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 

The correct answer is c. hope this