How is this pressure problem solved?

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1 Answer
May 25, 2016

Answer:

#-(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.