# For the reaction #"N"_2"O"_3(g) -> "NO"(g) + "NO"_2(g)#, given the following data, calculate the rate constant?

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#"N"_2"O"_3(g) -> "NO"(g) + "NO"_2(g)#

#["N"_2"O"_3]" "" Initial Rate"#

#"0.1 M" " "" "r(t) = 0.66# #"M/s"#

#"0.2 M" " "" "r(t) = 1.32# #"M/s"#

#"0.3 M" " "" "r(t) = 1.98# #"M/s"#

#"N"_2"O"_3(g) -> "NO"(g) + "NO"_2(g)#

#["N"_2"O"_3]" "" Initial Rate"#

#"0.1 M" " "" "r(t) = 0.66# #"M/s"#

#"0.2 M" " "" "r(t) = 1.32# #"M/s"#

#"0.3 M" " "" "r(t) = 1.98# #"M/s"#

##### 1 Answer

So, we have:

#"N"_2"O"_3(g) -> "NO"(g) + "NO"_2(g)#

#["N"_2"O"_3]" "" Initial Rate"#

#"0.1 M" " "" "r(t) = 0.66# #"M/s"#

#"0.2 M" " "" "r(t) = 1.32# #"M/s"#

#"0.3 M" " "" "r(t) = 1.98# #"M/s"#

Since you were given only the concentrations of the reactant, we are of course focusing on what happens when changing the reactant concentration.

Notice how **doubling** **doubles** the rate, and **tripling** **triples** the rate. i.e.

#2xx["N"_2"O"_3] harr 2 xx 0.66# #"M/s"#

#3xx["N"_2"O"_3] harr 3 xx 0.66# #"M/s"#

We can make the relationship saying that:

#((["N"_2"O"_3])/(["N"_2"O"_3]_0))^m = (r(t))/(r_0(t)# where

#(["N"_2"O"_3])/(["N"_2"O"_3]_0) = 2# for theratio of the new concentration to the old concentration of the reactantand#(r(t))/(r_0(t)) = 2# for theratio of the new rate to the old rate of reaction.#m# is the order of the reactant.

This is to compare how changing the concentration affects the rate. In examining trials 1 and 2, we see:

#2^m = 2 -> m = 1#

So the reactant is *first order*, and since it is the only reactant, the overall reaction is **first order**.

Now we can write the rate law as:

#\mathbf(r(t) = k["N"_2"O"_3])#

Thus, to get **same trial** as each other, and solve for

#0.66# #"M/s" = kxx"0.1 M"#

#color(blue)(k = 6.6)# #color(blue)("s"^(-1))#

And if you pick trials 2 or 3 for the calculation of