Consider the following reaction: #Xe(g) + 2F_2(g) -> XeF_4(g)#. A reaction mixture initially contains 2.24 atm #Xe# and 4.27 atm #F_2#. lf the equilibrium pressure of #Xe# is 0.34 atm, how do you find the equilibrium constant (#K_p#) for the reaction?
The equilibrium reaction given to you looks like this
#"Xe"_ ((g)) + color(red)(2)"F"_ (2(g)) rightleftharpoons "XeF"_ (4(g))#
You know that the mixture initially contains
It's worth mentioning that the problem provides you with pressures instead of moles because when the volume of the container remains unchanged, and the temperature of the reaction is kept constant, the pressure of a gas is directly proportional to the number of moles of gas present.
Now, notice that the pressure of xenon is decreasing to an equilibrium value of
In other words, the forward reaction is favored at the temperature at which the reaction takes place.
Consequently, you can expect the equilibrium pressure of fluorine gas to be significantly lower than its initial value.
So, use an ICE table to find the equilibrium pressures of fluorine gas and xenon tetrafluoride
#" ""Xe"_ ((g)) " "+" " color(red)(2)"F"_ (2(g)) " "rightleftharpoons" " "XeF"_ (4(g))#
Now, you know that the equilibrium concentration of xenon is
#2.24 - x = 0.34 implies x = 1.90#
The equilibrium pressures of fluorine gas and xenon tetrafluoride will thus be
#("F"_2) = 4.27 - color(red)(2) * 1.90 = "0.47 atm"#
#("XeF"_4) = "1.90 atm"#
#K_ p = (("XeF"_4))/(("Xe") * ("F"_2)^color(red)(2))#
Plug in your values to find
#K_p = (1.90 color(red)(cancel(color(black)("atm"))))/(0.34color(red)(cancel(color(black)("atm"))) * ("0.47 atm")^color(red)(2)) = "25.3 atm"^(-2)#
Equilibrium constant are usually given without added units, which means that your answer will be
#K_p = color(green)(|bar(ul(color(white)(a/a)color(black)(25.3)color(white)(a/a)|))) ->#rounded to three sig figs
As predicted, the equilibrium constant turned out to be grater than