A]s−1Radioactive decay; many first-order kinetics
Second (2)Rate = k[A]2 or k[A][B]mol−1 L s−1NO2 decomposition; many bimolecular reactions
Third (3)Rate = k[A]2[B] etc.mol−2 L2 s−1Rare; some termolecular gas reactions

Determining Orders from Initial Rates

Method of initial rates: use experiments where one concentration changes, others are constant. If [A] doubles and rate doubles: order w.r.t. A = 1 If [A] doubles and rate quadruples: order w.r.t. A = 2 If [A] doubles and rate stays same: order w.r.t. A = 0 General: rate₂/rate₁ = ([A]₂/[A]₁)ⁿ → n = log(rate₂/rate₁) / log([A]₂/[A]₁) Example: Exp 1: [A]=0.1, [B]=0.1, rate=2.0×10⁻³ Exp 2: [A]=0.2, [B]=0.1, rate=4.0×10⁻³ → doubling A doubles rate → order in A = 1 Exp 3: [A]=0.1, [B]=0.2, rate=8.0×10⁻³ → doubling B quadruples rate → order in B = 2 Rate = k[A][B]² (overall order = 3)
14.2

Integrated Rate Laws

Integrated Rate LawsExpress concentration as a function of time; allow determination of order from concentration-time data and calculation of concentration at any time.
OrderDifferential formIntegrated formLinear plotHalf-life (t½)
Zerorate = k[A] = [A]₀ − kt[A] vs t (slope = −k)t½ = [A]₀ / 2k
Firstrate = k[A]ln[A] = ln[A]₀ − ktln[A] vs t (slope = −k)t½ = ln2 / k = 0.693/k
Secondrate = k[A]²1/[A] = 1/[A]₀ + kt1/[A] vs t (slope = +k)t½ = 1 / (k[A]₀)

Determining Order from Graphs

Plot concentration data three ways simultaneously. The order is whichever gives a straight line:
• Zero order: [A] vs t is straight
• First order: ln[A] vs t is straight
• Second order: 1/[A] vs t is straight

Example (first-order): [A]₀ = 1.00 mol/L, k = 0.020 s⁻¹ After t = 50 s: ln[A] = ln(1.00) − 0.020×50 = 0 − 1.0 = −1.0 → [A] = e⁻¹·⁰ = 0.368 mol/L t½ = 0.693/0.020 = 34.7 s (same at every half-life for 1st order) After 2 half-lives: [A] = [A]₀/4 = 0.25 mol/L After 3 half-lives: [A] = [A]₀/8 = 0.125 mol/L

Half-Life Characteristics

Zero order: t½ = [A]₀/2k (depends on initial concentration) First order: t½ = 0.693/k (CONSTANT; independent of [A]₀) Second order: t½ = 1/(k[A]₀) (increases as concentration decreases) Diagnostic: If successive half-lives are equal → FIRST ORDER If successive half-lives increase → SECOND ORDER If successive half-lives decrease → ZERO ORDER
14.3

The Arrhenius Equation

Arrhenius Equationk = Ae−Ea/RT, where k = rate constant, A = pre-exponential (frequency) factor, Ea = activation energy (J/mol), R = 8.314 J mol−1 K−1, T = temperature in Kelvin. The logarithmic form: ln k = ln A − Ea/RT is used for graphical and two-point calculations.

Graphical Method for Ea

Plot ln k vs 1/T → straight line with: slope = −Eᵣ/R → Eᵣ = −slope × R y-intercept = ln A Example: Two data points: T₁ = 300 K, k₁ = 1.5×10⁻³ s⁻¹ T₂ = 320 K, k₂ = 5.3×10⁻³ s⁻¹ Two-point form: ln(k₂/k₁) = (Eᵣ/R)(1/T₁ − 1/T₂) ln(5.3/1.5) = (Eᵣ/8.314)(1/300 − 1/320) 1.259 = (Eᵣ/8.314)(2.083×10⁻⁴) Eᵣ = 1.259 × 8.314 / (2.083×10⁻⁴) = 50,300 J/mol = 50.3 kJ/mol

Physical Meaning of A and Ea

A (frequency factor): related to collision frequency and the fraction of collisions with correct orientation. Larger A → more collisions per second or better orientation requirement satisfied.

Ea (activation energy): energy barrier. Low Ea → fast reaction even at low T; high Ea → slow reaction, very sensitive to temperature. Catalyst lowers Ea, increasing k at same T.

14.4

Reaction Mechanisms

Reaction MechanismA sequence of elementary steps (each with its own rate law and molecularity) that together account for the overall balanced equation. The rate-determining step (RDS) is the slowest step; it controls the overall rate.

Elementary Steps and Molecularity

Unimolecular: A → products rate = k[A] (1st order) Bimolecular: A + B → products rate = k[A][B] (2nd order) Termolecular: A + B + C → products rate = k[A][B][C] (3rd order, rare) The rate law for an elementary step IS predicted by stoichiometry. But for OVERALL reactions, the rate law comes from the RDS only.

Mechanism Consistency with Rate Law

Example: Overall: 2NO₂ → 2NO + O₂ Proposed mechanism: Step 1 (slow, RDS): NO₂ + NO₂ → NO₃ + NO rate = k₁[NO₂]² Step 2 (fast): NO₃ → NO₂ + O fast Step 3 (fast): O + O → O₂ fast Net rate = k[NO₂]² — consistent with experiment (observed 2nd order). If experiment shows rate = k[A][B] but proposed mechanism has RDS as unimolecular (rate = k[A]), the mechanism is INCONSISTENT → rejected. If RDS involves an intermediate, substitute equilibrium expression for intermediate in terms of original reactants.

SN1 vs SN2 Mechanisms (from Unit 4)

SN2: one step (bimolecular) RO⁻ + R-X → [RO···R···X]⁻ → R-OR + X⁻ rate = k[R-X][RO⁻] (2nd order overall) SN1: two steps (rate-determining ionisation) Step 1 (slow): R-X → R⁺ + X⁻ rate = k[R-X] Step 2 (fast): R⁺ + RO⁻ → R-OR rate = k[R-X] (1st order; does NOT depend on [RO⁻]) These are real-world examples where the observed rate law reveals the mechanism.
14.5

Experimental Rate Measurement

Methods for Following Reactions

MethodMeasured quantityApplicable to
Colorimetry / spectrophotometryAbsorbance (Beer-Lambert: A = εcl)Coloured reactant or product
Gas syringe / manometryVolume or pressure of gas producedGas-producing reactions
Titrimetry (quench + titrate)Moles of reactant/product at time tAcid-base, redox reactions
ConductometryElectrical conductanceIonic reactions; change in ion type
Clock reactions (cross disappears)Time for visible changeThiosulfate + acid; iodine clock
PolarimetryOptical rotationChiral molecules: inversion of sucrose

The Iodine Clock Reaction

No videos added yet for this unit.

S₂O₈²⁻ + 2I⁻ → 2SO₄²⁻ + I₂ (rate-determining, produces I₂) I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻ (fast; consumes I₂ before it can react with starch) When S₂O₃²⁻ is exhausted: I₂ accumulates → sudden blue-black colour with starch Rate measured: 1/t (t = time for colour to appear) Rate ∝ 1/t → can investigate orders in [S₂O₈²⁻] and [I⁻]
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Exercises

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Quiz

Unit 14: Rate Laws & Measurements

25 Qs
Q1

The rate law for a reaction MUST be determined by:

Experiment. Orders cannot be read from stoichiometry of the overall equation. They reflect the rate-determining step. Only for elementary steps does stoichiometry predict order.
Q2

For rate = k[A]2[B], the units of k are:

Overall order = 3. Units of k: mol L−1 s−1 / (mol L−1)3 = mol−2 L2 s−1. General rule: units of k = (mol L−1)1−n s−1 where n = overall order.
Q3

A reaction is first order. Its half-life is:

First order: t½ = ln2/k = 0.693/k β€” constant, independent of [A]. This is the unique property of first-order kinetics exploited in radioactive dating. Each successive half-life is identical.
Q4

The integrated rate law for a second-order reaction (rate = k[A]2) is:

1/[A] = 1/[A]0 + kt. Plotting 1/[A] vs t gives a straight line with slope = k. Half-life = 1/(k[A]0) β€” depends on initial concentration.
Q5

The Arrhenius equation predicts that a plot of ln k vs 1/T gives:

ln k = ln A − Ea/RT. ln k = ln A − (Ea/R)(1/T). Comparing y = c + mx: slope = −Ea/R (negative). Higher T (smaller 1/T) gives larger k (larger ln k). Ea = −slope × R.
Q6

In a reaction mechanism, the rate-determining step is:

The slowest step is the bottleneck. The overall rate cannot be faster than the RDS. The rate law for the overall reaction reflects the RDS (and any pre-equilibria before it).
Q7

A reaction is zero order with respect to A. If [A] doubles:

Zero order: rate = k (constant). Changing [A] has no effect on rate. Example: enzyme-catalysed reaction at saturation: all active sites occupied regardless of substrate concentration. Half-life decreases as reaction proceeds.
Q8

Which graph indicates a first-order reaction?

ln[A] vs t linear (slope = −k) indicates first order. From integrated form: ln[A] = ln[A]0 − kt. [A] vs t linear indicates zero order; 1/[A] vs t linear indicates second order.
Q9

For a first-order reaction with k = 0.100 s−1, after 3 half-lives the fraction of reactant remaining is:

After n half-lives, fraction remaining = (1/2)n. After 3 half-lives: (1/2)3 = 1/8. So 87.5% has reacted. This applies to all first-order reactions (including radioactive decay).
Q10

The pre-exponential factor A in the Arrhenius equation represents:

A = collision frequency × orientation (steric) factor. It is the rate constant if Ea = 0 (every collision leads to reaction). A is approximately temperature-independent (varies slowly with T compared to the exponential factor). Units same as k.
Q11

An overall reaction is second order. The proposed mechanism has a first-order RDS. This mechanism is:

Inconsistent. A first-order RDS gives rate = k[A] (first order overall). If experiment shows second order, this mechanism cannot explain it. A different mechanism (second-order RDS or a fast pre-equilibrium giving an intermediate) must be proposed.
Q12

In the iodine clock reaction, the time t measured corresponds to:

S2O32− scavenges I2 until exhausted. At that moment, I2 accumulates → blue colour. Smaller t → faster rate. Rate ∝ 1/t. t is inversely proportional to the rate of the I2-producing RDS.
Q13

As a second-order reaction proceeds, the successive half-lives:

Second order: t½ = 1/(k[A]0). As [A] decreases with time, t½ increases. The reaction slows down dramatically. Contrast: first order has constant t½; zero order has decreasing t½.
Q14

Which of the following is an example of a molecularity statement (not order)?

Molecularity = number of species in a single elementary step. "Bimolecular" means 2 molecules collide in that step. Order is determined experimentally for overall reactions. Molecularity is a theoretical concept for elementary steps only.
Q15

A catalyst increases rate constant k by:

Catalyst provides pathway with lower Ea. In k = Ae−Ea/RT: smaller Ea → larger exponent magnitude shrinks → e−Ea/RT increases → k increases. A may also change (different mechanism), but lowering Ea is the primary effect.
Q16

For the reaction 2NO(g) + O2(g) → 2NO2(g), observed rate = k[NO]2[O2]. A proposed mechanism has Step 1 (fast): 2NO ↔ N2O2. Step 2 (slow): N2O2 + O2 → 2NO2. Derive the rate law from this mechanism.

RDS (step 2): rate = k2[N2O2][O2]. From fast equilibrium (step 1): K = [N2O2]/[NO]2 → [N2O2] = K[NO]2. Substitute: rate = k2K[NO]2[O2] = k[NO]2[O2]. Consistent with experiment! Mechanism is plausible.
Q17

The units of the rate constant for a zero-order reaction are:

Zero order: rate = k. Rate has units mol L−1 s−1; k must have same units to make equation work: mol L−1 s−1. Using formula (mol L−1)1−n s−1 with n=0: mol1 L−1 s−1 βœ“
Q18

Colorimetry is used to follow reactions because:

Beer-Lambert: A = εcl. Absorbance proportional to concentration → by measuring absorbance vs time, concentration-time profile is obtained. Only works for coloured species. Example: bromine (brown), permanganate (purple), NO2 (brown gas).
Q19

If Ea = 50 kJ/mol, which value of T gives a larger rate constant?

100°C (373 K). Higher T → larger e−Ea/RT (as −Ea/RT becomes less negative) → larger k. Rate constants always increase with temperature for reactions with positive Ea (virtually all ordinary reactions).
Q20

An intermediate in a reaction mechanism:

Intermediate: produced and consumed within the mechanism; not in overall equation; appears as a valley (local minimum) on the energy profile. Transition state is a peak (maximum), cannot be isolated. Intermediates can sometimes be detected spectroscopically.
Q21

Which statement correctly compares SN1 and SN2 mechanisms?

SN1: ionisation RDS (rate = k[RX], first order, carbocation intermediate, favoured by tertiary halides). SN2: concerted bimolecular mechanism (rate = k[RX][Nu], second order, inversion of configuration, favoured by primary halides). These are textbook examples of rate laws reflecting mechanisms.
Q22

The method of initial rates works by:

At t=0: concentrations are known and products haven't built up (no complications from reverse reaction). By changing [A] while keeping [B], [C] constant, the effect of [A] on rate is isolated. Comparing rates from 2+ experiments gives order in each reactant.
Q23

A reaction has rate = k[A][B]. If [A] is doubled and [B] is halved, the rate:

New rate = k(2[A])(0.5[B]) = k[A][B] = original rate. Rate unchanged. The two changes cancel exactly. This illustrates why knowing the rate law is important — the combined effect depends on the orders.
Q24

To determine Ea from two temperature-rate constant pairs, you use:

Two-point Arrhenius: ln(k2/k1) = (Ea/R)(1/T1 − 1/T2). Derived by writing Arrhenius at T1 and T2, then subtracting. Rearrange to find Ea. Note: 1/T1 − 1/T2 must use Kelvin temperatures.
Q25

Which statement about the integrated rate law for a zero-order reaction is correct?

Zero order: [A] = [A]0 − kt. [A] decreases linearly with time (straight line on [A] vs t graph). Rate = k = constant (doesn't change as [A] changes). Once [A] hits zero, reaction stops. Examples: surface-limited reactions; enzyme kinetics at substrate saturation.

Unit 14 Quiz — Rate Laws (25 Questions)

Select one answer each
Q1

The rate equation for A + B β†’ products is rate = k[A]Β²[B]. The overall order is:

Overall order = sum of all exponents = 2 + 1 = 3 (third order overall).
Q2

The units of k for a second-order reaction are:

Rate = k[A]Β². Units: mol dm⁻³ s⁻¹ = k Γ— (mol dm⁻³)Β². k = mol⁻¹ dmΒ³ s⁻¹.
Q3

A reaction is zero order with respect to a reactant when:

Rate = k[A]⁰ = k. Rate does not change even if [A] changes. Often seen when a surface or enzyme is saturated.
Q4

The half-life of a first-order reaction is:

For first-order: [A] = [A]β‚€e^(-kt). At t₁/β‚‚: [A] = [A]β‚€/2 β†’ t₁/β‚‚ = ln2/k. Same half-life regardless of starting [A].
Q5

The Arrhenius equation is:

k = Ae^(-Ea/RT). A = pre-exponential (frequency) factor; Ea = activation energy; R = 8.314 J mol⁻¹ K⁻¹; T in K.
Q6

A graph of ln k vs 1/T (Arrhenius plot) gives a straight line with slope:

ln k = ln A – Ea/RT. Plotting ln k vs 1/T: slope = –Ea/R. Ea calculated from slope: Ea = –slope Γ— R.
Q7

Doubling [A] doubles the rate. The order with respect to A is:

Rate ∝ [A]ⁿ. If [A] doubles and rate doubles: 2 = 2ⁿ β†’ n = 1. First order with respect to A.
Q8

Doubling [A] quadruples the rate. The order with respect to A is:

4 = 2ⁿ β†’ n = 2. Rate ∝ [A]Β². Second order with respect to A.
Q9

The integrated rate law for a first-order reaction is:

For A β†’ products, first order: d[A]/dt = –k[A]. Integration gives ln[A] = ln[A]β‚€ – kt (or [A] = [A]β‚€e^(–kt)).
Q10

A first-order reaction has t₁/β‚‚ = 10 min. After 30 min, the fraction remaining is:

30 min = 3 half-lives. After each: Β½ β†’ ΒΌ β†’ β…› remains. (1/2)Β³ = 1/8.
Q11

The pre-exponential factor A in the Arrhenius equation represents:

A = Z Γ— p (collision frequency Γ— probability of correct orientation). It is approximately constant with temperature.
Q12

A rate-determining step (RDS) in a mechanism is:

The RDS is the bottleneck. The rate law reflects the concentration of species in the RDS and all steps before it.
Q13

If the mechanism is: Step 1 (slow): A + B β†’ C; Step 2 (fast): C + D β†’ E. The rate law is:

Rate determined by slow step: rate = k₁[A][B]. Fast step doesn't affect overall rate.
Q14

Radioactive decay is an example of a ___-order reaction.

Rate = λN (∝ number of nuclei). This is first order: constant half-life, exponential decay, independent of chemical environment.
Q15

The relationship between half-life and initial concentration for a second-order reaction is:

For second order: t₁/β‚‚ = 1/(k[A]β‚€). As [A]β‚€ decreases, half-life increases (unlike first order where t₁/β‚‚ is constant).
Q16

A catalyst increases k in the Arrhenius equation by:

k = Ae^(-Ea/RT). Lower Ea β†’ less negative exponent β†’ larger exponential β†’ larger k β†’ faster reaction rate.
Q17

The order of reaction must be determined:

Stoichiometric coefficients β‰  rate law orders (except for elementary reactions). Orders must be found from experiments.
Q18

Initial rate experiments determine order by:

Initial rate removes complications from changing concentrations. Comparing initial rates for different [A] gives the order in A.
Q19

Using ln(kβ‚‚/k₁) = (Ea/R)(1/T₁ – 1/Tβ‚‚), if Ea = 50 kJ/mol and T increases from 300 to 310 K, k increases by approximately:

A 10 K rise with Ea β‰ˆ 50-60 kJ/mol approximately doubles the rate β€” consistent with the Q₁₀ = 2 approximation.
Q20

A plot of [A] vs t gives a straight line for a:

Zero order: rate = k (constant), so d[A]/dt = –k β†’ [A] = [A]β‚€ – kt. Linear [A] vs t with slope = –k.
Q21

A plot of ln[A] vs t gives a straight line for a:

First order: ln[A] = ln[A]β‚€ – kt. Linear ln[A] vs t with slope = –k. This is the test for first-order kinetics.
Q22

The steady-state approximation assumes that:

For short-lived intermediates: d[I]/dt β‰ˆ 0. This simplifies derivation of the overall rate law for complex mechanisms.
Q23

Pseudo-first-order conditions are created by:

If rate = k[A][B] and [B] >> [A]: rate β‰ˆ k'[A] where k' = k[B]. Observed as first order in A. Used to measure rate constants.
Q24

The temperature dependence of rate is stronger for reactions with:

d(lnk)/dT = Ea/RTΒ². Larger Ea β†’ larger sensitivity to temperature. High Ea reactions have strongly temperature-dependent rates.
Q25

Carbon-14 dating works because ¹⁴C decays by first-order kinetics with t₁/β‚‚ = 5730 years. An artefact with 25% of original ¹⁴C is:

After 1 t₁/β‚‚: 50%. After 2 t₁/β‚‚: 25%. 2 Γ— 5730 = 11460 years old.
πŸ“ Go to Unit Test β†’
πŸ“

Unit Test — 50 marks

Section A

30 marks
Q1 [6 marks]

The following data were collected for the reaction A + B → C at 25°C:
Exp 1: [A]=0.10 mol/L, [B]=0.10 mol/L, rate=6.0×10−4 mol/L/s
Exp 2: [A]=0.20 mol/L, [B]=0.10 mol/L, rate=2.4×10−3 mol/L/s
Exp 3: [A]=0.10 mol/L, [B]=0.30 mol/L, rate=6.0×10−4 mol/L/s
(a) Determine the order with respect to A and B. [3] (b) Write the rate equation and calculate k with units. [3]

(a) Order in A: Exp2/Exp1: (0.20/0.10)n = 2.4×10−3/6.0×10−4 = 4; 2n=4; n=2 (second order in A). Order in B: Exp3/Exp1: [B] triples but rate unchanged; zero order in B. (b) Rate equation: rate = k[A]2. k = rate/[A]2 = 6.0×10−4/(0.10)2 = 6.0×10−4/0.010 = 0.060 mol−1 L s−1.
Q2 [6 marks]

A first-order reaction has [A]0 = 2.00 mol/L and k = 5.0×10−3 s−1. (a) Calculate the half-life. [2] (b) Calculate [A] after 5 minutes. [2] (c) How long does it take for [A] to reach 0.25 mol/L? [2]

(a) t½ = 0.693/k = 0.693/(5.0×10−3) = 138.6 s (about 2.3 min). (b) t = 5 min = 300 s. ln[A] = ln(2.00) − 5.0×10−3×300 = 0.693 − 1.500 = −0.807; [A] = e−0.807 = 0.446 mol/L. (c) ln(0.25/2.00) = −kt; ln(0.125) = −2.079; t = 2.079/(5.0×10−3) = 416 s (3 half-lives: 0.25 = 2.00/8 = 2.00/(23) → 3×138.6 = 415.8 s βœ“).
Q3 [6 marks]

The rate constants for a reaction were measured at different temperatures: k = 3.00×10−5 s−1 at 300 K and k = 2.20×10−4 s−1 at 330 K. (a) Calculate Ea. [4] (b) Calculate the rate constant at 315 K. [2] (R = 8.314 J mol−1 K−1)

(a) ln(k2/k1) = (Ea/R)(1/T1 − 1/T2). ln(2.20×10−4/3.00×10−5) = ln(7.33) = 1.992. 1/300 − 1/330 = 3.333×10−3 − 3.030×10−3 = 3.03×10−4 K−1. Ea = 1.992 × 8.314 / 3.03×10−4 = 54,700 J/mol = 54.7 kJ/mol. (b) At 315 K: ln(k/3.00×10−5) = (54700/8.314)(1/300 − 1/315) = 6580 × 1.587×10−4 = 1.044; k = 3.00×10−5 × e1.044 = 3.00×10−5 × 2.84 = 8.52×10−5 s−1.
Q4 [6 marks]

Explain how to distinguish between zero, first, and second order reactions using: (a) concentration-time graphs [3]; (b) half-life analysis [3].

(a) Graphs: Plot [A] vs t, ln[A] vs t, and 1/[A] vs t simultaneously. Zero order: [A] vs t is linear (constant slope = −k). First order: ln[A] vs t is linear (slope = −k). Second order: 1/[A] vs t is linear (slope = +k). Only one plot will be straight; that determines the order. (b) Half-life: measure successive half-lives (time for [A] to halve each time). Zero order: successive t½ decrease (t½ = [A]0/2k; smaller [A]0 each half-life). First order: all t½ are identical (t½ = 0.693/k, independent of concentration). Second order: successive t½ increase (t½ = 1/(k[A]0); smaller [A]0 → larger t½).
Q5 [6 marks]

For the decomposition of H2O2: 2H2O2 → 2H2O + O2, the following mechanism is proposed:
Step 1: H2O2 + I− → H2O + IO− (slow)
Step 2: H2O2 + IO− → H2O + O2 + I− (fast)
(a) Identify the catalyst and the intermediate. [2] (b) Write the rate law predicted by this mechanism. [2] (c) What type of catalysis is this? [2]

(a) Catalyst: I− (consumed in step 1, regenerated in step 2 — not in overall equation). Intermediate: IO− (produced in step 1, consumed in step 2 — not in overall equation). (b) RDS is step 1 (slow): rate = k1[H2O2][I−]. Both reactants in RDS are original reactants (no intermediate in rate law). Rate law: rate = k[H2O2][I−] (first order in each, second order overall). (c) I− is in same phase as H2O2 (both aqueous) → homogeneous catalysis.

Section B

20 marks
Q6 [10 marks]

(a) Derive the integrated rate law for a first-order reaction from the differential form rate = −d[A]/dt = k[A]. Hence show that t½ = ln2/k. [5] (b) A radioactive isotope (first-order decay) has a half-life of 8.1 days. A sample initially contains 5.00 g. Calculate: (i) k; (ii) mass remaining after 30 days; (iii) time for the mass to fall to 0.10 g. [5]

(a) Derivation: rate = −d[A]/dt = k[A]. Separating variables: d[A]/[A] = −k dt. Integrating both sides: ∫d[A]/[A] = −k∫dt; ln[A] = −kt + C. At t=0: ln[A]0 = C. Therefore: ln[A] = ln[A]0 − kt, or ln([A]/[A]0) = −kt, or [A] = [A]0e−kt. For half-life: [A] = [A]0/2; ln(1/2) = −kt½; −ln2 = −kt½; t½ = ln2/k = 0.693/k. ✓
(b)(i) k = ln2/t½ = 0.693/8.1 = 0.0856 day−1. (ii) After 30 days: m = 5.00 × e−0.0856×30 = 5.00 × e−2.568 = 5.00 × 0.0768 = 0.384 g. (iii) ln(0.10/5.00) = −0.0856t; ln(0.020) = −3.912; t = 3.912/0.0856 = 45.7 days (about 5.6 half-lives).
Q7 [10 marks]

Discuss the significance of the Arrhenius equation in chemistry and industry. Include: (a) how it accounts for the temperature-dependence of rate; (b) how Ea and A affect reaction rate; (c) how catalysts can be understood through the Arrhenius framework; (d) one industrial example where Arrhenius considerations determine process conditions. [10]

(a) k = Ae−Ea/RT. As T increases, the exponent −Ea/RT becomes less negative → e−Ea/RT increases exponentially → k increases rapidly. This accounts for the well-known observation that reactions speed up dramatically with temperature. The exponential dependence means even a small T increase can double or triple k (e.g. for Ea=50 kJ/mol, 10 K rise near 300 K increases k by factor ~2.2). [2]
(b) Ea: reactions with large Ea are very temperature-sensitive (steep Arrhenius plot) and slow at low T. Reactions with small Ea (e.g. ionic reactions) are fast even at low T and less sensitive to T changes. A: reflects how often molecules collide in the right orientation. High A = many collisions with correct geometry. Even with low Ea, if A is very small (e.g. requires very precise alignment), rate can be slow. A is approximately constant with T (varies as T1/2 only). [3]
(c) Catalyst: provides alternative mechanism with lower Ea'. In Arrhenius: k' = Ae−Ea'/RT. Since Ea' < Ea: k' > k at same T. The ratio k'/k = e(Ea−Ea')/RT. For Ea−Ea' = 30 kJ/mol at 300 K: k'/k = e12 ≈ 163,000. A catalyst can increase rate by many orders of magnitude. Also, since reaction can proceed faster at lower T, side reactions may be avoided. [3]
(d) Haber process: N2+3H2↔2NH3. Ea for N≡N bond breaking is enormous (~300+ kJ/mol without catalyst). Without catalyst: need very high T to get acceptable k → but high T shifts equilibrium to reactants (exothermic reaction). Fe catalyst lowers Ea dramatically → acceptable k at 400-500°C where equilibrium yield is ~15-25%. Promoters (Al2O3, K2O) increase A by maximising surface area and improving active site geometry. The temperature choice (400-500°C) is a deliberate compromise between kinetics (Arrhenius: need enough T for k) and thermodynamics (Le Chatelier: lower T favours NH3). [2]
← Unit 13: Factors Affecting RateUnit 15: Radioactivity →

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