# How is this chemical Kinetics problem solved? (Deals with pressure along with the concentration)

May 24, 2016

The answer is (B) $\text{2 atm}$

#### Explanation:

Start by writing the balanced chemical equation that describes this synthesis reaction

$\textcolor{red}{2} {\text{NO"_ (2(g)) + "F"_ (2(g)) -> 2"NO"_ 2"F}}_{\left(g\right)}$

Notice that the reaction consumes color(red)(2 moles of nitrogen dioxide, ${\text{NO}}_{2}$, for every mole of fluorine gas, ${\text{F}}_{2}$, that takes part in the reaction, and produces $2$ moles of nitryl fluoride, $\text{NO"_2"F}$.

Now, you don't really need to know the rate of the reaction to solve this problem. As you know, the rate of a reaction gives you information about the speed at which the reaction proceeds.

Also, you don't need to work with concentration because the volume of the vessel is constant. All you need to use here is the moles of reactants and moles of product.

In your case, you must find out the pressure in the reaction vessel after the reaction is complete, so you're not really interested in how fast the reaction will proceed.

All you need to know is that it reaches completion.

So, the two reactants are placed in the vessel in a $2 : 1$ mole ratio. If you take $V$ to be the volume of the reaction vessel and $T$ to be the temperature, you can use the ideal gas law equation to write

$\textcolor{P u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{P}_{1} \cdot V = {n}_{1} \cdot R T} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ before the reaction takes place

The initial pressure in the reaction vessel is equal to $\text{3 atm}$, so

${P}_{1} = \text{3 atm}$

If you take $x$ to be the initial number of moles of nitrogen dioxide, you can say that the reaction vessel initially contains

xcolor(white)(a)color(red)(cancel(color(black)("moles NO"_2))) * "1 mole F"_2/(2color(red)(cancel(color(black)("moles NO"_2)))) = (x/2)color(white)(a)"moles F"_2

The total number of moles of gas initially present in the reaction vessel will be

${n}_{1} = x + \frac{x}{2} = \left(\frac{3 x}{2}\right) \textcolor{w h i t e}{a} \text{moles gas}$

Now, the two reactants are present in a $2 : 1$ mole ratio and they are consumed by the reaction in a $\textcolor{red}{2} : 1$ mole ratio.

This tells you that when the reaction is finished, both reactants will be completely consumed. Since nitryl fluoride is produced in a $\textcolor{red}{2} : 2$ mole ratio with nitrogen dioxide, the vessel will contain

${n}_{2} = x \textcolor{w h i t e}{a} {\text{moles NO"_2"F}}_{2} \to$ produced by the reaction

If you take ${P}_{2}$ to be the pressure in the vessel after the reaction is complete, you can say that

$\textcolor{P u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{P}_{2} \cdot V = {n}_{2} \cdot R T} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ after the reaction is complete

Keep in mind that the volume of the vessel and the temperature are constant.

Divide these two equations to get

$\frac{{P}_{1} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{V}}}}{{P}_{2} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{V}}}} = \frac{{n}_{1} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{R T}}}}{{n}_{2} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{R T}}}}$

Rearrange to find

${P}_{2} = {n}_{2} / {n}_{1} \cdot {P}_{1}$

This will get you

P_2 = (xcolor(red)(cancel(color(black)("moles"))))/((3x)/2color(red)(cancel(color(black)("moles")))) * "3 atm"

P_2 = (2color(red)(cancel(color(black)(x))))/(3color(red)(cancel(color(black)(x)))) * "3 atm" = color(green)(|bar(ul(color(white)(a/a)"2 atm"color(white)(a/a)|)))