Question #cce86

1 Answer
Apr 16, 2017

Answer:

Here's what I got.

Explanation:

This is a great example of why units are so important in any problem.

You know that

#2"NO"_ ((g)) + "Br"_ (2(g)) rightleftharpoons 2"NOBr"_ ((g))#

By definition, #K_p# is equal to

#K_ p= ( ("NOBr")^2)/(("NO")^2 * ("Br"_2))#

Now, let's work with torr as the unit for pressure. Using just the units, you will have

#K_p = color(red)(cancel(color(black)("torr"^2)))/(color(red)(cancel(color(black)("torr"^2))) * "torr") = "torr"^(-1)#

The problem doesn't provide you with units for #K_p#, so you assumed that you have

#K_ p = "28.4 torr"^(-1)#

However, you can actually have something like this

#K_ p = "28.4 atm"^(-1)#

Let's go with this possibility. In order to be able to work with the pressure in torr, you must convert #K_p# from #"atm"^(-1)# to #"torr"^(-1)# by using the fact that

#"1 atm = 760 torr"#

You will have

#28.4color(white)(.)1/"atm" = 28.4 color(white)(.)1/"760 torr" = 28.4/760# #"torr"^(-1)#

You will now have

#28.4/760 color(red)(cancel(color(black)("torr"^(-1))))= (("NOBr")^2 )/(114^2 "torr"^2 * 154 color(red)(cancel(color(black)("torr"))))#

This will get you

#("NOBr") = sqrt(28.4/760 * 114^2color(white)(.)"torr"^2 * 154) = "273 torr"#

The answer is rounded to three sig figs.

Similarly, you can have

#K_p = "28.4 atm"^(-1)#

and convert the partial pressure from torr to atm to get

#28.4 color(red)(cancel(color(black)("atm"^(-1)))) = ("NOBr")^2/(114^2/(760^2) "atm"^2 * 154/760 color(red)(cancel(color(black)("atm"))))#

You will end up with

#("NOBr") = sqrt(28.4 * 114^2/760^2color(white)(.)"atm"^2 * 154/760) = "0.35984 atm"#

which will, once again, get you

#0.35984 color(red)(cancel(color(black)("atm"))) * "760 torr"/(1color(red)(cancel(color(black)("atm")))) = "273 torr"#

Now, my guess would be that the problem wanted to exploit the fact that we often list and use equilibrium constants without added units.

However, you must always be aware of the fact that sometimes, equilibrium constants do have units. In such cases, you must make sure that the units you're using for the partial pressures match the units used in the expression of #K_p#.