chemical equilibria and kinetics

0.0(0)
Studied by 0 people
call kaiCall Kai
learnLearn
examPractice Test
spaced repetitionSpaced Repetition
heart puzzleMatch
flashcardsFlashcards
GameKnowt Play
Card Sorting

1/40

encourage image

There's no tags or description

Looks like no tags are added yet.

Last updated 4:17 AM on 10/10/26
Name
Mastery
Learn
Test
Matching
Spaced
Call with Kai
Chat

No analytics yet

Send a link to your students to track their progress

41 Terms

1
New cards

dynamic equilibria

Reaction doesn't stop when equilibrium is reach

  • Rate of product and reactant formation and reaction is equal

  • So that culmativily, reaction is constant


<p>Reaction doesn't stop when equilibrium is reach</p><ul><li><p><span>Rate of product and reactant formation and reaction is equal</span></p></li><li><p><span>So that culmativily, reaction is constant</span></p></li></ul><p></p>
2
New cards

equilibrium constant

ratio of products: reactants

K = [products]p / [reactants]r

  • Little letter=where big number goes

K constant value at equilibrium

  • Change in concentration will not change K

Size of K indicates extent of the reaction

  • Change in temperature will change K


If K >> 1 equilibrium is product favoured

If K << 1 equilibrium is reactant favoured

<p>ratio of products: reactants</p><p>K = [products]p / [reactants]r</p><ul><li><p><span>Little letter=where big number goes</span></p></li></ul><p>K constant value at equilibrium</p><ul><li><p>Change in concentration will not change K</p></li></ul><p>Size of K indicates extent of the reaction</p><ul><li><p>Change in temperature will change K</p></li></ul><p></p><p>If K &gt;&gt; 1 equilibrium is product favoured</p><p>If K &lt;&lt; 1 equilibrium is reactant favoured</p>
3
New cards

Do not include pure liquids in K

Do not include solid components in K

 conc= amount in given volume

  • If pure, made entirly of that

  • If some taken away, volume decreases => conc is constant

Tehrefore don’t incude

4
New cards

K expression for calculations

when deltarG=0, reaction is at equilibrium

At equilibrium, there are special conditions, so use K

  • G<0=forwards reaction favour

  • G>0=reverse reaction favoured

Q for when not at equilibrium

5
New cards

comparing Q and K

can determie direction of reaction

  • Q < K favours forward reaction, towards products

  • Q > K favours reserves reaction, towards reactants


<p>can determie direction of reaction</p><ul><li><p>Q &lt; K favours forward reaction, towards products</p></li></ul><ul><li><p>Q &gt; K favours reserves reaction, towards reactants</p></li></ul><p></p>
6
New cards

thermodynamics of equilibria

∆𝑟𝐺 = ∆𝑟𝐺𝑜 + 𝑅𝑇 𝑙𝑛 [𝐶]𝑐[𝐷]𝑑 / [𝐴]𝑎[𝐵]𝑏

at equilibrium:

  • ∆rG = 0 & K = Q

  • 0 = ∆𝑟𝐺° + 𝑅𝑇 ln 𝐾

∆𝒓𝑮° = −𝑹𝑻 𝐥𝐧 𝑲

when using in calculations, convert celcius to kelvin. +273.15 for temp

<p>∆𝑟𝐺 = ∆𝑟𝐺𝑜 + 𝑅𝑇 𝑙𝑛 [𝐶]𝑐[𝐷]𝑑 / [𝐴]𝑎[𝐵]𝑏</p><p>at equilibrium:</p><ul><li><p>∆rG = 0 &amp; K = Q</p></li><li><p>0 = ∆𝑟𝐺° + 𝑅𝑇 ln 𝐾</p></li></ul><p>∆𝒓𝑮° = −𝑹𝑻 𝐥𝐧 𝑲 </p><p>when using in  calculations, convert celcius to kelvin. +273.15 for temp</p>
7
New cards

van’t hoff isochore

𝐥𝐧 𝑲 = − ∆𝑯°/ 𝑹𝑻 + ∆𝑺° / 𝑹

tehrefore estimating K at different temperature

  • if we know K1 at T1 we can estimate K2 at T2 (because of the linear relationship

  • n 𝐾2 − ln 𝐾1 = − ∆𝑟𝐻° / 𝑅 (1/ 𝑇2 − 1/ 𝑇1)


<p>𝐥𝐧 𝑲 = − ∆𝑯°/ 𝑹𝑻 + ∆𝑺° / 𝑹</p><p>tehrefore estimating K at different temperature</p><ul><li><p>if we know K1 at T1 we can estimate K2 at T2 (because of the linear relationship</p></li><li><p>n 𝐾2 − ln 𝐾1 = − ∆𝑟𝐻° / 𝑅 (1/ 𝑇2 − 1/ 𝑇1) </p></li></ul><p></p>
8
New cards

acid base chemistry is just equilbira

𝐾𝑎 = [𝐴−][𝐵𝐻+]/ [𝐻𝐴]

  • Ka tells us how much dissociate has occurred

  • Larger Ka stronger the acid

pKa = -log10(Ka)

  • smaller pka= stronger acid


<p><strong>𝐾𝑎 = [𝐴−][𝐵𝐻+]/ [𝐻𝐴]</strong></p><ul><li><p>Ka tells us how much dissociate has occurred</p></li></ul><ul><li><p>Larger Ka stronger the acid</p></li></ul><p><strong>pKa = -log10(Ka)</strong></p><ul><li><p>smaller pka= stronger acid</p></li></ul><p></p>
9
New cards

ph scale

PH=-log [H3O+]

Simple way of expressing concentration of H3O+ ions in a solution. Measure of acidity, but also basicity.

in acid

Identify if a strong or weak acid

• Determined by the value of Ka

Calculate [H3O+]

• If strong acid [H3O+] = [HA]

• If weak acid [H3O+] = √(Ka x [HA])

Convert to pH scale

• pH = -log10[H3O+]

in base

dentify if a strong or weak base

• Determined by the value of Kb

Calculate [OH-]

• If strong acid [OH-] = [base]

• If weak acid [OH-] = √(Kb x [base])

Convert to pH scale

• pOH = -log10[OH-] then

• pH = 14 - pOH


10
New cards

buffers in biology - breathing

Respiration reaction produces CO2

This then reacts with H2O in the blood forming carbonic acid (H2CO3)

Do more exercise, produce more CO2, pushes the reaction forward.

Second equilibrium occurring, forming bicarbonate (HCO3-) and protons (H+)

  • decreasing blood ph

Breathe harder, expels more CO2, pushes the equilibrium backwards

  • increae blood ph


<p>Respiration reaction produces CO2</p><p>This then reacts with H2O in the blood forming carbonic acid (H2CO3)</p><p>Do more exercise, produce more CO2, pushes the reaction forward.</p><p>Second equilibrium occurring, forming bicarbonate (HCO3-) and protons (H+)</p><ul><li><p>decreasing blood ph</p></li></ul><p>Breathe harder, expels more CO2, pushes the equilibrium backwards</p><ul><li><p>increae blood ph</p></li></ul><p></p>
11
New cards

blood as a buffer

Changes in pH controlled through a combination of

chemical and physical processes

Amount of CO2 controlled by respiration

Amount of HCO3- controlled by renal system

Buffering maintains the pH (keeps us alive!!!)

<p>Changes in pH controlled through a combination of</p><p>chemical and physical processes</p><p>Amount of CO2 controlled by respiration</p><p>Amount of HCO3- controlled by renal system</p><p>Buffering maintains the pH (keeps us alive!!!) </p>
12
New cards

what is a buffer

Aqueous solution that is resistant to pH change.

Made of weak acid [HA] & its conjugate base [A-]

Add strong acid,

  • addinsg prtons in, react with conjugate acid, so don’t get build up of protons

Add string base

  • OH reacts with weak base, forms equilibrium reaction forming h2o and conjugate base (ph stays stable


<p>Aqueous solution that is resistant to pH change.</p><p>Made of weak acid [HA] &amp; its conjugate base [A-]</p><p>Add strong acid,</p><ul><li><p><span>addinsg prtons in, react with conjugate acid, so don’t get build up of protons</span></p></li></ul><p>Add string base</p><ul><li><p><span>OH reacts with weak base, forms equilibrium reaction forming h2o and conjugate base (ph stays stable</span></p></li></ul><p></p>
13
New cards

ph of a buffer

dependent on ratio of [A-]/[HA]

  • When [HA] = [A-] pH = pKa

    When [HA] > [A-] (more acid is present)

    pH (of buffer) < pKa - buffer has greater capacity to neutralise base

    When [HA] < [A-] (more base is present)

    pH (of buffer) > pKa and buffer has base capacity to neutralise acid

  • Ka=acid dissociation constant

    The biger the pka the stronger aicd/ph

     

    When doing covid test, must put swab into buffer, as antibodies and nasal mucus have different ph


14
New cards

buffering capacity

The best buffering will occur when the ratio of [HA] to [A-] is almost 1:1.

  • When pH = pKa. Buffers are effective with [HA] to [A-] ranges between ± 10%

Total concentration of the buffer impacts capacity. More mols of buffer available to absorb additional H+/OH- ions.

Although too strong can affect the reaction. By controlling the [HA] to [A-] ratio, we can control the pH of the buffer

15
New cards

relating pH, pKa, concentration ratio - Henderson-Hasselbalch Equation

at equlibrium, HA=A- therefore HA/A=1. log1=0, therefore PH=pka

𝑝𝐻 = 𝑝𝐾𝑎 + 𝑙𝑜𝑔10 ([𝐴−] / [𝐻𝐴])

  • [A-] = conjugate base

    [H+] = protons

    [HA] = weak acid

assumptions

  • Monoprotic acid

  • Salt completely dissociates

  • Ignoring self-ionisation of water (Kw)

Limitations

  • Most valid for pKa 4-9 range


<p>at equlibrium, HA=A- therefore HA/A=1. log1=0, therefore PH=pka</p><p><strong>𝑝𝐻 = 𝑝𝐾𝑎 + 𝑙𝑜𝑔10 ([𝐴−] / [𝐻𝐴])</strong></p><ul><li><p>[A-] = conjugate base</p><p>[H+] = protons</p><p>[HA] = weak acid </p></li></ul><p>assumptions</p><ul><li><p>Monoprotic acid</p></li></ul><ul><li><p>Salt completely dissociates</p></li><li><p>Ignoring self-ionisation of water (Kw)</p></li></ul><p>Limitations</p><ul><li><p>Most valid for pKa 4-9 range</p></li></ul><p></p>
16
New cards

kinetics vs thermodynamics

Depending on activation energy, can get thermodynamically favoured products or kintically favoured products

  • Just because a reaction is spontaneous DOES NOT mean we will see it occur.

  • If a reaction can produce different products (cis vs trans isomers) the thermodynamically most stable MAY NOT be the main product.

  • Understanding the kinetics of a reaction is clearly important


<p>Depending on activation energy, can get thermodynamically favoured products or kintically favoured products</p><ul><li><p>Just because a reaction is spontaneous DOES NOT mean we will see it occur.</p></li><li><p>If a reaction can produce different products (cis vs trans isomers) the thermodynamically most stable MAY NOT be the main product. </p></li><li><p>Understanding the kinetics of a reaction is clearly important</p></li></ul><p></p>
17
New cards

energy landscape

Energy lncscape of molecule=potential energy surface of moelcule

Molescule want to minimise energy, take lowest energy pathway to get from reaxtant to product

  • Can map out potentional energy surfaces of reactions


18
New cards

transition states

  • Very short-lived transient structure

  • Cannot measure the structure by “normal” experimental methods

  • Can determine the barrier height (Ea/∆H‡/∆G‡) by experimental and computational methods (highest energy point on pathway directly connecting A and B)


  •  as move around potential energy surface, take lowest energy pathways.

    When projected into 2d, see curve, with big lump where transition state as reaction progresses (between reactnt and final product)

    • Transition state is evry shortlived

    • Reaction coordinate: transition state is set of specific coordinates that link reactant to product

      • Reaction coordinate= Specific state of molecule at particular time

    • Cannot easily be detected, btu can get info about structure via chemical intuition

     

    Determine height of barrier= determine transition state= determine activation energy


<ul><li><p>Very short-lived transient structure</p></li><li><p>Cannot measure the structure by “normal” experimental methods</p></li><li><p>Can determine the barrier height (Ea/∆H‡/∆G‡) by experimental and computational methods (highest energy point on pathway directly connecting A and B)</p></li></ul><p></p><ul><li><p><span>&nbsp;</span>as move around potential energy surface, take lowest energy pathways.</p><p>When projected into 2d, see curve, with big lump where transition state as reaction progresses (between reactnt and final product)</p><ul><li><p><span>Transition state is evry shortlived</span></p></li><li><p><span>Reaction coordinate: transition state is set of specific coordinates that link reactant to product</span></p><ul><li><p><span>Reaction coordinate= Specific state of molecule at particular time</span></p></li></ul></li><li><p><span>Cannot easily be detected, btu can get info about structure via chemical intuition</span></p></li></ul><p>&nbsp;</p><p>Determine height of barrier= determine transition state= determine activation energy</p></li></ul><p></p>
19
New cards

activation energy

Rate of reaction is inversely proportional to activation energy

  • Reactions can be elementary (single-step) or complex (multi-step

  • The step with the highest barrier is known as the rate limiting step

Slowest step of the reaction, controls the rates

  • activation energy= energy required to activate reaction

Study kinitic to determine whetehr have single step reacton or multistep reaction with mulitple intermediates and transsition states

  • Can use this info to improve reaction,eg can stop one of byproducts reacting to ensure get one enantiomer, lower activation energy etc

 

One of barrier is bigger than the others, is slowest step= rate limiting step

20
New cards

three step mechanism of spontaneous reaction

  • determine rate limiting step

  • reaction energy diagram

  • consider brond breakage/formation, to preditc transition state structures


<ul><li><p>determine rate limiting step</p></li><li><p>reaction energy diagram</p></li><li><p>consider brond breakage/formation, to preditc transition state structures</p></li></ul><p></p>
21
New cards

studying kinetics in a lab - rate of reaction

  rate of reaction=change in conc as funciton of time

Products increases, reagents decrease (so negative sign)

 Interested in change in concentration of species during a reaction.

𝑟𝑎𝑡𝑒 = 𝑑[𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠] / 𝑑𝑡

𝑟𝑎𝑡𝑒 = − 𝑑[𝑅𝑒𝑎𝑔𝑒𝑛𝑡𝑠] / 𝑑𝑡

  • d=reaction differential= change of concentration

    Rate=change in conc/change in time

Any experimental method that can determine concentration (even inderictly eg beer lamberts law) can be used to study kinetics


22
New cards

experimental rate law

Rate of reaction is often found to be proportional to the oncentrations of the reactants raised to a simple power

Rate = kr[A]a[B]b

  • r is the experimental rate constant

    [A] is the concentration of reactant A

    a is the power term, partial orders

  • overall, order of reaction is sum of partial orders

partial order

  • power term called partial orders

    • As rate of reaction control by whats in there, more often than not are integer values

    • Sum of partial order= overall order of reaction, can then determine mechanism and rate limiting step of that reaction


23
New cards

partial order

  • The partial orders relate the concentration of the reactants and the rate of the reaction.

  • Usually simple numbers (0, ½, 1, 2) and they may be positive or negative.

  • Classify reactions based on kinetics – similar behaviou

similar behaviour=molecularity (unimolecular, bi molecular etc)

  • 1st order: rate of reacrion on depend on A, proportioanl to A conc

  • 2nd order, bimolecular: 1st order with respect to A and B, 2nd order overall, requires molecule A and B to react together

  • 2nd order: with respect to A only


24
New cards

determining orders and rate laws

Initial rate method

  • Start of the reaction, we’ll know the exact concentration of reagents and slope is will be steepest.

  • if we vary the concentration and measure initial rates, we can see the effect of that reagent of the rate of reaction.

  • Repeating for all reagents, we determine the rate law

If we double [A] and we find that:

  • rate does not change, then alpha = 0, zeroth order in A

  • rate doubles, then alpha = 1, or first order in A

  • rate quadruples, then alpha = 2, or second order in A


start is most accurate point, so if vary conc and see effect on rate of reaction, can identify how important componenets are to reactions, then can isolate to see specific compounds effect


25
New cards

mechanism of reactions via rate law and stoichiometry

You cannot write experimental rate law based only on reaction stoichiometry. All comes down to mechanisms…

By comparing the experimental rate law and the stoichiometric reaction we can get some understanding of the mechanism

  •     compare numbers between stoichiometry reaction (big number) and rate of law

If rate law matches stoichiometry= elementary reaction

  • But if have rate law different to reaction stoichiometry, must be complex mechanism


<p>You cannot write experimental rate law based only on reaction stoichiometry. All comes down to mechanisms…</p><p>By comparing the experimental rate law and the stoichiometric reaction we can get some understanding of the mechanism</p><ul><li><p><span>&nbsp;&nbsp;&nbsp; </span>compare numbers between stoichiometry reaction (big number) and rate of law</p></li></ul><p>If rate law matches stoichiometry= elementary reaction</p><ul><li><p>But if have rate law different to reaction stoichiometry, must be complex mechanism </p></li></ul><p></p>
26
New cards

Determining orders & rate laws - initial rate method

  • interested in variation in rate with [reactant]

Start of the reaction, we’ll know the exact concentration of reagents and slope is will be steepest.

If we vary the concentration and measure initial rates, we can see the effect of that reagent of the rate of reaction.

Repeating for all reagents, we determine the rate law


If we double [A] and we find that:

  • rate does not change, then a = 0, zeroth order in A

  • rate doubles, then a = 1, or first order in A

  • rate quadruples, then a = 2, or second order in A


27
New cards

Determining orders & rate laws - integrated rate law

looks at relationship between [reactant] an time

Write the rate law in a form that can be integrated:

  • r = k[A][B]→ y=m x +c

Apply to data and plot graphically, Which equation does the data fit?

  • Looking for straight line

  • Gradient is significantly more accurate than y intercept, therefore intergrated rate law is more accurate


<p>looks at relationship between [reactant] an time</p><p>Write the rate law in a form that can be integrated:</p><ul><li><p>r = k[A][B]→ y=m x +c</p></li></ul><p>Apply to data and plot graphically, Which equation does the data fit?</p><ul><li><p>Looking for straight line</p></li><li><p>Gradient is significantly more accurate than y intercept, therefore intergrated rate law is more accurate</p></li></ul><p></p>
28
New cards

zeroth order reaction

[A]= -kt +[A]o

can use equation to plot [A] vs time (t)

  • slope=-k

  • intercept= [A]o

  • negative straight line slope, starts high gets lowers


<p>[A]= -kt +[A]o</p><p>can use equation to plot [A] vs time (t)</p><ul><li><p>slope=-k</p></li><li><p>intercept= [A]o</p></li><li><p>negative straight line slope, starts high gets lowers </p></li></ul><p></p>
29
New cards

first order reaction

ln[A]= -kt +ln[A]o

  • plot ln[A] vs time (t)

  • slope=-k

  • intercept=ln[A]o

  • straight line, negative relationship (starts high, gets lower)

exponential decay


<p>ln[A]= -kt +ln[A]o</p><ul><li><p>plot ln[A] vs time (t)</p></li><li><p>slope=-k</p></li><li><p>intercept=ln[A]o</p></li><li><p>straight line, negative relationship (starts high, gets lower) </p></li></ul><p>exponential decay</p><p></p>
30
New cards

second order reaction

1/[A]=kt + 1/[A]o

  • plots 1/[A] vs time (t)

  • slope= k

  • intercept= 1/[A]o

  • positive straight line relationship

non exponential decay


<p>1/[A]=kt + 1/[A]o</p><ul><li><p>plots 1/[A] vs time (t)</p></li><li><p>slope= k</p></li><li><p>intercept= 1/[A]o</p></li><li><p>positive straight line relationship </p></li></ul><p>non exponential decay</p><p></p>
31
New cards

half lives: first order reactions

t1/2= - (ln1/2/k)=ln2/k

  • independent of concentration


32
New cards

radioactive decay

All radioactive decay process follow 1st order kinetics

  • Independent of initial concentration

  • Ideal of archaeological analysis

eg radiocarbon dating

  • Radiocarbon dating

    14C isotope has half life 5730 yrs.

    Converted to CO2

    Analysed using mass spectrometry


<p>All radioactive decay process follow 1st order kinetics</p><ul><li><p>Independent of initial concentration</p></li><li><p>Ideal of archaeological analysis</p></li></ul><p>eg radiocarbon dating</p><ul><li><p>Radiocarbon dating</p><p>14C isotope has half life 5730 yrs.</p><p>Converted to CO2</p><p>Analysed using mass spectrometry</p></li></ul><p></p>
33
New cards

half lives: second order reactions

1/k[A]o=t1/2

  •   half life of second order reaction is dependent onn intial conc as nonexponential decay curve, cannt use for carbon dating


34
New cards

second order reaction, two reactants

at any point in time, [A]=[A]o - x and [B]=[B]o - x

so ln ([A]o[B]/[A][B]o)= k([B]o-[A]o)t

<p>at any point in time, [A]=[A]o - x and [B]=[B]o - x</p><p>so ln ([A]o[B]/[A][B]o)= k([B]o-[A]o)t </p>
35
New cards

Kr: rate constant

   rate constant only constant at a given temp (as tellls us how many molecule are going over the barrier at a given temp, change temp, changes rate, changes rate constant)

  • Temp dependence of the reaction rate, dependent on activation energy

  • Rate is the measured experimentally

    Determine partial orders and overall order

    Last thing to consider is kr – the rate constant

    Directly links the reagent concentrations to the rate

    ONLY a constant at a given temperature. Because determined

    by transition state barrier height

    Explains why reactions vary with temperature (change temp, changes Kr)


36
New cards

determining Kr (from integrated rate plots)

Units of kr

  • Rate = mol L-1 s-1

  • Conc = mol L-1

1st order reaction

  • 𝑘𝑟 = 𝑚𝑜𝑙 𝐿−1𝑠−1 / 𝑚𝑜𝑙 𝐿−1 = 𝒔−𝟏

second order reaction

  • 𝑚𝑜𝑙 𝐿−1𝑠−1 / 𝑚𝑜𝑙 𝐿−1 𝑚𝑜𝑙 𝐿−1 = 𝒎𝒐𝒍−𝟏 𝑳 𝒔−𝟏

can also use equation Kr=rate / [A]a[B]b

  • for 1st and 2nd order, value of gradient is same as value of rate law when detemrining rate constant from integrated rate plots


37
New cards

how do molecules react

collide in correct orientation wth enough energy

  • for a reaction, collisions must have correct energy, and orientation when colliding

     

     

    Most particles do not have enough energy  overcome the energy barrier, pink is proportion of particles with enough energy to react

    • The larger the rate constant, the more reaction is happening, the higher the proportio of molecules that have enoug energy

     

    If increase temp, more particles have high enough energy, therefore rection happens quicker (more reaction happens as a function of time - rate increases)

    • Numerically determine within rate constant, as directly relates to reaction rate


38
New cards

determining Ea from rate constant (Kr)

eg here with  1st order reaction, as temp increase, rate constant also increases

  • Exopoential relationship between rate cosntant and temp, governed by activation energy (height of barrier

    • This relationship describes why reactions vary with temp

    • If plot values A with function of time, get curved graph, difficult to extract data from due to exponential term, can instead use natural log to give different equation

 

A= number before exponential term, at this level= what rate constant would be if reaction was barrierless

  • Tells max rate that could happen if no barrier 

lnK= -Ea/RT + lnA

plot lnKr vs 1/temp (Kelvin)

slope = -Ea/R

therefore Ea=-slope x R (slope given by excel)


<p>eg here with<span>&nbsp; </span>1st order reaction, as temp increase, rate constant also increases</p><ul><li><p><span>Exopoential relationship between rate cosntant and temp, governed by activation energy (height of barrier</span></p><ul><li><p><span>This relationship describes why reactions vary with temp</span></p></li><li><p><span>If plot values A with function of time, get curved graph, difficult to extract data from due to exponential term, can instead use natural log to give different equation </span></p></li></ul></li></ul><p>&nbsp;</p><p>A= number before exponential term, at this level= what rate constant would be if reaction was barrierless</p><ul><li><p><span>Tells max rate that could happen if no barrier&nbsp;</span></p></li></ul><p>lnK= -Ea/RT + lnA</p><p>plot lnKr vs 1/temp (Kelvin)</p><p>slope = -Ea/R</p><p>therefore Ea=-slope x R (slope given by excel)</p><p></p>
39
New cards

types of catalysts

Homogeneous (organic, metal ion, complexes, salts)

• Heterogeneous (solid surfaces, porous materials; metals, oxides)

• Enzymes

40
New cards

catalysts

Catalyst allows for a new reaction pathway, with a lower Ea, and is not consumed during the reaction.

  • lower Ea, therefore under same conditions more reactions occur

catalytic converter eg as in automobiles

  • Converts convert harmful exhaust gases into less harmful ones.

    Heterogeneous reaction – Pt, Pd, Rh on an Al2O3 porous structure


<p>Catalyst allows for a new reaction pathway, with a lower Ea, and is not consumed during the reaction.</p><ul><li><p>lower Ea, therefore under same conditions more reactions occur</p></li></ul><p>catalytic converter eg as in automobiles</p><ul><li><p>Converts convert harmful exhaust gases into less harmful ones.</p><p>Heterogeneous reaction – Pt, Pd, Rh on an Al2O3 porous structure</p></li></ul><p></p>
41
New cards

enzyme catalyst

Enzyme active sites are complex 3D structures, consisting of hydrogen bonding interactions, hydrophobic regions, vdw interactions, that lead to an optimised fit for the substrates.

Product dissociates from active site, allowing for catalytic turnover.

But active site is very sensitive to its environment.