# Question b23fd

Oct 6, 2015

${K}_{p} = {\text{0.0407 atm}}^{- 1}$

#### Explanation:

You can derive an expression for ${K}_{p}$ using the ideal gas law and ${K}_{c}$.

Your equilibrium reaction looks like this

${\text{PbCl"_text(3(g]) + "Cl"_text(2(g]) rightleftharpoons "PbCl}}_{\textrm{5 \left(g\right]}}$

By definition, what is ${K}_{c}$ equal to?

${K}_{c} = \left(\left[{\text{PbCl"_5])/( ["PbCl"_3] * ["Cl}}_{2}\right]\right)$

The equilibrium concentrations featured in the expression of ${K}_{c}$ can be written as a ratio between the number of moles of each species that takes part in the rection and the volume of the container in which said reaction takes place.

If you take $V$ to be the volume of the container, you can say that

$\left[P b C {l}_{3}\right] = {n}_{\textrm{P b C {l}_{3}}} / V \text{ }$ and $\text{ } \left[C {l}_{2}\right] = {n}_{C {l}_{2}} / V$

$\left[P b C {l}_{5}\right] = {n}_{P b C {l}_{5}} / V$

Now take a look at the expression for ${K}_{p}$

${K}_{p} = \frac{\left(P b C {l}_{5}\right)}{\left(P b C {l}_{3}\right) \cdot \left(C {l}_{2}\right)}$

The partial pressure of each component of the mixture can be expressed using the ideal gas law

$P V = n R T \implies P = \frac{n R T}{V} = \frac{n}{V} \cdot R T$

This means that you have

$\left(P b C {l}_{5}\right) = {\underbrace{{n}_{P b C {l}_{5}} / V}}_{\textcolor{b l u e}{= \left[P b C {l}_{5}\right]}} \cdot R T$

$\left(C {l}_{2}\right) = {\underbrace{{n}_{C {l}_{2}} / V}}_{\textcolor{b l u e}{= \left[C {l}_{2}\right]}} \cdot R T$

$\left(P b C {l}_{3}\right) = {\underbrace{{n}_{P b C {l}_{3}} / V}}_{\textcolor{b l u e}{= \left[P b C {l}_{3}\right]}} \cdot R T$

This means that you can write

${K}_{p} = \frac{\left[P b C {l}_{5}\right] \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{R T}}}}{\left[C {l}_{2}\right] \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{R T}}} \cdot \left[P b C {l}_{3}\right] \cdot R T}$

${K}_{p} = {\overbrace{\frac{\left[P b C {l}_{5}\right]}{\left[P b C {l}_{3}\right] \cdot \left[C {l}_{2}\right]}}}^{\textcolor{g r e e n}{= {K}_{c}}} \cdot \frac{1}{R T}$

${K}_{p} = {K}_{c} \cdot {\left(R T\right)}^{- 1}$

If you take $R = 0.082 \left(\text{atm" * "L")/("mol" * "K}\right)$, and knowing that your reaction takes plac at $\text{500 K}$, you get

${K}_{p} = {K}_{c} \cdot {\left[0.082 \left(\text{atm" * "L")/("mol} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{K}}}\right) \cdot 500 \textcolor{red}{\cancel{\textcolor{b l a c k}{K}}}\right]}^{- 1}$

K_p = K_c * 0.0244"mol"/("atm" * "L")

Well, if you look at the expression for ${K}_{c}$, you will see that it is not unitless, because
K_c = ( [PbCl_5]color(red)(cancel(color(black)("M"))))/( [PbCl_3]color(red)(cancel(color(black)("M"))) * [Cl_2]" M") = [1/"M"]
SInce $\text{M}$ is $\text{mol"/"L}$, you get that
K_p = 1.67 color(red)(cancel(color(black)("L")))/color(red)(cancel(color(black)("mol"))) * 0.0244 color(red)(cancel(color(black)("mol")))/("atm" * color(red)(cancel(color(black)("L")))) = color(green)("0.0407 atm"^(-1))#