Experimental rate law, PLEASE HELP?
The experimental rate law for the reaction 3A + B -> 3C is
rate = k[A]^2[B].
What will happen to the reaction rate if the concentration of A is doubled?
More Questions From This User See All
Answers & Comments
Verified answer
rate = k[A]^2[B]
If concentration of A is doubled then the reaction rate = k[2A]^2[B]
rate = 4k[A]^2[B]
Thus, the reaction rate will increase 4 times greater than the initial rate
very confusing aspect. look into from search engines like google. this could actually help!