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Rates of Reactions

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Factors affecting rates

• Temperature– Increasing temperature increases rate.

– More molecules have sufficient energy to react.

• Reactant concentrations– Dependence on concentration must be determined experimentally.

– Can be used to deduce mechanism.

• Catalysts– speed up reactions

– heterogeneous (e.g. solids) or homogeneous (same phase)

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Dependence of concentration on time, in solution

For the simple reaction A C,

starting with [A] = 1.0 and [C] = 0

• Concentration of A decreases.

• Concentration of C increases at the same rate.

• Reaction slows, but continues until A runs out, or until equilibrium is established.

• At completion, or equilibrium, concentrations of A and C are constant.

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Dependence of concentration on time, in solution

Rate of reaction can be expressedas the rate of disappearance of A oras the rate of appearance of C.

For the reaction A C, Rate = -[A]/ t = +[C]/t

Once we see how the rate of reactiondepends on the concentrations, we willwrite mathematical expressions for theconcentrations as a function of time, these are the integrated rate laws.

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Dependence of concentration on time, in solution

For the reaction A 2C, The rate of appearance of C is twice the rate of disappearance of A

+ [C]/ t = 2(- [A]/ t )

In general, for any reactiona A + b B c C

- [A] = - [B] = + [C] a t b t c t

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Measuring the rate of a reaction

• The rate is often measured as Xt, where X may be a reactant or product.

• Depending on the nature of X, the change in concentration may be monitored by a change in – colour (intensity of some wavelength)

– pressure (for gases)

– pH (for OH- or H3O+)

– conductivity (ions)

– radioactivity, etc.

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Measuring the initial rate

• The rate of the forward reaction depends on the concentration of reactants, not products.

• The dependence may be linear, quadratic, etc., this must be determined experimentally.

• The rate is measured at the beginning of the reaction (the initial rate), as a function of the initial reactant concentrations

• This determines the reaction order.

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Initial reaction rates2 NO (g) + 2 H2 (g) N2 (g) + 2 H2O (g)

[NO] /M [H2] /M Rate /Ms-1 factor 0.210 0.122 0.0339 baseline

0.420 double NO

0.122 0.136 4 = 22

R [NO]2

0.210 0.244 double H2

0.0678 2 = 21 R [H2]

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Rate = k [NO]2 [H2]1

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Initial reaction rates reaction order

[A] /M [B] /M Rate /Ms-1 factor 1 1 1 baseline

2 double A

1 4 4 = 22

[A]2

1 2 double B

2 2 = 21 [B]1

Rate = k [A]2 [B]1

Reaction is second order in A,first order in Band third order overall

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CONSIDER THE RATE DATA FOR THE REACTION:

2NO + O2 2NO2

[NO]mol/L

[O2]mol/L

-[NO]/ tmol/L.s

0.01 0.01 2.5 x 10-5

0.02 0.01 1.0 x 10-4

0.01 0.02 5.0 x 10-5

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Reaction order & the rate law

• The rate law is: Rate = k[A]x[B]y

• The order of a reaction is equal to the value of the exponent in the rate law.

• The reaction has an order with respect to each reactant (and each catalyst).

• The overall reaction order is the sum of the individual orders.

• k is the rate constant, which depends on T.

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The rate constant

• Once the form of the rate law is known, we can fill in the data from any one run of our determination to find the rate constant.

e.g. 2 NO (g) + 2 H2 (g) N2 (g) + 2 H2O (g)

Rate = 0.0339 M·s-1 = k (.210 M)2(.122 M)

k = 6.30 M-2 ·s-1

• The units of k depend on the order of the reaction

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The rate constant

• The value of k depends on the nature of the reactants and on the temperature.

• Arrhenius found that the temperature dependence could be expressed as

k = Ae-Ea / RT

• The preexponential factor A, and the activation energy, Ea, are relatively independent of temperature.

• What are these parameters? • Why does k have this dependence?

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•Consider the combination reaction of NO and OConsider the combination reaction of NO and O22 to produce NO to produce NO2 2 : :

•2 NO2 NO(g) (g) + O+ O2(g)2(g) 2 NO 2 NO2 (g)2 (g)

•Determination of the Rate Law (via Methods of Initial Rates)Determination of the Rate Law (via Methods of Initial Rates)

• Initial Concentrations • (mol/ L) Initial Rate Experiment [NO] [O2] (mol/L • s)

•1 0.020 0.010 0.028

•2 0.020 0.020 0.057

•3 0.020 0.040 0.114

•4 0.040 0.020 0.227

•5 0.010 0.020 0.014

•Based on these data, what is the rate equation? What is the value of the rate constant k?

Rate Law Determination

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Rate Law; Solving for rate ConstantThe general rate law is:

Rate law = k [NO]2 [O2]

the rate constant k is determine by selecting one of the experiments and solving the equation.

Consider experiment#1:

Rate = 0.028 = k [0.020]2 [0.010]

k = 0.028 / (4•10-4)(0.010) = 7.1•103 M-

2 s-1

Rate Law: Rate = 7.1•103 [NO]2 [O2]

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Microscopic view

• In order to understand our macroscopic observations about temperature and concentration dependence, we should look at the reaction microscopically - on the size scale of atoms and molecules.

• The rates of chemical reactions are explained by collision theory, which is based on kinetic theory.

• Collision theory views a reaction as the result of a ‘successful’ collision between two or more reactants and/or catalysts.

• A few reactions occur without any collision.

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Collision Theory

• The number of collisions between two or more species is proportional to the product of their concentrations.

• When the reaction is the result of a single collision – an elementary step – then the concentration dependence is directly related to the stoichiometry of that collision.

• The probability that A will collide with B is proportional to [A][B].• The probability that A will collide with A is proportional to [A]2

• For more complicated processes, the rate law is some combination of these elementary steps.

• In order to react, the molecules must collide in a favourable orientation and with sufficient energy.

• These factors are accounted for in the rate constant.

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Molecularity of elementary steps

• For an elementary step (arising from one collision), the rate law depends on the stoichiometry of the ‘collision’.

• A step involving only one molecule is called unimolecular.

Rate = k[A]

• A step involving two molecules is called bimolecular.

Rate = k[A][B], or Rate = k[A]2

• A step involving three molecules is called termolecular.

Rate = k[A][B][C], etc.

• Very few elementary steps involve more than 3 molecules.

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Reaction progress & Ea

• For an elementary process we can plot the potential (chemical) energy of the molecules as they approach each other, collide, react and move apart.

• For an elementary process which involves only one molecule, we can plot the potential energy as some internal coordinate, such as bond length or angle, changes.

• This plot is sometimes called a reaction coordinate diagram, or an energy plot. There is typically a maximum near the ‘collision’.

• Molecules move along this reaction coordinate with some initial kinetic energy. K.E. is converted to P. E. to overcome the energy barrier, the activation energy.

• Those molecular ‘collisions’ starting with enough kinetic energy can overcome the barrier and react.

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Arrhenius and Boltzmann

• We saw in chapter 13 that only a certain proportion of molecules had enough energy to remain in the gas phase. The same type of energy distribution is at play here.

The Boltzmann distribution tells us that at any particular temperature a certain percentage of the molecules are above some energy cut-off.

• The cut-off of interest in this case is the activation energy.

• The percentage of molecules with energy above Ea is related to the factor exp(-Ea/RT) in the Arrhenius expression in the rate.

• As the temperature increases, so does the percentage of molecules above the cut-off.

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Calculations with Ea & T

k = Ae-Ea / RT ln k = ln A – (Ea/RT)

• Increasing the temperature from 300 K to 310 K increases the rate by a factor of 2. What is the activation energy?

• Given a set of T and k data, a plot of ln k vs. 1/T has a slope of -Ea/R

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Rate determining step

• When the reaction is a series of elementary steps, rather than a single step, the rate of reaction is determined by the slowest step, which is typically the step with the highest activation barrier.

• This step is called the rate determining step, and the rate law for a known mechanism can be written in terms of the rate for this step.

• If the rate determining step is not the first step, the rate may depend on some species which do not appear as reactants in the overall reaction equation.

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Reaction mechanism

• Chemists often study reaction rates in order to deduce or confirm a reaction mechanism – the stepwise progress of the reaction.

• A proposed mechanism is written as a sum of elementary steps, which may be reversible.

• If the rate law derived from the proposed mechanism matched the observed rate law, then we are more confident in our proposal, but still unsure.

• If the rate laws do not match, we must come up with a different proposal.

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2 NO (g) + Br2 (g) 2 BrNO (g)

• Step 1 Rate = k1[Br2][NO]

Br2 (g) + NO (g) NOBr2 (g)

• Step 2 Rate 2 = k2[Br2][NOBr2]

NOBr2 (g) + NO (g) 2 BrNO (g)

• NOBr2 is an intermediate – it is formed and then used up.

• The overall rate will depend on which step is rate determining, and on whether either step is reversible.

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2 NO2 (g) + F2 (g) 2 FNO2 (g)

• Step 1 rate = k1[NO2][F2]

NO2 + F2 FNO2 + F slow

• Step 2 rate = k2[NO2][F]

NO2 + F FNO2 fast

Overall rate = k1[NO2][F2]

Rate of reaction = rate of the slowest step

k2 >> k1

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2 NO (g) + O2 (g) 2 NO2 (g)

• Step 1 is reversible K1 = [NO3] / [NO][O2]

NO + O2 NO3 fast equilibrium

• Step 2 rate = k2[NO3][NO]

NO3 + NO 2 NO2 slow

• Overall rate = k2 [NO3][NO]

• Rate of reaction = rate of the slowest step, but NO is an intermediate – difficult to determine its concentration. Want to replace [NO] with known quantities:

K1 = [NO3] / [NO][O2] [NO3] = K1 [NO] [O2]

Rate = k2(K1 [NO] [O2]) [NO] = k’ [NO]2[O2]

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Equilibria in reaction mechanisms

• Note that this topic is not covered in Kotz and Treichel• In principle all reaction are reversible, but only some are reversible on

a time scale relevant to the overall process.• A reaction, or step, which is fast in both the forward and reverse

direction will come to equilibrium rapidly.• Dynamic equilibrium is reached when the rate of the forward reaction

equals the rate of the reverse reaction.• For an elementary step 2A B + C, at equilibrium

rate forward = k1[A]2 = k-1[B][C] = rate reverse

= equilibrium constant. KA

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