Question

A reaction has the experimental rate law "Rate = K[A]^2[B]. If the concentration of A is doubled and the concentration of B is halved, what happens to the reaction rate?

Answer 1

suppose the concentrations of A and B both start at 1.0M. That makes the rate law \[Rate = K[1M]^2[1M]\]or \[Rate = K\] Now change the concentrations the way the problems asks for, and see what that does to the value of R\[Rate = K[2M]^2[.5M]\]