# For a reaction whose mechanism is given below with initial rate #"0.0030 M/s"#, #(a)# what is the new rate if #["A"]# is doubled, #(b)# if #["B"]# is doubled, and #(c)# if both #["A"]# and #["B"]# are multiplied by #5#?

##
Overall reaction:

#A + C -> D#

Mechanism:

#A => B# (#"slow"# )

#B + C => D# (#"fast"# )

Overall reaction:

Mechanism:

##### 1 Answer

Well, we know that the rate-limiting step is unimolecular with

Single mechanistic steps are **elementary reactions**, for which the stoichiometric coefficients give the order with respect to the reactants involved.

If such a mechanistic step is *slow*, then it gives you the rate law directly. This rate law has already incorporated the behavior of

So, the **rate law** can be written as:

#r(t) = k[A]#

Thus, if

#color(red)(2) xx r(t) = k xx color(red)(2) xx [A]#

#color(blue)(r(t)' = 2r(t) = "0.0060 M/s")#

If **zero order** with respect to

Hence, doubling

#color(blue)(r(t)' ~~ r(t) = "0.0030 M/s")#

given the original concentrations.

Again, as in

#color(red)(5) xx r(t) = k xx color(red)(5) xx [A]cancel([B]^(0))^(1)#

Evidently the reaction rate then multiplies by

#color(blue)(r(t)' = 5r(t) = "0.0150 M/s")#