What is the order for the reaction 2"N"_2"O"_5(g) -> 4"NO"_2(g) + "O"_2(g)?

What is the order of reaction? What is the concentration of ${\text{O}}_{2}$ after $10$ minutes? What is $\frac{\Delta \left[{\text{NO}}_{2}\right]}{\Delta t}$? What is the half-life of ${\text{N"_2"O}}_{5}$? What is the concentration of ${\text{N"_2"O}}_{5}$ after $100$ minutes?

Jun 11, 2017

Yes, graphing is the easiest way to do this (or this could be visually challenging!). In fact, learning how to use Excel is a very important skill. I've put a relatively short video on how to use Excel for chemistry here (which goes into graphs near the end):

1)

If you JUST put time in minutes on the $x$ axis and the concentration of ${\text{N"_2"O}}_{5}$ in $\text{M}$ on the $y$ axis in Excel, it would turn out to be not linear.

But with some trial and error...

• graphing the second order integrated rate law does not work; while it does give a positive slope, the graph is curved (not linear), and thus is of the wrong order.

• graphing the first order integrated rate law works to give a linear graph, confirming it is first order:

I've included the best fit line equation, which is given by:

bb(underbrace(overbrace(ln["N"_2"O"_5])^(y))_"ln of current conc." = underbrace(overbrace(-k)^(m))_"rate constant"overbrace(t)^(x) + underbrace(overbrace(ln["N"_2"O"_5]_0)^(b))_"ln of initial conc.")

where $k$ is the rate constant, $\left[{\text{N"_2"O}}_{5}\right]$ is the concentration of ${\text{N"_2"O}}_{5}$ in $\text{M}$, and ${\left[\text{ }\right]}_{0}$ means initial concentration. You know that $t$ means time in $\text{min}$.

From looking at the best fit line equation, here is what you can immediately get from it:

$\text{slope} = - k$

$\implies k = - \text{slope}$

$\implies \textcolor{b l u e}{k = {\text{0.0301 min}}^{- 1}}$

$\text{y-int.} = - 4.3894$

=> color(blue)(["N"_2"O"_5]_0) = e^("y-int.") = e^(-4.3894) = color(blue)("0.0124 M")

And furthermore, the rate law for this would therefore be (from knowing what the order is):

$r \left(t\right) = k \left[{\text{N"_2"O}}_{5}\right]$

2) At $t = \text{10 min}$, to get the concentration of ${\text{O}}_{2}$, we can examine the concentration of ${\text{N"_2"O}}_{5}$ remaining, $\text{0.0092 M}$, and determine the amount of ${\text{O}}_{2}$ currently made.

This may or may not be a new concept, but let's consider something called an ICE Table (initial, change, equilibrium), which is normally a way to track the changes in concentration in a reaction on its way to "equilibrium", the point when the reaction has the same rate forwards and backwards.

In the ICE Table below, the coefficients in front of $x$ correspond to the coefficients in the chemical reaction.

$2 {\text{N"_2"O"_5(g) -> 4"NO"_2(g) + "O}}_{2} \left(g\right)$

$\text{I"" ""0.0124 M"" "" ""0 M"" "" "" ""0 M}$
$\text{C"" "-2x" "" "" "+4x" "" "" } + x$
$\text{E"" ""0.0092 M"" "" "4x" "" "" "" } x$

Well, we can track the progress of the reaction using the reaction quotient, $Q$, even though it hasn't reached equilibrium yet. It's the same kind of definition, but we are using the current concentrations instead.

Since we know that the change in concentration of ${\text{N"_2"O}}_{5}$ was $2 x = \text{0.0032 M}$, we know that $\textcolor{b l u e}{x = \text{0.0016 M}}$ is the change in concentration of ${\text{O}}_{2}$.

And since ${\text{O}}_{2}$ started at $\text{0 M}$ (i.e. the reaction started with only reactant), that is its concentration at $\boldsymbol{\text{10 min}}$!

3) The initial rate of production of ${\text{NO}}_{2}$ comes from the rate law, since we know the rate constant and the starting concentration of ${\text{N"_2"O}}_{5}$:

$r \left(t\right) = k \left[{\text{N"_2"O}}_{5}\right]$

$r \left(t\right) = 3.735 \times {10}^{- 4} \text{M/min" = |overbrace(-)^"consumption"1/2 overbrace((Delta["N"_2"O"_5])/(Deltat))^"Initial rate} |$

$= {\overbrace{+}}^{\text{production"1/4 overbrace((Delta["NO"_2])/(Deltat))^"Initial rate}}$

So, it is:

color(blue)((Delta["NO"_2])/(Deltat)) = 4/2 xx 3.735 xx 10^(-4) "M/min"

$= \textcolor{b l u e}{7.47 \times {10}^{- 4} \text{M/min}}$

4) The half-life is known to be at the point where ${\left[{\text{N"_2"O"_5] = 1/2["N"_2"O}}_{5}\right]}_{0}$, i.e. when the reactant concentration has halved. We would obtain this from the integrated rate law:

$\ln {\left[{\text{N"_2"O"_5] = -kt + ln["N"_2"O}}_{5}\right]}_{0}$

First, plug in ${\left[{\text{N"_2"O"_5] = 1/2["N"_2"O}}_{5}\right]}_{0}$:

ln(1/2["N"_2"O"_5]_0) = -kt_"1/2" + ln["N"_2"O"_5]_0

Now use the property that $\ln a - \ln b = \ln \left(\frac{a}{b}\right)$:

ln\frac(1/2cancel(["N"_2"O"_5]_0))(cancel(["N"_2"O"_5]_0)) = -kt_"1/2"

$\implies - \ln 2 = - k {t}_{\text{1/2}}$

Thus, the first-order half-life is given by:

$\implies \textcolor{g r e e n}{{t}_{\text{1/2}} = \frac{\ln 2}{k}}$

And now, we can get:

$\textcolor{b l u e}{{t}_{\text{1/2") = (ln2)/("0.0301 min}}^{- 1}}$

$=$ $\textcolor{b l u e}{\text{23.03 min}}$

5) At $\text{100 min}$, then, we should expect that

$\text{100 mins passed"/"23.03 min half-life} \approx 4.34$ half-lives passed.

Therefore, the initial concentration ${\left[{\text{N"_2"O}}_{5}\right]}_{0}$ should have halved $4.34$ times, or been multiplied by $\frac{1}{2} \cdots \frac{1}{2} = {\left(\frac{1}{2}\right)}^{4.34} = \frac{1}{{2}^{4.34}}$.

This means the concentration after $\text{100 min}$ is approximately:

color(blue)(["N"_2"O"_5]_("100 min.")) = "0.0124 M" xx 1/(2^(4.34)) ~~ color(blue)(6.11 xx 10^(-4) "M")

(or $0.0611 \times {10}^{- 2} \text{M}$.)

Jun 11, 2017

Kinetic Rate Law for Decomposition of ${N}_{2} {O}_{5}$
$\text{Rate} = 0.100 {\min}^{-} 1 {\left[{N}_{2} {O}_{5}\right]}^{1}$
Kinetics of ${N}_{2} {O}_{5}$ Decomposition: