# How is this pressure problem solved?

##### 1 Answer

#### Explanation:

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

#"A"_ (2(g)) -> "B"_ ((g)) + 1/2"C"_ ((g))#

Notice the **for every mole** of *consumed* by the reaction, you get **mole** of **moles** of

Now, let's take **initial number of moles** of **consumed** by the reaction in **minutes**.

Assuming that at the start of the reaction the vessel only contains **after**

#n_(A_2) = (n_0 - x)color(white)(a)"moles A"_2#

#n_B = 0 + x = xcolor(white)(a)"moles B"#

#n_C = 0 + 1/2x = (1/2x)color(white)(a)"moles C"#

When **volume and temperature** are kept constant, pressure and number of moles have a **direct relationship**. If you take **after**

#P_0/n_0 = P_f/( (n_0 -x) + x + 1/2x)#

#P_0/n_0 = P_f/(n_0 + 1/2x)#

This can be rearranged as

#n_0/(n_0 + 1/2x) = P_0/P_f#

Now, I'm not really sure what pressures of

#n_0/(n_0 + 1/2x) = (100 color(red)(cancel(color(black)("mm"))))/(120color(red)(cancel(color(black)("mm"))))#

This will get you

#12n_0 = 10n_0 + 5x#

#x = 2/5n_0#

So, the **number of moles** of *decreases* by **minutes**, which means that in **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

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

Since you're supposed to express the rate of disappearance in *moles*, i.e. for

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 **decreasing** as the reaction proceeds.