## Writing rate law: the 2 questions

Miriam Ifegwu 1D
Posts: 17
Joined: Fri Sep 20, 2013 3:00 am

### Writing rate law: the 2 questions

Hey,
I have 2 questions about writing the rate law:
1. I noticed that the examples used in lecture were with reactants that had 1 as its stoichiometric coefficient. How do you write the rate law when a reactant has 2 (or a number other than 1) as a coefficient? Example: 2A + B --> 2C
2. My second questions pertains to reaction mechanisms:
a) For the example No2 (g) + CO2 (g) --> NO (g) + CO2 (g). You stated that step 1 is the rate determining step because of the given experimental rate law (since the slowest step determines the rate of overall reaction). For similar questions where you are asked to find the rate determining step, will the experimental rate law be given? If not, how can one tell which intermediate reaction is the rate determining step?
b) My second question is more of a clarification. Since the slowest step determines the rate of the overall reaction. Is the reactant(s) in the overall rate law of a reaction the reactants of the rate determining step? Is this always the case?
***I apologize for the long questions. I am just really curious ^_^***

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

### Re: Writing rate law: the 2 questions

In response to your first question, page 569 states that the rate law can only be determined from experimental data and cannot be determined from a chemical equation. I believe that means that regardless of what the coefficients are, you need to have experimental data to form a rate law and you would write it in the same way if the coefficients were one or not.

I'm not completely sure about the answers to your second question, but I believe we will be given the experimental rate law in that case and the reactants in the overall rate law should be the reactants in the rate determining / slow step. (But again, I'm not entirely sure.)

Sonia Kumar 2A
Posts: 6
Joined: Fri Sep 26, 2014 2:02 pm

### Re: Writing rate law: the 2 questions

When you say overall rate law of the reaction, do you mean once the steps of an equation have been added together?

Chem_Mod
Posts: 17810
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 405 times

### Re: Writing rate law: the 2 questions

If 2A + B were the reactants in an elementary reaction, then Rate = k*[A]^2 * [B] so you would raise to the power of coefficients. If the reaction is not elementary, you would say (1/2)dA/dt = -k*[A]^n * [B]^m where the orders of the reactants are yet unknown. However the rate has (1/2) in front of it since you divide by the coefficient of A.

The overall (=experimental) rate law and rate determining step are dependent on each other, so you would be given at least one of those in any problem.

The reactants of the rate-determining step will lead to the overall rate law, however, any intermediates have to be cancelled out using the pre-equilibrium method.

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

### Re: Writing rate law: the 2 questions

Remember that when you are working with kinetics, the exponent is not always equal to the stoichiometric power like we did for equilibrium constants (K and Q).

Miriam Ifegwu 1D
Posts: 17
Joined: Fri Sep 20, 2013 3:00 am

### Re: Writing rate law: the 2 questions

Wow, i finally understand it. Thank you for clarifying it.