Question

How is this pressure problem solved?

Answer 1

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

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 `"A"_2` consumed by the reaction, you get `1` mole of `"B"` and `1/2` moles of `"C"`.

Now, let's take `n_0` to be the initial number of moles of `"A"_2`. Let's take `x` to be the number of moles of `"A"_2` consumed by the reaction in `5` minutes.

Assuming that at the start of the reaction the vessel only contains `"A"_2`, you can say that after `5` minutes, the reaction vessel will contain

`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 `P_f` to be the pressure in the vessel after `5` minutes, you can say that

`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 `"100 mm"` and `"120 mm"` are supposed to mean, but I'll use them as-given

`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 `"A"_2` decreases by `2/5n_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 `"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 `"mm min"^(-1)`, I think that the problem wants you to use `"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 `"A"_2` is decreasing as the reaction proceeds.