Question

Question #b1be6

Answer 1

`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.