For the following reaction, find the rate law and rate constant? #2"NO"(g) + "O"_2(g) -> 2"NO"_2(g)#
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- Truong-Son
#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)))])#
Be sure to expand this description!
- Truong-Son
1 Answer
Alright, let's just copy this down here:
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
Now, let's look at trials
That means
#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
Similarly, we can do this for
Looking at trials
The one-to-one relationship of that change says that
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
#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"))#
#color(blue)(k ~~ "7.11 "1/("M"^2*"s"))#