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 $P_1$?

You know that you have

$"A"_ ((s)) rightleftharpoons color(red)(2)"B"_ ((g)) + "C"_ ((g)) + color(blue)(3)"D"_ ((g))$

This tells you that for every $color(blue)(3)$ moles of $"D"$ that are produced by the reaction, you also get

• $color(red)(2)$ $"moles of B"$
• $"1 mole of C"$

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 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 $->$ this is known as Dalton's Law of Partial Pressures.

If you take $P_"total"$ to be the total pressure of the mixture, i.e. of $"B"$, $"C"$, and $"D"$, you can say that the partial pressure of $"D"$, let's say $P_"D"$, is equal to

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

$P_"D"= 3/6 * P_"total"$

$P_"D" = 1/2 * P_"total"$

Rearrange to get the value of $P_"total"$ in terms of $P_1$ and use the fact tha $P_"D" = P_1$

$P_"total" = 2 * P_1$

Next, use this value to find an expression for the partial pressure of $"B"$, let's say $P_"B"$, and the partial pressure of $"C"$, let's say $P_"C"$.

You will have

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

$P_"B" = 2/6 * P_"total"$

$P_"B" = 1/3 * P_"total"$

And so

$P_"B" = 1/3 * (2 * P_1) = 2/3 * P_1$

Similarly, you will have

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

$P_"C" = 1/6 * P_"total"$

And so

$P_"C" = 1/6 * (2 * P_1) = 1/3 * P_1$

You can thus say that you have

${(P_"B" = 2/3 * P_1), (P_"C" = 1/3 * P_1), (P_"D" = P_1) :}$

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

$K_p = (P_"B")^color(red)(2) * P_"C" * (P_"D")^color(blue)(3)$

Plug in your values to find

$K_p = (2/3 * P_1)^color(red)(2) * 1/3 * P_1 * (P_1)^color(blue)(3)$

$color(darkgreen)(ul(color(black)(K_p = 4/27 * (P_1)^6)))$