Question

Super lost on this question and need some major help!

Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimental data provided.

Q + X products

Trial [Q] [X] Rate
1 0.12 M 0.10 M 1.5 × 10-3 M/min
2 0.24 M 0.10 M 3.0 × 10-3 M/min
3 0.12 M 0.20 M 12.0 × 10-3 M/min

Answer 1

You need to determine the exponent of each reactant separately. In order to do this, compare two of the trials such that only 1 of the reactants is changing in concentration (the other's concentration remains constant), thus you can attribute any change in the rate to the change in concentration to that reactant.

For example, if you compare trial 1 and 2, where [Q] doubles in concentration and the rate doubles, we observe a linear relationship.

Trial [Q] [X] Rate
1 0.12 M 0.10 M 1.5 × 10-3 M/min
2 0.24 M 0.10 M 3.0 × 10-3 M/min


to make any calculation simple, we can choose trial 1 concentration to be 1, and trial to to be 2 (concentration doubled, right?). Same with the rates.

Trial 1: \(\sf 1 =[1]^n\)

Trial 2: \(\sf 2 =[2]^n\)

The above is only true when \(\sf n=1\), so the rate law so far is:

\(\sf rate =k[Q][X]^m\)

once you find the order (exponent) of X, plug in all the values from the concentrations and rate to find k.