# Question c4b96

Jan 17, 2016

The reaction is second order.

#### Explanation:

As you know, the rate of a reaction is simply a measure of how the concentrations or the partial pressures of the reactants (or of the products) change per unit of time.

The rate law of the reaction, which establishes a relationship between the rate constant, $k$, the rate of the reaction, and, in your case, the partial pressure of the gas, will look like this

color(blue)("rate" = k * (P_a)^n)" ", where

${P}_{a}$ - the partial pressure of the gas
$n$ - the order of the reaction

Now, the problems provides you with information about the rate of the reaction at two different partial pressure values.

You know that the rate of the reaction is equal to

• ${\text{rate"_1 = "1.07 torr s}}^{- 1} \to$ when 5% of the reactant underwent decomposition

• ${\text{rate"_2 = "0.760 torr s}}^{- 1} \to$ when 20% of the reactant underwent decomposition

Let's take ${P}_{0}$ to be the initial partial pressure of the gas - you don't actually need to use the given value of $\text{363 torr}$.

After $5$ of the gas undergoes decomposition, you will be left with

${P}_{1} = {P}_{0} - \frac{5}{100} \cdot {P}_{0} = \frac{95}{100} \cdot {P}_{0}$

Likewise, after 20%# of the gas undergoes decomposition, you will be left with

${P}_{2} = {P}_{0} - \frac{20}{100} \cdot {P}_{0} = \frac{80}{100} \cdot {P}_{0}$

You can now write

$\text{rate"_1 = k * (95/100 * P_0)^color(blue)(n)" }$ and ${\text{ ""rate}}_{2} = k \cdot {\left(\frac{80}{100} \cdot {P}_{0}\right)}^{\textcolor{b l u e}{n}}$

Divide these two equations to get rid of the rate constant

${\text{rate"_1/"rate}}_{2} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{k}}} \cdot {\left(\frac{95}{100} \cdot {P}_{0}\right)}^{\textcolor{b l u e}{n}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{k}}} \cdot {\left(\frac{80}{100} \cdot {P}_{0}\right)}^{\textcolor{b l u e}{n}}}$

This will be equivalent to

${\text{rate"_1/"rate}}_{2} = {\left(\frac{95}{100} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{P}_{0}}}} \cdot \frac{100}{80} \cdot \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{P}_{0}}}}}\right)}^{\textcolor{b l u e}{n}}$

${\text{rate"_1/"rate}}_{2} = {\left(\frac{95}{80}\right)}^{\textcolor{b l u e}{n}}$

Take the log from both sides to get

$\log \left({\text{rate"_1/"rate}}_{2}\right) = \log \left[{\left(\frac{95}{80}\right)}^{\textcolor{b l u e}{n}}\right]$

$\log \left({\text{rate"_1/"rate}}_{2}\right) = \textcolor{b l u e}{n} \cdot \log \left(\frac{95}{80}\right)$

Therefore,

$\textcolor{b l u e}{n} = \log \frac{{\text{rate"_1/"rate}}_{2}}{\log} \left(\frac{95}{80}\right)$

Plug in your values to get

$\textcolor{b l u e}{n} = \log \frac{\left(1.07 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{torr s"^(-1)))))/(0.760 color(red)(cancel(color(black)("torr s}}^{- 1}}}}\right)}{\log \left(\frac{95}{80}\right)} = 1.991 \approx \textcolor{g r e e n}{2} \to$ the reaction is second order