Overall Order

$aR \to bP, Rate = -\frac{1}{a} \frac{d[R]}{dt} = \frac{1}{b}\frac{d[P]}{dt}$

FrancoNancy_Sec1L
Posts: 26
Joined: Fri Sep 20, 2013 3:00 am

Overall Order

Hi,
I am not sure how to find the overall order. When do we consider the coefficient of other reactants? For example, on page 569 the redox reaction S2O8^2- (aq)+3I-(aq)---> 2SO42- (aq)+I3-(aq)

The rate of conception of S2O8^2-=k[S2O8^2-][I-] and the overall order of the reaction is 2. However, the coefficient of I- is 3 and it is not included, when trying to find the rate.
Help. Thanks

Neil DSilva 1L
Posts: 70
Joined: Fri Sep 26, 2014 2:02 pm

Re: Overall Order

The overall order of a reaction is determined by adding the powers in a rate law expression.

The coefficients are not used in determining rate laws such as those on page 569 ("The rate law for a reaction is experimentally determined and cannot in general be inferred from the chemical equation for the reaction"). We only use the coefficients when dealing with average rates / unique average rates / instantaneous rates as described in the first two sections. The rate law is different from those rates.

Neil DSilva 1L
Posts: 70
Joined: Fri Sep 26, 2014 2:02 pm

Re: Overall Order

Another instance where we would use the coefficients to come up with a rate law is if we're given the elementary steps of the reaction mechanism and we're told what the slow step is. But since that's not the case with the example given, again, we don't deal with coefficients.

Justin Le 2I
Posts: 142
Joined: Fri Sep 26, 2014 2:02 pm

Re: Overall Order

Adding on to what Neil said, don't confuse kinetics constants with equilibrium constants. When we did equilibrium in 14A, we used stoichiometric coefficients for the exponents.

It makes sense to use coefficients when working with elementary steps because the order depends on the molecularity of the elementary reaction (i.e. unimolecular or bimolecular).