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Chemical kinetics or dynamics. 3 lectures leading to one exam question Texts: “Elements of Physical Chemistry” Atkins & de Paula Specialist text in Hardiman Library “Reaction Kinetics” by Pilling & Seakins, 1995

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chemical kinetics or dynamics
Chemical kinetics or dynamics

3 lectures leading to one exam question

  • Texts: “Elements of Physical Chemistry”

Atkins& de Paula

  • Specialist text in Hardiman Library
    • “Reaction Kinetics” by Pilling & Seakins, 1995

These notes available On NUI Galway web pages athttp://www.nuigalway.ie/chem/degrees.htm

What is kinetics all about?

academic ozone chemistry
Academic? Ozone chemistry
  • Ozone; natural formation (l» 185-240 nm)
    • O2 + hu ® 2O
    • O + O2® O3
  • Ozone; natural destruction (l» 280-320 nm) Thomas Midgely
    • O3 + hu ® O + O21922 TEL; 1930 CFCs
    • O + O3® 2O2
  • ‘Man-made’ CCl2F2 + hu ® Cl + CClF2
    • Cl + O3® Cl̶ O + O2
    • Cl̶ O + O ® Cl + O2
    • -----------------------------
    • Net result is: O + O3® 2 O2
  • 1995 Nobel for chemistry: Crutzen, Molina & Rowland
  • 1996 CFCs phased out by Montreal protocol of 1987
chemical kinetics
Thermodynamics

Direction of change

Kinetics

Rate of change

Key variable: time

What times?

1018 s age of universe

10-15 s atomic nuclei

108 to 10-14 s

Ideal theory of kinetics?

structure, energy

calculate fate

Now?

compute rates of elementary reactions

most rxnsnot elementary

reduce observed rxn. to series of elementary rxns.

Chemical kinetics
slide4

Thermodynamics vs kinetics

Kinetics determines the rate at which change occurs

kinetics and equilibrium
Kinetics and equilibrium

kinetics

equilibrium

hierarchical structure
Hierarchical structure

C.K. Westbrook and F.L. Dryer

Prog. Energy Combust. Sci., 10 (1984) 1–57.

slide8

Reactionmechanism

k = A Tn exp(-Ea/RT)

Reaction AfnfEaf ArnrEar

rate of reaction symbol r v
Rate of reaction {symbol: R, v, ..}

Stoichiometric equation

  • m A + n B = p X + q Y
    • Rate =-(1/m) d[A]/dt
    • =- (1/n) d[B]/dt
    • =+ (1/p) d[X]/dt
    • =+ (1/q) d[Y]/dt
    • Units: (concentration/time)
    • in SI mol/m3/s, more practically moldm–3 s–1
rate of reaction symbol r n
Rate of reaction {symbol: R,n, ..}

5Br- + BrO3- + 6H+ = 3Br2 + 3H2O

Rate?conc/time or in SI mol dm-3 s-1

– (1/5)(d[Br-]/dt) = – (1/6) (d[H+]/dt) =

(1/3)(d[Br2]/dt) = (1/3)(d[H2O]/dt)

Rate law?Comes from experiment

Rate = k [Br-][BrO3-][H+]2

where k is the rate constant (variable units)

rate law
Rate Law
  • How does the rate depend upon [ ]s?
  • Find out by experiment

The Rate Law equation

  • R = kn [A]a [B]b … (for many reactions)
    • order, n = a + b + … (dimensionless)
    • rate constant, kn (units depend on n)
    • Rate = kn when each [conc] = unity
experimental rate laws
Experimental rate laws?

CO + Cl2®COCl2

  • Rate = k [CO][Cl2]1/2
    • Order = 1.5 or one-and-a-half order

H2 + I2®2HI

  • Rate = k [H2][I2]
    • Order = 2 or second order

H2 + Br2®2HBr

  • Rate = k [H2][Br2] / (1 + k’ {[HBr]/[Br2]} )
    • Order = undefined or none
determining the rate law
Determining the Rate Law
  • Integration
    • Trial & error approach
    • Not suitable for multi-reactant systems
    • Most accurate
  • Initial rates
    • Best for multi-reactant reactions
    • Lower accuracy
  • Flooding or Isolation
    • Composite technique
    • Uses integration or initial rates methods
integration of rate laws
Integration of rate laws
  • Order of reaction

For a reaction aA products, the rate law is:

rate of change in the

concentration of A

first order reaction1
First-order reaction

A plot of ln[A] versus t gives a straight

line of slope ̶ kAif r = kA[A]1

half life first order reaction
Half life: first-order reaction
  • The time taken for [A] to drop to half its original value is called the reaction’s half-life, t1/2. Setting [A] = ½[A]0and t = t1/2 in:
when is a reaction over
When is a reaction over?
  • [A] = [A]0exp{-kt}

Technically [A]=0 only after infinite time

second order reaction1
Second-order reaction

A plot of 1/[A] versus t gives a straight

line of slope kA if r = kA[A]2

second order test a a p
Second order test: A + A®P

2nd Order reaction

Slope = k

kinetics and equilibrium1
Kinetics and equilibrium

kinetics

equilibrium

initial rate method
Initial Rate Method

5 Br- + BrO3- + 6 H+® 3 Br2 + 3 H2O

General example: A + B +…® P + Q + …

  • Rate law: rate = k [A]a[B]b…??

log R0 = alog[A]0 + (log k+ blog[B]0 +…)

y = mx + c

  • Do series of expts. in which all [B]0, etc are constant and only [A]0 is varied; measure R0
  • Plot log R0 (Y-axis) versus log [A]0 (X-axis)
  • Slope Þ a
example r 0 k no a h 2 b
Example: R0 = k [NO]a[H2]b

2 NO + 2H2®N2 + 2H2O

  • Expt. [NO]0 [H2]0R0
    • 1 25 10 2.4×10-3
    • 2 25 5 1.2×10-3
    • 3 12.5 10 0.6×10-3

Deduce orders wrt NO and H2 and calculate k.

  • Compare experiments #1 and #2 Þ b
  • Compare experiments #1 and #3 Þ a

Now, solve for k from k = R0 / ([NO]a[H2]b)

how to measure initial rate
How to measure initial rate?
  • Key:- (d[A]/dt) » - (d[A]/dt) » (d[P]/dt)

A + B + … ® P + Q + …

t=0 100 100 ® 0 0 mol m-3

10s 99 99 ® 1 1 ditto

  • Rate?

- (100-99)/10 = -0.10 mol m-3 s-1

+(0-1)/10 = -0.10 mol m-3 s-1

  • Conclusion? Use product analysis for best accuracy.
isolation flooding
Isolation / flooding

IO3- + 8 I- + 6 H+® 3 I3- + 3 H2O

  • Rate = k [IO3-]a[I-]b[H+]g…
    • Add excess iodate to reaction mix
    • Hence [IO3-] is effectively constant
    • Rate = k¢[I-]b[H+]g…
    • Add excess acid
    • Therefore [H+] is effectively constant
  • Rate » k² [I-]a
  • Use integral or initial rate methods as desired
rate law for elementary reaction
Rate law for elementary reaction
  • Law of Mass Action applies:
    • rate of rxnµproduct of active masses of reactants
    • “active mass” molar concentration raised to power of number of species
  • Examples:
    • A ® P + Q rate = k1 [A]1
    • A + B ® C + D rate = k2 [A]1 [B]1
    • 2A + B ® E + F + G rate = k3 [A]2 [B]1
molecularity of elementary reactions
Molecularity of elementary reactions?
  • Unimolecular (decay) A ® P

–(d[A]/dt) = k1 [A]

  • Bimolecular (collision) A + B ® P

–(d[A]/dt) = k2 [A] [B]

  • Termolecular (collision) A + B + C ® P

–(d[A]/dt) = k3 [A] [B] [C]

  • No other are feasible! Statistically highly unlikely.
co cl 2 x cocl 2
Experimental rate law: –(d[CO]/dt) = k [CO] [Cl2]1/2

Conclusion?: reaction does not proceed as written

“Elementary” reactions; rxns. that proceed as written at the molecular level.

Cl2®Cl + Cl(1)

Cl + CO ®COCl(2)

COCl + Cl2® COCl2 + Cl(3)

Cl + Cl® Cl2(4)

Steps 1 thru 4 comprise the “mechanism” of the reaction.

decay

collisional

collisional

collisional

CO + Cl2XCOCl2
d co dt k 2 cl co
- (d[CO]/dt) = k2 [Cl] [CO]

If steps 2 & 3 are slow in comparison to 1 & 4

then, Cl2⇌2Cl or K= [Cl]2 / [Cl2]

So[Cl] =ÖK × [Cl2]1/2

Hence:

  • - (d[CO] / dt) = k2 × ÖK × [CO][Cl2]1/2

Predict that: observed k = k2 × ÖK

  • Therefore mechanism confirmed (?)
h 2 i 2 2 hi
H2 + I2® 2 HI
  • Predict: + (1/2) (d[HI]/dt) = k [H2] [I2]
  • But if via:
    • I2® 2 I
    • I + I + H2® 2 HI rate = k2 [I]2 [H2]
    • I + I ® I2

Assume, as before, that 1 & 3 are fast cf. to 2

Then: I2⇌2 IorK = [I]2 / [I2]

  • Rate =k2 [I]2 [H2] = k2 K [I2] [H2] (identical)

Check?I2 + hn® 2 I (light of 578 nm)

problem
Problem
  • In the decomposition of azomethane, A, at a pressure of 21.8 kPa & a temperature of 576 K the following concentrations were recorded as a function of time, t:

Time, t /mins 0 30 60 90 120

[A] / mmol dm-3 8.70 6.52 4.89 3.67 2.75

  • Show that the reaction is 1st order in azomethane & determine the rate constant at this temperature.
slide38
Recognise that this is a rate law question dealing with the integral method.

- (d[A]/dt) = k [A]? = k [A]1

Re-arrange & integrate (bookwork)

  • Test: ln [A] = - k t + ln [A]0

Complete table:

Time, t /mins 0 30 60 90 120

ln [A] 2.16 1.88 1.59 1.30 1.01

  • Plot ln [A] along y-axis; t along x-axis
  • Is it linear? Yes. Conclusion follows.

Calc. slope as: -0.00959 so k = + 9.6×10-3 min-1

more recent questions
More recent questions …
  • Write down the rate of rxn for the rxn:

C3H8 + 5 O2 = 3 CO2 + 4 H2O

  • for both products & reactants [8 marks]

For a 2nd order rxn the rate law can be written:

- (d[A]/dt) = k [A]2

What are the units of k ? [5 marks]

  • Why is the elementary rxn NO2 + NO2 N2O4 referred to as a bimolecular rxn? [3 marks]