In an equation element "A"(s)rightleftharpoons 2"B"(g)+"C"(g)+3"D"(g). If the partial pressure of "D" at equilibrium is P_1, calculate the partial pressures of "B" and "C" . Also, calculate the value of K_p in terms of P_1?

Jun 1, 2017

Here's what I got.

Explanation:

You know that you have

${\text{A"_ ((s)) rightleftharpoons color(red)(2)"B"_ ((g)) + "C"_ ((g)) + color(blue)(3)"D}}_{\left(g\right)}$

This tells you that for every $\textcolor{b l u e}{3}$ moles of $\text{D}$ that are produced by the reaction, you also get

• $\textcolor{red}{2}$ $\text{moles of B}$
• $\text{1 mole of C}$

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

Consequently, you can say that the partial pressure of a gas that's part of a gaseous mixture depends on the mole fraction of said gas and on the total pressure of the mixture $\to$ this is known as Dalton's Law of Partial Pressures.

If you take ${P}_{\text{total}}$ to be the total pressure of the mixture, i.e. of $\text{B}$, $\text{C}$, and $\text{D}$, you can say that the partial pressure of $\text{D}$, let's say ${P}_{\text{D}}$, is equal to

${P}_{\text{D" = overbrace( (color(blue)(3)color(red)(cancel(color(black)("moles"))))/((color(red)(2) + 1 + color(blue)(3))color(red)(cancel(color(black)("moles")))))^(color(purple)("the mole fraction of D")) * P_"total}}$

${P}_{\text{D"= 3/6 * P_"total}}$

${P}_{\text{D" = 1/2 * P_"total}}$

Rearrange to get the value of ${P}_{\text{total}}$ in terms of ${P}_{1}$ and use the fact tha ${P}_{\text{D}} = {P}_{1}$

${P}_{\text{total}} = 2 \cdot {P}_{1}$

Next, use this value to find an expression for the partial pressure of $\text{B}$, let's say ${P}_{\text{B}}$, and the partial pressure of $\text{C}$, let's say ${P}_{\text{C}}$.

You will have

${P}_{\text{B" = overbrace( (color(red)(2)color(red)(cancel(color(black)("moles"))))/((color(red)(2) + 1 + color(blue)(3))color(red)(cancel(color(black)("moles")))))^(color(purple)("the mole fraction of B")) * P_"total}}$

${P}_{\text{B" = 2/6 * P_"total}}$

${P}_{\text{B" = 1/3 * P_"total}}$

And so

${P}_{\text{B}} = \frac{1}{3} \cdot \left(2 \cdot {P}_{1}\right) = \frac{2}{3} \cdot {P}_{1}$

Similarly, you will have

${P}_{\text{C" = overbrace( (1color(red)(cancel(color(black)("mole"))))/((color(red)(2) + 1 + color(blue)(3))color(red)(cancel(color(black)("moles")))))^(color(purple)("the mole fraction of C")) * P_"total}}$

${P}_{\text{C" = 1/6 * P_"total}}$

And so

${P}_{\text{C}} = \frac{1}{6} \cdot \left(2 \cdot {P}_{1}\right) = \frac{1}{3} \cdot {P}_{1}$

You can thus say that you have

$\left\{\begin{matrix}{P}_{\text{B" = 2/3 * P_1 \\ P_"C" = 1/3 * P_1 \\ P_"D}} = {P}_{1}\end{matrix}\right.$

Finally, the equilibrium constant for this equilibrium can be written using the equilibrium partial pressures of the three gases

${K}_{p} = {\left({P}_{\text{B")^color(red)(2) * P_"C" * (P_"D}}\right)}^{\textcolor{b l u e}{3}}$

Plug in your values to find

${K}_{p} = {\left(\frac{2}{3} \cdot {P}_{1}\right)}^{\textcolor{red}{2}} \cdot \frac{1}{3} \cdot {P}_{1} \cdot {\left({P}_{1}\right)}^{\textcolor{b l u e}{3}}$

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{K}_{p} = \frac{4}{27} \cdot {\left({P}_{1}\right)}^{6}}}}$