Initial rates

• For some reactions it may not be feasible to measure the rate of reaction for the full length of time of the reaction

• As the initial concentrations are known it is possible to measure the initial rate of reaction

• If we have a reaction with more than 1 reactant we can experiment with different initial concentrations to find out the order wrt each reactant

• We vary the concentration of 1 reactant and keep the concentration of all the others constant

• From this data we can then work out the order wrt each reactant.

Calculating order from inital rates data

What is the order with respect to each reactant in the following example.

Using expt no 1 and 2, the rate of reaction goes up by 4, [X] doubles, and the 2nd order wrt [X]

When [Y] doubles the rate stays the same… 0th order wrt Y

When [Z] doubles the rate doubles… 1st oder wrt Z

Calculating k from inital rates data

• Once you know the order for each reactant you can write a rate equation

  Rate = k[X]²[Z]

• Rearrange the euqation in terms of k

  k = Rate / [X]²[Z]

• Using the data from one row in the table calculate k

• Finally work out what the units of rate constant are

• k = 0.0765 dm⁶mol^-2s^-1

0th order wrt A

1st order wrt B

2nd order wrt C

Rate = k[C]²[B]

k = 27.8 dm⁶mol^-2s^-1

Different steps in chemical reactions

• If we know the stoichiometry of.a reaction can we work out the order for each reaction without doing any experiments?

  • NO!

• Chemical reactions often take place in more than 1 step

• ONLY if the reaction takes place in 1 step will the rate equation be directly related to the stoichiometric equation

• Consider the following example

  HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

• in this case rate = k[HCl][NaOH]

• This reaction takes place in only one step

• In this reaction each HCl needs to collide with a NaOH (with enough energy) for a reaction to take place

• A collision between 2 reactants is relatively common (consider car accidents, the majority are between 2 cars only)

Multi-step reactions

• Consider the following example

  4NO(g) + O2(g) → 2N2O3(g)

• If this reaction takes place in just 1 step it will involve 4 NO molecules and 1 O2 molecule all colliding together at the SAME time

• This is extremely unlikely to occur (compare with a car accident with 5 cars all colliding with each other at ONCE)

• In actual fact this reaction doesn’t occur with just one step, it is considered to take place in 2 steps (please note this reaction isn’t well understood)