# How is this pressure problem solved?

## May 25, 2016

-(d["A"_2])/(dt) = "8 mm min"^(-1)

#### Explanation:

Your starting point here will be the balanced chemical equation that describes this reaction

${\text{A"_ (2(g)) -> "B"_ ((g)) + 1/2"C}}_{\left(g\right)}$

Notice the for every mole of ${\text{A}}_{2}$ consumed by the reaction, you get $1$ mole of $\text{B}$ and $\frac{1}{2}$ moles of $\text{C}$.

Now, let's take ${n}_{0}$ to be the initial number of moles of ${\text{A}}_{2}$. Let's take $x$ to be the number of moles of ${\text{A}}_{2}$ consumed by the reaction in $5$ minutes.

Assuming that at the start of the reaction the vessel only contains ${\text{A}}_{2}$, you can say that after $5$ minutes, the reaction vessel will contain

${n}_{{A}_{2}} = \left({n}_{0} - x\right) \textcolor{w h i t e}{a} {\text{moles A}}_{2}$

${n}_{B} = 0 + x = x \textcolor{w h i t e}{a} \text{moles B}$

${n}_{C} = 0 + \frac{1}{2} x = \left(\frac{1}{2} x\right) \textcolor{w h i t e}{a} \text{moles C}$

When volume and temperature are kept constant, pressure and number of moles have a direct relationship. If you take ${P}_{f}$ to be the pressure in the vessel after $5$ minutes, you can say that

${P}_{0} / {n}_{0} = {P}_{f} / \left(\left({n}_{0} - x\right) + x + \frac{1}{2} x\right)$

${P}_{0} / {n}_{0} = {P}_{f} / \left({n}_{0} + \frac{1}{2} x\right)$

This can be rearranged as

${n}_{0} / \left({n}_{0} + \frac{1}{2} x\right) = {P}_{0} / {P}_{f}$

Now, I'm not really sure what pressures of $\text{100 mm}$ and $\text{120 mm}$ are supposed to mean, but I'll use them as-given

${n}_{0} / \left({n}_{0} + \frac{1}{2} x\right) = \left(100 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mm"))))/(120color(red)(cancel(color(black)("mm}}}}\right)$

This will get you

$12 {n}_{0} = 10 {n}_{0} + 5 x$

$x = \frac{2}{5} {n}_{0}$

So, the number of moles of ${\text{A}}_{2}$ decreases by $\frac{2}{5} {n}_{0}$ in $5$ minutes, which means that in $1$ minute it decreased by

1color(red)(cancel(color(black)("minute"))) * ((2/5n_0)color(white)(a)"moles")/(5color(red)(cancel(color(black)("minutes")))) = (2/25n_0)color(white)(a)"moles"

This tells you that the rate of disappearance of ${\text{A}}_{2}$ is equal to

-(d["A"_2])/(dt) = (2/25n_0)color(white)(a)"moles min"^(-1)

Since you're supposed to express the rate of disappearance in ${\text{mm min}}^{- 1}$, I think that the problem wants you to use $\text{100 mm}$ as a replacement for moles, i.e. for ${n}_{0}$. . I don't really get how that would work, but let's assume that this is the case.

You will thus have

-(d["A"_2])/(dt) = 2/25 * "100 mm" = color(green)(|bar(ul(color(white)(a/a)"8 mm min"^(-1)color(white)(a/a)|)))

The negative sign is used here because the concentration of ${\text{A}}_{2}$ is decreasing as the reaction proceeds.