# 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?

Jul 24, 2016

${K}_{p} = 25.3$

#### Explanation:

The equilibrium reaction given to you looks like this

${\text{Xe"_ ((g)) + color(red)(2)"F"_ (2(g)) rightleftharpoons "XeF}}_{4 \left(g\right)}$

You know that the mixture initially contains $\text{2.24 atm}$ of xenon, $\text{Xe}$, and $\text{4.27 atm}$ of fluorine gas, ${\text{F}}_{2}$.

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 $\text{0.34 atm}$. This is a significant decrease, which can only mean that the equilibrium constant, ${K}_{p}$, is greater than $1$.

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

${\text{ ""Xe"_ ((g)) " "+" " color(red)(2)"F"_ (2(g)) " "rightleftharpoons" " "XeF}}_{4 \left(g\right)}$

color(purple)("I")color(white)(aaaaacolor(black)(2.24)aaaaaaaacolor(black)(4.27)aaaaaaaaaaacolor(black)(0)
color(purple)("C")color(white)(aaacolor(black)((-x))aaaaaacolor(black)((-color(red)(2)x))aaaaaaaaacolor(black)((+x))
color(purple)("E")color(white)(aaacolor(black)(2.24-x)aaaacolor(black)(4.27-color(red)(2)x)aaaaaaaaacolor(black)(x)

Now, you know that the equilibrium concentration of xenon is $\text{0.34 atm}$. This means that you have

$2.24 - x = 0.34 \implies x = 1.90$

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"

By definition, ${K}_{p}$ will be equal to

${K}_{p} = \left({\left({\text{XeF"_4))/(("Xe") * ("F}}_{2}\right)}^{\textcolor{red}{2}}\right)$

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)

${K}_{p} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{25.3} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ rounded to three sig figs
As predicted, the equilibrium constant turned out to be grater than $1$.