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#?

Question

Olivia Anton · Accepted Answer

Here's what I got.

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 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 #-># 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)))#

Ellie Andrews · Answer

undefined To find the partial pressures of B and C, you can use the stoichiometry of the reaction. Since 1 mole of A produces 3 moles of D, the partial pressure of D at equilibrium (P1) is related to the initial pressure of A (PA0) by P1 = 3PA0. Therefore, the partial pressures of B and C are both equal to half of P1, which is P1/2.

The expression for Kp is calculated using the partial pressures of the products and reactants raised to their stoichiometric coefficients. For this reaction, Kp = (P(B)^2 * P(C)^1) / (P(D)^3). Substituting the partial pressures of B, C, and D, you get Kp = ((P1/2)^2 * (P1/2)^1) / (P1^3) = P1^(-4).