Unit 13 · Physical Chemistry

Factors Affecting Rate of Reactions

Collision theory, factors affecting rate, Maxwell-Boltzmann distribution, activation energy, and catalysis.

13.1

Collision Theory

Collision TheoryA chemical reaction occurs when reactant particles collide with: (1) sufficient energy ≥ activation energy Eᵣ and (2) correct orientation. Rate depends on frequency of effective collisions.

Activation Energy

Eᵣ = minimum energy required for a collision to result in reaction. Maxwell-Boltzmann distribution: area under curve to RIGHT of Eᵣ = fraction of molecules with enough energy. Rate = Z × f = collision frequency × fraction with E ≥ Eᵣ f = e^(−Eᵣ/RT) (Boltzmann factor)
13.2

Factors Affecting Rate

FactorEffect on RateReason (Collision Theory)
Concentration (solutions)Increase → Rate increasesMore particles per unit volume → more frequent collisions
Pressure (gases)Increase → Rate increasesCompresses gas → higher particle density → more collisions (equivalent to concentration)
Surface area (solids)Increase → Rate increasesMore surface exposed → more collision sites for reactant particles
TemperatureIncrease → Rate increases significantlyParticles move faster (more KE): more frequent AND more energetic collisions; more molecules exceed Eᵣ
CatalystRate increasesProvides alternative pathway with lower Eᵣ; more molecules have E ≥ lower Eᵣ
Light (photochemical)Rate increasesPhotons provide energy to initiate radical reactions (e.g. halogenation of alkanes)
13.3

Effect of Temperature in Detail

Maxwell-Boltzmann Distribution

At higher temperature: distribution shifts to higher energies, peak lowers and broadens. The area under the curve to the right of Eᵣ (fraction of molecules with E ≥ Eᵣ) increases dramatically. Even a small temperature rise (e.g. 10 K) roughly doubles the rate for many reactions.

Rule of thumb: rate roughly doubles for every 10°C rise (for many reactions near room temperature). Exact relationship given by Arrhenius equation (Unit 14): k = Ae^(−Eᵣ/RT) ln k = ln A − Eᵣ/RT

Effect on Maxwell-Boltzmann Graph

At T₂ > T₁: the curve peak shifts right and lowers (area stays constant = same number of molecules); more molecules are to the right of Eᵣ. The shaded area (molecules with E ≥ Eᵣ) increases greatly even for a small temperature increase.

13.4

Catalysis

CatalystA substance that increases the rate of a reaction without being consumed. It provides an alternative reaction pathway with a lower activation energy.
TypeDescriptionExamples
HomogeneousSame phase as reactantsH⁺ in ester hydrolysis; Fe²⁺/Fe³⁺ in persulfate reaction; NO in SO₂→SO₃ (lead chamber)
HeterogeneousDifferent phase (usually solid catalyst, gas/liquid reactants)Fe in Haber process; V₂O⁵ in Contact process; Pt in catalytic converter; Ni in hydrogenation
Biological (enzymes)Protein catalysts; highly specific; operate at low TAmylase (starch→sugar), lactase, catalase (H₂O₂→H₂O+O₂)
AutocatalyticProduct catalyses its own formationMnO₄²⁻ in KMnO₄/oxalic acid; Ce³⁺ in oscillating reactions

Heterogeneous Catalysis: Mechanism

Steps: (1) Adsorption of reactants onto catalyst surface (chemisorption weakens bonds) (2) Surface reaction (lower Eᵣ pathway) (3) Desorption of products from surface (4) Products diffuse away; surface available again Catalyst is NOT consumed overall, but may be slowly poisoned by impurities (catalyst poisoning). Example: Pt catalyst in catalytic converter poisoned by lead → unleaded petrol essential.
13.5

Energy Profile Diagrams

Reading Energy Diagrams

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For: A + B → C + D Energy profile shows: Reactants at starting energy level Products at final energy level Transition state at the peak (highest energy point) = activated complex Eᵣ(forward) = peak − reactant energy Eᵣ(reverse) = peak − product energy ΔH = product energy − reactant energy Catalyst: lowers the peak (reduces Eᵣ(forward) AND Eᵣ(reverse) by same amount) Does NOT change ΔH (thermodynamics unchanged) May provide a different-shaped profile with an intermediate
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Exercises

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Quiz

Unit 13: Factors Affecting Rate

25 Qs
Q1

Collision theory states that a reaction occurs when particles collide with:

E ≥ Eᵣ AND correct orientation. Both conditions must be met. A head-on collision with E < Eᵣ does not result in reaction. Even energetic collisions may fail if orientation is wrong for bond formation.
Q2

Increasing pressure in a gas-phase reaction increases rate because:

Compression increases particle density (higher concentration of gas molecules per unit volume) → more frequent collisions → higher rate. At constant temperature, gas molecule speed doesn't change with pressure.
Q3

Why does finely powdered calcium carbonate react faster with HCl than large lumps?

Greater surface area: more of the solid is exposed to acid. Rate ↑ proportionally to surface area. Same total amount of CaCO₃ reacts eventually, but faster. Used in industrial processes (ore roasting, combustion).
Q4

On a Maxwell-Boltzmann distribution at temperature T₂ > T₁, compared to T₁:

At higher T: peak shifts right (higher energy) and lowers (same area = same number of molecules). The tail extends further → much larger fraction of molecules above Eᵣ → dramatic rate increase.
Q5

A catalyst increases rate by:

Lower Eᵣ alternative pathway. More molecules have E ≥ lower Eᵣ → greater fraction of effective collisions. ΔH and equilibrium position are NOT changed by catalyst (thermodynamics unchanged).
Q6

The transition state in an energy profile diagram is:

The transition state (activated complex) is at the maximum energy point. It is a transient species that cannot be isolated. Eᵣ = energy to reach transition state from reactants.
Q7

Platinum in a catalytic converter is deactivated by:

Lead (Pb) atoms from leaded petrol adsorb irreversibly onto Pt surface, blocking active sites. Catalyst poisoning. Hence all modern car engines use unleaded petrol to preserve the catalytic converter.
Q8

Which statement about a catalyst is correct?

Catalyst lowers both Eᵣ(forward) and Eᵣ(reverse) by the same amount (the peak is lowered; start and end points unchanged). ΔH unchanged. Equilibrium position unchanged. Catalyst is regenerated (not consumed).
Q9

An enzyme is described as highly specific because:

Lock-and-key (or induced fit): enzyme active site has a precise 3D shape complementary to one substrate (or class). Only that substrate binds with the right geometry for catalysis. This specificity allows metabolic pathways to operate without interference.
Q10

The rule of thumb that rate doubles for every 10°C rise is because:

The exponential Boltzmann factor e^(−Ea/RT) is very sensitive to T. For typical Ea (~50–80 kJ/mol) near 25°C, a 10 K rise roughly doubles this factor → doubles rate. (Collision frequency increases only ~2% per 10 K by √T dependence.)
Q11

Which factor does NOT increase the rate of a solid-liquid reaction?

Increasing mass without changing surface area (e.g. using a bigger lump of same dimensions): more solid is present but no more exposed surface → rate unchanged. All other factors increase rate.
Q12

Autocatalysis is when:

Product catalyses the reaction. Example: MnO₄²⁻ produced in KMnO₄/C₂O₄²⁻ reaction catalyses further reaction → S-shaped rate-time graph (slow start, then accelerates). Also: Fe²⁺ in certain reactions.
Q13

Vanadium pentoxide (V₂O⁵) in the Contact process acts as a:

V₂O⁵ is solid; SO₂ and O₂ are gases → heterogeneous catalyst. Mechanism: V⁵⁺ oxidises SO₂ to SO₃ while being reduced to V⁴⁺; V⁴⁺ reoxidised to V⁵⁺ by O₂. Net: catalyst regenerated.
Q14

A reaction profile shows two maxima (two peaks) with a valley between them. This means:

Two peaks = two transition states = two elementary steps. The valley = a reaction intermediate (relatively stable, unlike the transition state). This is common in SN1 reactions (carbocation intermediate) and enzyme-catalysed reactions.
Q15

The activation energy of a reaction can be determined experimentally from:

Using Arrhenius: ln k = ln A − Ea/RT. Plot ln k vs 1/T → straight line with slope = −Ea/R. Gradient × (−R) = Ea.
Q16

Light initiates some chemical reactions (photochemical reactions) because:

Photons (hν) have enough energy to break weak bonds homolytically, producing free radicals. Example: Cl₂ + hν → 2Cl•. These highly reactive radicals initiate chain reactions. Photographic film, vitamin D synthesis, and ozone layer reactions are photochemical.
Q17

Iron acts as a heterogeneous catalyst in the Haber process. The mechanism involves:

Adsorption mechanism: N₂ adsorbs on Fe surface (chemisorption weakens N≢N triple bond), H₂ also adsorbs, N and H react on surface forming NH₃, which then desorbs. Fe surface provides lower Eᵣ pathway.
Q18

Which of these reactions is NOT affected by changing the pressure?

Aqueous solution reactions: liquids are essentially incompressible → changing pressure has negligible effect on concentration of dissolved species → negligible effect on rate. Gas-phase reactions are strongly affected by pressure.
Q19

The fraction of molecules with energy ≥ Ea is given by the Boltzmann factor:

f = e^(−Ea/RT). This fraction is always between 0 and 1. Increases as T increases or Ea decreases. At high Ea or low T, f is very small (slow reaction). Forms the basis of the Arrhenius equation.
Q20

Catalyst poisoning in heterogeneous catalysis occurs when:

Irreversible adsorption of impurity onto active sites prevents reactant adsorption → catalyst deactivated. Examples: Pb poisons Pt converter; CO poisons Ni hydrogenation catalyst; S-compounds poison many metal catalysts. Regeneration may require chemical treatment or replacement.
Q21

An inhibitor is the opposite of a catalyst because:

Inhibitors slow reactions: some compete with reactants for active sites (competitive inhibition); some bind elsewhere and distort active site (non-competitive inhibition); some react with intermediates or radicals to terminate chains. Used in food preservation (antioxidants), medicine (enzyme inhibitors as drugs), and corrosion prevention.
Q22

The energy profile of a catalysed reaction compared to uncatalysed shows:

Catalyst lowers the transition state energy (peak). Reactant and product energies (and thus ΔH) are unchanged. The difference Eᵣ(fwd) is smaller → more molecules can reach transition state → faster rate.
Q23

In the Contact process: 2SO₂ + O₂ ⇌ 2SO₃. The catalyst V₂O⁵ works via:

V⁵⁺ ↔ V⁴⁺ cycle: V⁵⁺ oxidises SO₂ to SO₃ (reduced to V⁴⁺); V⁴⁺ reoxidised by O₂ back to V⁵⁺. This is homogeneous catalysis in the gas-solid system (both gases). Net: catalyst regenerated, SO₂+½O₂→SO₃.
Q24

Temperature increases both collision frequency and the Boltzmann factor. Which effect is larger for most reactions?

The Boltzmann factor e^(−Ea/RT) is exponentially sensitive to T → dominates. Collision frequency increases only as √T (very slowly). For most reactions, the dramatic rate increase with T is almost entirely due to the exponential increase in the fraction of molecules exceeding Eᵣ.
Q25

Surface area is important for heterogeneous reactions. Which industrial process relies most critically on maximising catalyst surface area?

Heterogeneous catalysis: reaction occurs at catalyst surface. Catalysts are designed with maximum surface area per gram (porous, finely divided): Fe with alumina promoter in Haber; Pt gauze in Ostwald; Pt/Pd/Rh as microparticles on ceramic honeycomb in catalytic converter.
Q26

A student adds more solid KMnO₄ to a reaction mixture but does not change its surface area (large lumps). What happens to the rate?

For a solid reactant, rate depends on surface area, not mass. Adding more of the same large lumps does not increase exposed surface. Rate is unchanged. Only grinding to finer powder or dissolving (if soluble) would increase rate.

Unit 13 Quiz — Reaction Rates (25 Questions)

Select one answer each
Q1

Reaction rate is defined as:

Rate = Δ[concentration]/Δtime (mol dm⁻³ s⁻¹). Rate of disappearance of reactant or appearance of product.
Q2

According to collision theory, a reaction occurs when:

Both conditions needed: kinetic energy ≥ activation energy AND molecules properly aligned to form transition state.
Q3

Increasing temperature increases reaction rate because:

Higher T: the Maxwell-Boltzmann distribution broadens and shifts right — larger fraction of molecules exceed Ea.
Q4

The activation energy (Ea) is:

Ea = energy difference between reactants and the transition state (activated complex). Lower Ea → faster reaction.
Q5

A catalyst increases reaction rate by:

Catalyst forms intermediates with lower Ea. More molecules have sufficient energy → greater fraction of effective collisions.
Q6

Increasing concentration increases rate because:

Higher concentration → greater collision frequency → more effective collisions per second → faster rate.
Q7

The Maxwell-Boltzmann distribution curve shows that:

The curve peaks at the most probable energy. Adding a catalyst shifts the Ea threshold left — more molecules exceed it.
Q8

Increasing surface area of a solid reactant increases rate because:

Crushing or powdering a solid increases surface area dramatically → more collision sites → faster reaction.
Q9

A catalyst is not consumed in the overall reaction because:

Catalyst lowers Ea by providing a new pathway and forming intermediates, but is regenerated as products form.
Q10

The effect of pressure on gaseous reaction rate is similar to concentration because:

P ∝ n/V at constant T. Higher pressure = more gas molecules per unit volume = higher effective concentration.
Q11

A homogeneous catalyst is in:

Homogeneous: same phase (e.g. acid catalysing ester hydrolysis in solution). Heterogeneous: different phase (solid Pt in gas reaction).
Q12

The Haber process uses an iron catalyst which is an example of:

Gases adsorb onto Fe surface → bonds weaken → reaction occurs on surface → products desorb. Different phases = heterogeneous.
Q13

An inhibitor (negative catalyst) works by:

Inhibitors can poison catalysts (occupy active sites) or interrupt chain reactions by scavenging free radicals.
Q14

Enzymes are biological catalysts that are highly specific because:

Only the correct substrate fits the enzyme's active site. High specificity due to 3D shape and H-bonding of active site.
Q15

The rate of a reaction generally doubles for every 10°C rise in temperature. This means:

Average speed ∝ √T — increases by only ~1.7% per 10°C. But fraction above Ea can double → reaction doubles.
Q16

Light increases the rate of photochemical reactions by:

Photons absorbed by molecules promote electrons to higher energy states — effectively 'bypassing' the activation energy barrier.
Q17

In the iodine clock reaction, the sudden appearance of blue-black colour indicates:

While Na₂S₂O₃ is present, I₂ is instantly reduced back to I⁻. When S₂O₃²⁻ is exhausted, I₂ builds up → starch → blue-black.
Q18

Acid-base catalysis is common because:

H⁺ catalyses ester hydrolysis by protonating the carbonyl oxygen, making the carbon more electrophilic → easier nucleophilic attack.
Q19

A negative activation energy would mean:

Some multi-step mechanisms have apparent negative Ea — overall rate decreases at higher T (equilibrium shifts unfavourably).
Q20

Transition state theory states that reactants must pass through:

The transition state is the highest-energy, unstable arrangement of atoms as old bonds break and new bonds form simultaneously.
Q21

Autocatalysis occurs when:

Example: MnO₄⁻ oxidation of oxalate — Mn²⁺ formed catalyses the reaction. Characteristic S-shaped rate-time curve.
Q22

Chain reactions involve:

Initiation: creates radicals. Propagation: radicals react and regenerate. Termination: two radicals combine, ending the chain.
Q23

Ozone depletion by CFCs involves catalysis because:

Cl is regenerated → catalytic cycle. One Cl atom can destroy ~100,000 O₃ molecules before termination.
Q24

Temperature coefficient Q₁₀ = 2 means:

Q₁₀ = rate(T+10)/rate(T). Q₁₀ = 2 means a 10°C increase doubles the rate — common approximation for many reactions.
Q25

Measuring rate by colorimetry works when:

Colorimeter measures light absorbance ∝ [coloured species]. Tracking absorbance vs time gives rate without sampling.
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Unit Test — 50 marks

Section A

30 marks
Q1 [6 marks]

Using collision theory, explain the effect of each of the following on the rate of the reaction: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g). (a) Doubling HCl concentration [2]; (b) Halving particle size of Mg [2]; (c) Increasing temperature by 20°C [2].

(a) More HCl molecules per unit volume → collision frequency between HCl and Mg surface increases proportionally → rate approximately doubles. (b) Smaller particles → much greater total surface area of Mg → more collision sites per second → rate increases significantly (approximately quadruples for half particle diameter → 8× surface area). (c) Molecules have higher kinetic energy; distribution broadens (Maxwell-Boltzmann shift); much larger fraction of molecules exceeds Eᵣ; also collision frequency increases slightly. Rate roughly quadruples (double for each 10°C).
Q2 [5 marks]

Describe the Maxwell-Boltzmann distribution and use it to explain why a small temperature increase causes a disproportionately large increase in reaction rate. [5]

M-B distribution: shows the distribution of molecular energies in a gas at a given temperature. x-axis: kinetic energy; y-axis: number of molecules (or fraction). Curve starts at origin, rises to a peak (most probable energy), then falls with a long tail (some molecules have very high energy). Area under entire curve = total number of molecules (constant). At higher T: peak shifts to right and lowers; tail extends further right. Eᵣ is a fixed point on the x-axis. The shaded area to the right of Eᵣ (fraction of molecules with E ≥ Eᵣ) increases dramatically even for small T rise because the tail is exponential. Even a 10 K rise (3.3% of 300 K) can double the number of molecules above Eᵣ for typical Ea values.
Q3 [5 marks]

Compare the mechanism of action of (a) a heterogeneous metal catalyst (e.g. Ni in hydrogenation); (b) a homogeneous catalyst (e.g. H⁺ in esterification); (c) a biological enzyme (e.g. catalase). [5]

(a) Heterogeneous Ni: H₂ and alkene adsorb onto Ni surface (chemisorption weakens C=C and H–H bonds); hydrogenation occurs on surface; product desorbs. Lower Eᵣ because bonds partially weakened before transition state. Easy to separate product from catalyst (filter). (b) Homogeneous H⁺: protonates C=O in esterification, forming oxocarbenium ion intermediate (alternative lower-Eᵣ pathway); H⁺ regenerated. Same phase as reactants; harder to recover catalyst. (c) Enzyme catalase: H₂O₂ binds to active site (lock-and-key); enzyme-substrate complex forms; O₂ and H₂O released; enzyme regenerated. Highly specific; works at 37°C; denatured by heat/pH extremes.
Q4 [4 marks]

Sketch and label energy profile diagrams for: (a) an uncatalysed endothermic reaction and (b) the same reaction with a catalyst. Explain what changes and what stays the same. [4]

(a) Endothermic: reactants at lower energy than products. Curve rises to peak (transition state), then falls to products (at higher energy than reactants). ΔH positive (products higher). Eᵣ(fwd) = large (from reactant to peak). Eᵣ(rev) = small (from product to peak). (b) Catalysed: same reactant and product energy levels (ΔH unchanged). Peak is lower (smaller Eᵣ(fwd) and Eᵣ(rev)). May show two peaks with an intermediate. Changes: Eᵣ decreases. Same: ΔH, reactant energy, product energy.
Q5 [5 marks]

Explain three industrial applications where catalysts are essential. For each: name the catalyst, the reaction, and explain why the catalyst is needed (what happens without it). [5]

1. Haber process: Fe catalyst. N₂+3H₂⇌2NH₃. Without catalyst: N≢N triple bond (945 kJ/mol) has enormous Eᵣ; reaction too slow at any practical temperature. Fe adsorbs N₂, weakening the bond. Economically essential for fertiliser production. 2. Contact process: V₂O⁵ catalyst. 2SO₂+O₂⇌2SO₃. Without catalyst: SO₂+O₂ reaction extremely slow at ambient T. V₂O⁵ cycles V⁵⁺↔V⁴⁺ to provide lower-Eᵣ pathway. Essential for H₂SO₄ production. 3. Catalytic converter: Pt/Pd/Rh. CO+O₂→CO₂; NOx→N₂; hydrocarbons combustion. Without catalyst: these reactions too slow at exhaust temperatures to convert pollutants. Saves millions of lives by reducing air pollution.
Q6 [5 marks]

A student investigates the rate of reaction between sodium thiosulfate and hydrochloric acid: Na₂S₂O₃ + 2HCl → 2NaCl + S + SO₂ + H₂O. Sulfur precipitates and makes the solution cloudy. The student measures time for a cross to disappear through the solution. Describe how this experiment can be used to investigate (a) effect of concentration and (b) effect of temperature. Include controls and how rate is measured. [5]

(a) Concentration: keep T, volume constant. Vary [Na₂S₂O₃]: e.g. 0.05, 0.10, 0.15, 0.20, 0.25 mol/L (dilute with water, keep total volume constant). Measure time t for cross to disappear; rate ∝ 1/t. Plot rate vs [Na₂S₂O₃]. Controls: same volume; same HCl concentration; same temperature (water bath). (b) Temperature: keep concentrations constant. Use water bath to control T; test at 20, 30, 40, 50, 60°C. Allow solution to equilibrate before mixing. Measure t; rate ∝ 1/t. Plot rate vs T or ln(rate) vs 1/T (Arrhenius). Controls: same volume, concentration. Limitation: cross judgment is subjective → use light sensor for consistency.
Q5 [5 marks]

Explain three industrial applications where catalysts are essential. For each: name the catalyst, the reaction, and explain why the catalyst is needed. [5]

1. Haber: Fe, N₂+3H₂→2NH₃. Without catalyst: enormous Eᵣ for N≢N (945 kJ/mol); reaction too slow. 2. Contact: V₂O⁵, 2SO₂+O₂→2SO₃. Without: too slow at practical T. V cycles V⁵/V⁴ for lower Eᵣ route. 3. Catalytic converter: Pt/Pd/Rh. Without: exhaust pollutants (CO, NOx, HC) not converted at exhaust gas temperature. Essential for clean air.

Section B

20 marks
Q7 [10 marks]

(a) Explain using collision theory and Maxwell-Boltzmann distribution why temperature has such a dramatic effect on rate compared to concentration. Include a description of the exponential (Arrhenius) relationship. [5] (b) A reaction has Ea = 60 kJ/mol. Calculate the Boltzmann factor at 25°C and 35°C. Show that the rate approximately doubles. (R = 8.314 J/mol/K) [5]

(a) Concentration: increasing concentration increases collision frequency linearly (e.g. double concentration → double collisions/second → double rate). This is linear. Temperature: (1) slightly increases collision frequency (∝√T, small effect). (2) Much more importantly: shifts Maxwell-Boltzmann distribution to higher energies → exponentially larger fraction of molecules exceed Eᵣ. Fraction = e^(−Ea/RT): this is exponential in 1/T. For Ea=60 kJ/mol at 298 K: exponent = −60000/(8.314×298) = −24.2. A 10 K rise changes exponent to −60000/(8.314×308) = −23.4. Ratio = e^(−23.4)/e^(−24.2) = e^0.8 ≈ 2.2. So rate more than doubles. Concentration would need to double to double rate (linear). Temperature rise of just 3.3% (10/300 K) gives >2× rate increase.
(b) At 25°C (298 K): f₁ = e^(−60000/(8.314×298)) = e^(−24.21) = 3.0×10⁻¹⁰¹. At 35°C (308 K): f₂ = e^(−60000/(8.314×308)) = e^(−23.42) = 6.6×10⁻¹⁰¹. Ratio f₂/f₁ = 6.6/3.0 = 2.2. Rate at 35°C is 2.2× rate at 25°C — confirming the rule of thumb.
Q8 [10 marks]

Evaluate the use of catalysts in modern chemistry and industry. Discuss: (a) economic and environmental benefits of catalytic processes; (b) catalyst deactivation and regeneration; (c) the emerging role of green chemistry and biocatalysis. [10]

(a) Economic: catalysts enable reactions at lower T and P → less energy needed → lower production costs. NH₃ production (Haber) without Fe catalyst would require extreme T to achieve adequate rate → uneconomic. H₂SO₄ production (V₂O⁵): world’s largest chemical by volume — catalyst essential. Environmental: (i) Catalytic converters convert >99% of CO, NOx, HC → prevent air pollution in cities (estimate: save 200,000+ lives/year in Europe). (ii) Selective catalysts reduce by-products → less waste. (iii) Lower temperature reactions → less CO₂ from energy use. (iv) Atom economy: catalytic routes often 100% (no atoms lost to by-products). [3]
(b) Deactivation: (i) Poisoning — irreversible adsorption of impurity (Pb, S, CO); (ii) Sintering — heating causes small particles to fuse → reduced surface area; (iii) Coking — carbon deposits from hydrocarbons block surface. Regeneration: (i) Oxidative regeneration — burn off carbon deposits (catalyst crackers regenerated by burning coke in air); (ii) Chemical washing — remove impurities; (iii) For Pt converters: not regeneratable (replace); (iv) Zeolite catalysts: regenerated by heating. Replacement costs are major industrial expense — research into more robust catalysts ongoing. [4]
(c) Green chemistry biocatalysis: enzymes catalyse reactions at 37°C, pH 7 in water — ideal green conditions (no hazardous solvents, no high T or P). Pharmaceutical industry: enzymes used for enantioselective synthesis (one mirror image only) — conventional catalysts produce racemic mixtures. Examples: lipases in biodiesel production; amylases in bioethanol production; nitrile hydratase in acrylamide production (replaces toxic Cu catalyst at high T). Immobilised enzymes (attached to surface) reusable. Directed evolution (Nobel Prize 2018) engineers enzymes for non-natural reactions. Growing role: enzyme cascade reactions replacing multi-step synthetic routes. [3]
← Unit 12: Electrochemical CellsUnit 14: Rate Laws →

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