CHMA11 Lecture Notes Chemistry UTSC

Chemistry lec 3

Molecularity

■ The number of reactant particles in an elementary step is called its molecularity

 – This is the reaction order for an elementary reaction

■ A unimolecular step involves one particle

■ A bimolecular step involves two particles – though they may be the same kind of particle

■ A termolecular step involves three particles – though these are exceedingly rare in elementary steps

Rate Laws for Elementary Steps

■ Each step in the mechanism is like its own little reaction – with its own rate law

■ The rate law for an overall reaction must be determined experimentally

■ But the rate law of an elementary step can be deduced from the equation of the step

H2(g) + 2 ICl(g) ® 2 HCl(g) + I2(g)

1) H2(g) + ICl(g) ® HCl(g) + HI(g) Rate = k1[H2][ICl]

2) HI(g) + ICl(g) ® HCl(g) + I2(g) Rate = k2[HI][ICl]

Rate Determining Step

■ In most mechanisms, one step occurs slower than the other steps

■ The result is that product production cannot occur any faster than the slowest step

– the step determines the rate of the overall reaction

■ We call the slowest step in the mechanism the rate determining step

■ The rate law of the rate determining step determines the rate law of the overall reaction

Another Reaction Mechanism

NO₂(g) + CO(g) -> NO(g) + CO₂(g) Rate_obs = k[NO₂] ²

1. NO₂(g) + NO₂(g) -> NO₃(g) + NO(g) Rate = k₁[NO₂] ² Slow

2. NO₃(g) + CO(g) -> NO₂(g) + CO₂(g) Rate = k₂[NO₃][CO] Fast

SUM: NO₂(g) + CO(g) -> NO(g) + CO₂(g)

The first step is slower than the second step

The first step in this mechanism is the rate determining step.

The rate law of the first step is the same as the rate law of the overall reaction.

Validating a Mechanism

 ■ To validate (not prove) a mechanism, two conditions must be met:

 1. The elementary steps must sum to the overall reaction

2. The rate law predicted by the mechanism must be consistent with the experimentally observed rate law

Mechanisms with a Fast Initial Step

■ When a mechanism contains a fast initial step, the rate limiting step may contain intermediates ■ When a previous step is rapid and reaches equilibrium, the forward and reverse reaction rates are equal – so the concentrations of reactants and products of the step are related

– and the product is an intermediate

 ■ Substituting into the rate law of the RDS will produce a rate law in terms of just reactants

                          The Effect of Temperature on Rate

■ Changing the temperature changes the rate constant of the rate law

■ Arrhenius investigated this relationship and found that:

where T is the temperature in Kelvin

R is the gas constant in energy units, 8.314 J/(mol•K)

A is called the frequency factor, the rate the reactant molecules collide “correctly”

E_a is the activation energy, the extra energy needed to start the molecules reacting

Arrhenius Plots

■ The Arrhenius Equation can be expressed in the following form:

This looks like the Claussius-Clapeyron equation – that talks about vapour pressure!

a graph of ln(k) vs. (1/T) is a straight line

Example: Determine the activation energy and frequency factor for the reaction O₃(g) -> O₂(g) + O(g)

given the following data:

slope, m = 1.12 x 10⁴ K                                                                                                             y–intercept, b = 26.8