Chem 101 Homework 8: Kinetics Question 11

Question

A reaction has a rate law of Rate = (1.25 M⁻¹s⁻¹)[A][B]. What is the rate of the reaction if [A] = 0.503 M and [B] = 0.280 M?

Answer = ____M/s

Answer 1

Cay,

Plug the concentrations into the rate law given so that you now have

Rate = 1.25 M^-1s ^-1 x (0.503M) (0.280M) = 0.176 M/s

Answer 2

Let’s first think about what this equation means. It’s asking for a rate, and gives a rate constant. The question is asking for molarity produced and not molarity consumed. The units for the two different processes will not be the same! Also, the rate at which something is formed must be >0 and the rate at which it's consumed must be <0. More on this later.

Chemists like to think in moles, more than molarity. What do we do about molarity vs. moles? We fiddle with the units for the rate constant! [A] and [B] are in molarity (or they could have been in moles). Molarity produced is what it says it is. That means the units of the rate constant must cancel out enough to give us molarity produced is what it says it is. Its units must be: per molarity per second, or 1/(M • s) ….

Note: 1/(M • s) is just L/(mole • s) ...... units of moles!

Rate = (1.25 /(M • s)) • (0.503 M)(0.280 M)

Rate = 0.176 M/s, or 0.176 L/(mole • s)

NOTE: If this reaction took place in solution, then the volume probably wouldn’t change. So we actually would not care about the volume, and we would actually have a rate which would have units in moles/second (remember: chemists usually count molecules, i.e. moles). But if this reaction occurred in the gas phase, then the volume could change. So then it would be good to use moles/liter. :-)