Question #9ef8a

1 Answer
Mar 17, 2017

Answer:

Here;s what I got.

Explanation:

You can't really calculate the value of #K_p# because you're missing the temperature at which the reaction takes place.

You can calculate the value of #K_c#, but you must know the temperature at which the reaction takes place in order to be able to calculate the value of #K_p#.

So, for the equilibrium reaction

#2"SO"_ (2(g)) + "O"_ (2(g)) rightleftharpoons 2"SO"_ (3(g))#

you have

#K_c = (["SO"_3]^2)/( ["SO"_2]^2 * ["O"_2])#

As you know, the expression of #K_c# uses the equilibrium concentrations of the three chemical species.

In your case, you will have

#K_c = ("0.259 M")^2/( ("0.59 M")^2 * "0.05 M")#

#K_c = (0.259^2 color(red)(cancel(color(black)(("mol L"^(-1))^2))))/(0.59^2 color(red)(cancel(color(black)(("mol L"^(-1))^2))) * "0.05 mol L"^(-1))#

#K_c = "3.854 mol"^(-1)"L"#

Now, you should know that

#color(blue)(ul(color(black)(K_p = K_c * (RT)^(Deltan))))#

Here

  • #R# is the universal gas constant, equal to #0.0821("atm" * "L")/("mol" * "K")#
  • #T# is the absolute temperature at which the reaction takes place
  • #Deltan# is the difference between the total number of moles on the products' side and the total number of moles on the reactants' side

In your case, you have

  • #"2 moles SO"_2 + "1 mole O"_2 = "3 moles gas"#

You have two moles of sulfur dioxide reacting with one mole of oxygen gas on the reactants' side

  • #"2 moles of SO"_3 = "2 moles of gas"#

You have two moles of sulfur trioxide being produced on the products' side

This means that

#Deltan = n_"total products" - n_"total reactants"#

will be equal to

#Deltan = 2 - 3 = -1#

You can now say that #K_p# will be equal to

#K_p = K_c * (RT)^(-1)#

If you take #T# #"K"# to be the temperature at which the reaction takes place, you can say that

#K_p = "3.854 mol"^(-1)"L" * (0.0821 ("atm" * "L")/("mol" * color(red)(cancel(color(black)("K")))) * Tcolor(white)(.) color(red)(cancel(color(black)("K"))))^(-1)#

#K_p = 3.854 color(red)(cancel(color(black)("mol"^(-1)))) color(red)(cancel(color(black)("L"))) * 1/(0.0821 * T) color(red)(cancel(color(black)("mol")))/("atm" * color(red)(cancel(color(black)("L"))))#

which gets you

#color(darkgreen)(ul(color(black)(K_p = (46.9/T)color(white)(.)"atm"^(-1))))#

All you have to do to get the actual value of #K_p# is to plug in the value you have for the absolute temperature at which the reaction takes place.

Now, does this result make sense?

Notice that in this case, #K_p# is equal to

#K_p = (("SO"_3)^2)/(("SO"_2)^2 * ("O"_2))#

Keep in mind that the expression for #K_p# uses the equilibrium partial pressures of the three gases.

If you measure the partial pressures of the three gases in atmospheres, you will have -- using only units

#K_p = ( color(red)(cancel(color(black)("atm"^2))))/(color(red)(cancel(color(black)("atm"^2))) * "atm") = "atm"^(-1)#

This means that, at least from a dimensional point of view, the answer makes sense.