For the reaction a → b, the rate law is: rate = k× [a]. over the course of 4 minutes, the concentration of a reduces by one-half, from 1.00 mto 0.50 m. how is the rate affected?
the rate law for the equation is as follows; Rate = k [A] the order with respect to A is 1, therefore this is a first order reaction as A is the only reactant. At t = 0 minutes [A] = 1.00 M rate - initial rate rate = k [1.00 M] ---1) At t = 4 minutes [A] = 0.50 M rate' - new rate rate' = k [0.50] --2) Divide 2)/1) rate'/rate = 0.5/1 rate' = 0.5 rate rate has been reduced by half
