16.1 Rate Expression & Reaction Mechanism

What does this mean?

With any luck you'll remember that:

Rate of reaction = Change of Concentration / Time

It doesn't matter whether we're talking about the change of concentration of a reactant or the change of concentration of a product.

So for HNO₃ + LiOH --> LiNO₃ + H₂O

Rate = [LiOH] / Time = [HNO₃]/ Time = [LiNO₃] / Time

This allowed us to develop the rate equation

Since Rate ∝ [ HNO₃]

And Rate ∝ [ LiOH ]

We could predict from the equation that Rate = k [ HNO₃ ] [ LiOH ]

But we wouldn't know this was correct for sure unless we knew that the reaction happened in one step.

It would be truer to say,

Rate = k [ HNO₃ ]^x [ LiOH ]^y

Where x and y could be 0, 1 or 2.

And we'd only find out for sure by studying the initial rate of reaction with different concentrations of the two reactants.

Comparing Trial 1 to Trial 2: Doubling [IO₃^-] only, doubles the rate

Therefore, 1st order wrt [IO₃^-]

Comparing Trial 4 to Trial 1: Doubling [H⁺] only, quadruples the rate

Therefore, 2nd order wrt [H⁺]

Comparing Trial 4 to Trial 3: 1.5 x [I^-] only, rate x 2.25

Therefore, 2nd order wrt [I^-]

Rate = k [I^-]² [H⁺]² [IO₃^-]

1. In a reaction between Bromine and Propanal the following data was found

Write the rate equation and find the units of k

2.

Write the rate equation and find the units of k

3.

A GCSE student can tell you that a catalyst "speeds up a reaction without _______________________".

A good GCSE student can say that this is because they "lower the _______________________"

An amazing GCSE student may say that they "lower the ________________________ by providing an alternative p________________".

In other words, the catalyst may actually react with one of the reactants (often in the first step) but is re-made during another step, leading to the same overall reaction but faster since the new step has a very low Activation Energy.

The Contact Process is often studied at GCSE but the details of how the catalyst works are not.

Step 1 is simply burning Sulphur - it is basically a one way reaction.

Step 2 is an equilibrium catalysed by Vanadium (V) Oxide which is a solid whereas both reactants are gases - this makes it a H___________________catalyst (different state than the reactants)

We could miss out Step 3 and just dissolve the SO₃ in water but this is too exothermic.

We don't need to know the exact details of Step 2, so we could summarise as;

This has replaced the one step equilibrium with two steps, both of which have lower Activation Energies.

The V₂O₄ is a reactive intermediate - a substance briefly formed and then destroyed during a reaction

If the catalyst was Homogeneous - in the same state as the reactants - then its concentration might affect the rate of the reaction and the catalyst might appear in the rate equation.

As far as the rate equation is concerned, only the number of reacting molecules in the Rate Determining Step (RDS) is important.

A Unimolecular Step will involve ________ molecule, usually breaking up into two or more

A Bimolecular Step will involve ________ molecules, which may combine into one, or form multiple products

A Termolecular Step will involve ________ molecules.

Why are Termolecular steps so uncommon?

_________________________________________________________________________________________________________

_________________________________________________________________________________________________________

The number of reacting molecules in the RDS determines the rate equation as below:

Example 1. The reaction of Propanone and Iodine happens in 4 steps.

If we cancel the products of one step that are destroyed by another

And then add the three steps together we get:

CH₃COCH₃ + I₂ --> CH₃COCH₂I + HI

Only Step 2 is slow.

It involves doesn't involve Iodine so that can't be in the rate equation.

It does involve CH₃COHCH₃⁺ which is effectively one CH₃COCH₃ molecule and one H⁺ joined together.

So the rate equation is Rate = k [CH₃COCH₃][H⁺]

Example 2

Nitrogen Monoxide can react with Hydrogen: 2NO + 2H₂ --> 2H₂O + N₂

This reaction happens in more than one step

If we cancel the products of one step that are destroyed by another

And then add the three steps together we get:

2NO + H₂ + H₂ --> H₂O + H₂O + N₂

_Or, 2NO + 2H₂ --> 2H₂O + N₂

Only Step 2 is slow.

It involves one H₂ molecule, and one N₂O₂ molecule which is effectively two NO molecules joined together.

So the rate equation is Rate = k [H₂][NO]²

1. Consider the mechanism of the reaction between BromoMethylpropane and Sodium Hydroxide.

What is the rate equation? What are the units of k?

2. Consider the mechanism of the reaction of Nitrogen Dioxide and Carbon Monoxide

What is the rate equation? What are the units of k?

3. Consider the mechanism for the decomposition of N₂O

What is the rate equation? What are the units of k?

4. Consider the mechanism for the reaction of NItrogen Dioxide and Fluorine

What is the rate equation? What are the units of k?

Be careful, these graphs look similar if you don't pay attention to the labels on the axes.

It pays to have a good think about what the labels mean.

If a reaction is zero order with respect to a reactant then it doesn't matter how much you change the initial concentration of that reactant, it won't alter the rate.

But since it is a reactant its concentration will go down over time, and the concentration of Zero order reactants decrease inversely proportionally with time

For a first order reaction, doubling the initial concentration will double the rate - the graph is directly proportional.

The concentration of the reactant will decrease exponentially though.

If the reactant is second order then doubling its initial concentration quadruples the rate, so the rate increases exponentially with concentration.

Unfortunately, the concentration time graph looks rather like the one for 1st order unless we can see the numbers on the axes, or the first and second order graphs next to each other.

Examiners sometimes give you a graph rather than initial rates data to establish the order of reaction with respect to one of the reactants.