# Question 07e8f

Jun 17, 2016

The reaction is second-order.

#### Explanation:

You're dealing with a decomposition reaction of the form

$\text{AB " -> " products}$

for which the differential rate law can be written as

"rate" = k * ["AB"]^n

Here

$k$ - the rate constant for the reaction
$n$ - the order of the reaction

According to experimental data, you have

Pick any two entries and plug the data into the differential rate law

2.75 * 10^(-8)"mol dm"^(-3)"s"^(-1) = k * ("0.20 mol dm"^(-3))^n

11.0 * 10^(-8) "mol dm"^(-3)"s"^(-1) = k * ("0.40 mol dm"^(-3)")^n#

You can get rid of the unknown rate constant by dividing these two equations

$\left(2.73 \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{10}^{- 8}}}} \textcolor{red}{\cancel{{\textcolor{b l a c k}{{\text{mol dm"^(-3)"s"^(-1)))))/(11.0 * color(red)(cancel(color(black)(10^(-8))))color(red)(cancel(color(black)("mol dm"^(-3)"s"^(-1))))) = color(red)(cancel(color(black)(k)))/color(red)(cancel(color(black)(k))) * (0.20)^n/(0.40)^n color(red)(cancel(color(black)(("mol dm"^(-3))^n)))/color(red)(cancel(color(black)(("mol dm}}^{- 3}}}^{n}}}\right)$

$\frac{2.75}{11.0} = {\left(\frac{0.20}{0.40}\right)}^{n}$

This is equivalent to

$\ln \left(\frac{2.75}{11.0}\right) = \ln \left[{\left(\frac{0.20}{0.40}\right)}^{n}\right]$

Solve for $n$ to get

$\ln \left(\frac{1}{4}\right) = n \cdot \ln \left(\frac{1}{2}\right)$

$\ln \left[{\left(\frac{1}{2}\right)}^{2}\right] = n \cdot \ln \left(\frac{1}{2}\right)$

$2 \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{\ln \left(\frac{1}{2}\right)}}} = n \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{\ln \left(\frac{1}{2}\right)}}}$

Therefore,

$n = 2$

and your decomposition reaction is second-order.

SIDE NOTE The order of the reaction must come out the same regardless of what data you chose to use in the calculations.

I recommend using all the possible combinations of data sets to see that this is the case.

$\ln \left(\frac{24.75}{11.0}\right) = n \cdot \ln \left(\frac{0.60}{0.40}\right)$

$\ln \left(\frac{2.75}{24.75}\right) = n \cdot \ln \left(\frac{0.20}{0.60}\right)$