Question

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

`"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"`

Answer 1

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 `["N"_2"O"_3]` doubles the rate, and tripling `["N"_2"O"_3]` 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 the ratio of the new concentration to the old concentration of the reactant and `(r(t))/(r_0(t)) = 2` for the ratio 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 `k`, simply pick a value of `r(t)` and `["N"_2"O"_3]`, both of which are in the same trial as each other, and solve for `k`:

`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 `k`, you will get the same thing.