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

##
*Be sure to expand this description!*

- 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 iskeeping 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

**doubled**. The rate, as a result, **quadrupled** (

That means **second-order reactant**. How do I know that? If

#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

Looking at trials **doubling** **doubles** the rate (

The one-to-one relationship of that change says that **first-order reactant**. You can do a similar calculation check as we did above for

**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"))#