Question #b1be6

1 Answer
Feb 12, 2016

Answer:

#K_p = 1.46#

Explanation:

Your goal here is to find the equilibrium constant expressed in terms of partial pressures, #K_p#, for this equilibrium reaction

#"C"_text((s]) + "CO"_text(2(g]) rightleftharpoons 2"CO"_text((g])" ", K_p = ?#

by using the equilibrium constants for these equilibrium reactions

#"C"_text((s]) + 2"H"_2"O"_text((g]) rightleftharpoons "CO"_text(2(g]) + 2"H"_text(2(g])" ", K_text(p1) = 3.65#

#"H"_text(2(g]) + "CO"_text(2(g]) rightleftharpoons "H"_2"O"_text((g]) + "CO"_text((g])" ", K_text(p2) = 0.633#

In order to do that, you must find a way to express the first reaction in terms of the two reactions for which the equilibrium constant is known - think Hess' Law.

Notice that happens when you multiply the second reaction by #color(blue)(2)#

#["H"_text(2(g]) + "CO"_text(2(g]) rightleftharpoons "H"_2"O"_text((g]) + "CO"_text((g])] xx color(blue)(2)#

#color(blue)(2)"H"_text(2(g]) + color(blue)(2)"CO"_text(2(g]) rightleftharpoons color(blue)(2)"H"_2"O"_text((g]) + color(blue)(2)"CO"_text((g])#

The equilibrium constant for this reaction is equal to

#K_(p2)^' = ( ("H"_2"O")^color(blue)(2) * ("CO")^color(blue)(2))/(("H"_2)^color(blue)(2) * ("CO"_2)^color(blue)(2))#

But since the original reaction had

#K_(2p) = (("H"_2"O") * ("CO"))/(("H"_2) * ("CO"_2))#

you can say that

#K_(2p)^' = [(("H"_2"O") * ("CO"))/(("H"_2) * ("CO"_2))]^color(blue)(2) = K_text(p2)^color(blue)(2)#

Now notice what happens when you add this reaction and the first reaction

#"C"_text((s]) + color(purple)(cancel(color(black)(2"H"_2"O"_text((g])))) rightleftharpoons "CO"_text(2(g]) + color(red)(cancel(color(black)(2"H"_text(2(g]))))#

#color(red)(cancel(color(black)(2"H"_text(2(g])))) + 2"CO"_text(2(g]) rightleftharpoons color(purple)(cancel(color(black)(2"H"_2"O"_text((g])))) + 2"CO"_text((g])#
#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaa)/(color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)#

#"C"_text((s]) + 2"CO"_text(2(g]) rightleftharpoons "CO"_text(2(g]) + 2"CO"_text((g])#

This is equivalent to

#"C"_text((s]) + "CO"_text(2(g]) rightleftharpoons 2"CO"_text((g]) -># the target reaction

The equilibrium constant will be equal to

#K_p = K_(p1) xx K_(p2)^'#

#K_p = K_(p1) xx K_(p2)^color(blue)(2)#

#K_p = 3.65 xx 0.633^2 = color(green)(1.46)#

The answer is rounded to three sig figs.