For the following reaction, find the rate law and rate constant? #2"NO"(g) + "O"_2(g) -> 2"NO"_2(g)#

Alright, let's just copy this down here:

#color(white)(aaaaaaaaa)2 "NO"(g) + "O"_2(g) stackrel(" "" "k" "" ")(->) 2 "NO"_2(g)#

#color(white)([(color(black)("Exp. No."), color(black)(["NO"]("M")), color(black)(["O"_2]("M")), color(black)("Init. Rate (""M/s)")),(color(black)(1),color(black)(0.0126),color(black)(0.0125),color(black)(1.41xx10^(-2))), (color(black)(2),color(black)(0.0252),color(black)(0.0250),color(black)(1.13xx10^(-1))), (color(black)(3),color(black)(0.0252),color(black)(0.0250),color(black)(5.64xx10^(-2)))])#

THE RELEVANT PART OF THE RATE LAW

The part of the rate law that is relevant is gotten from this relationship:

#\mathbf(r(t) = k[A]^x[B]^y)#

where #x# and #y# are the orders of the reactants #"A"# and #"B"#, and #k# is the rate constant in whatever units gives #r(t)#, the initial rate, the units of #"M/s"#.

We are considering this scenario: if you change the concentration, how does the rate change in response?

DETERMINING THE ORDER OF EACH REACTANT

You are given the concentrations for different trials and the corresponding rates, so this is already done for you.

What's helpful is keeping the concentration of one reactant constant while changing the other only. So, the trick is to look at the right trials.

CHANGE IN #["NO"]#

Now, let's look at trials #1# and #3# to understand the change in #["NO"]#.

#Delta["O"_2] = 0#, but #["NO"]# doubled. The rate, as a result, quadrupled (#1.41xx10^(-2) -> 5.64xx10^(-2)#).

That means #"NO"# is a second-order reactant. How do I know that? If #["NO"] = x#, then #2["NO"] = 2x#:

#r(t) = k(x^2)["O"_2]^y#

#=> color(green)(4)r(t) = k(2x)^2["O"_2]^y = color(green)(4)k(x^2)["O"_2]^y#

CHANGE IN #["O"_2]#

Similarly, we can do this for #"O"_2#.

Looking at trials #3# and #2#, we can see that #Delta["NO"] = 0#, but doubling #["O"_2]# pretty much doubles the rate (#5.64xx10^(-2) -> 1.13xx10^(-1)#).

The one-to-one relationship of that change says that #"O"_2# is a first-order reactant. You can do a similar calculation check as we did above for #"NO"#.

THE RATE LAW?

Now, we know that the rate law is:

#color(blue)(r(t) = k["NO"]^2["O"_2])#

As an aside, the part of the rate law that we don't need to use for this problem is:

#color(green)(r(t) = -1/2(d["NO"])/(dt) = -(d["O"_2])/(dt) = 1/2(d["NO"_2])/(dt))#

THE RATE CONSTANT

Now we know everything we need to determine #k#. Pick a trial and plug in numbers. I'm picking trial #1#.

#r(t) = k["NO"]^2["O"_2]#

#1.41xx10^(-2) "M/s" = k("0.126 M")^2("0.125 M")#

#k = (1.41xx10^(-2) cancel("M")"/s")/(("0.126 M")^2("0.125" cancel"M"))#

#k# therefore has units of #1/("M"^2*"s")#. Can you see why by cancelling the units yourself?

#color(blue)(k ~~ "7.11 "1/("M"^2*"s"))#