Question #c2cfc

You know that when the following reaction takes place at #"500 K"#

#color(blue)(2)"A"_ ((g)) rightleftharpoons "B"_ ((g))#

the equilibrium constant is equal to

#K_p = 2.77 * 10^(-5)#

Even without doing any calculations, the fact that you have #K_p < 1# tells you that you should expect the equilibrium partial pressure of #"B"# to be smaller than the equilibrium partial pressure of #"A"#.

Now, you know that when volume and temperature are kept constant, the pressure of a gas is directly proportional to the number of moles of gas present in the sample.

Notice that it takes #color(blue)(2)# moles of #"A"# to produce #1# mole of #"B"#. This is equivalent to saying that in order for the partial pressure of #"B"# to increase by a value #x#, the partial pressure of #"A"# must decrease by #color(blue)(2)x#.

You can thus say that once equilibrium is reached, you will have

#0 + x = x -># the equilibrium partial pressure of #"B"#

#5.40 - color(blue)(2)x -># the equilibrium partial pressure of #"A"#

By definition, the equilibrium constant for this reaction looks like this

#K_p = (P_"B")/(P_"A")^color(blue)(2)#

In your case, you will have

#2.77 * 10^(-5) = x/((5.40 - color(blue)(2)x)^color(blue)(2)#

Now, because the equilibrium constant is significantly smaller than the initial pressure of #"A"#, you can use the approximation

#5.40 - color(blue)(2)x ~~ 5.40#

The above equation becomes

#2.77 * 10^(-5) = x/5.40^color(blue)(2)#

Solve for #x# to find

#x = 2.77 * 10^(-5) * 5.40^color(blue)(2) = 8.08 * 10^(-4)#

Therefore, you can say that the equilibrium partial pressure of #"B"# will be

#P_ "B" = color(darkgreen)(ul(color(black)(8.08 * 10^(-4)color(white)(.)"atm")))#

The answer is rounded to three sig figs, the number of sig figs you have for the initial pressure of gas #"A"#.