J.R. S. answered • 05/22/17
Tutor
5.0
(145)
Ph.D. University Professor with 10+ years Tutoring Experience
Given that the rate constant has units of LM-1min-1, this indicates the reaction is 2nd order. Thus, one can write the rate equation as rate = k[C2H5I][OH-], and substituting the rate constant and the values for concentrations...
rate = 1.5x10-3 L•mol-1•min-1 (0.0331 mol/L)(0.0331 mol/L)
rate = 1.6x10-6 M/min
J.R. S.
tutor
First order is when the rate of reaction varies directly with the concentration of reactant. For example, if you double the concentration of a reactant, and the rate doubles, it is first order with respect to that reactant. If you double the concentration and the rate increases by FOUR fold (instead of doubling), it is 2nd order with respect to that reactant. You cannot tell the order from the reaction. It must be determined experimentally. Does this help, or do you have additional questions?
Report
05/22/17
J.R. S.
tutor
I just noticed that you asked how to tell from the rate equation. It will be whatever the exponent is. For example, if the rate equation is rate = k[A][B]2, then it is first order in A and 2nd order in B.
Report
05/22/17
Nicole N.
05/22/17