# Question b1be6

Feb 12, 2016

${K}_{p} = 1.46$

#### 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

$\text{C"_text((s]) + 2"H"_2"O"_text((g]) rightleftharpoons "CO"_text(2(g]) + 2"H"_text(2(g])" } , {K}_{\textrm{p 1}} = 3.65$

$\text{H"_text(2(g]) + "CO"_text(2(g]) rightleftharpoons "H"_2"O"_text((g]) + "CO"_text((g])" } , {K}_{\textrm{p 2}} = 0.633$

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 $\textcolor{b l u e}{2}$

$\left[{\text{H"_text(2(g]) + "CO"_text(2(g]) rightleftharpoons "H"_2"O"_text((g]) + "CO}}_{\textrm{\left(g\right]}}\right] \times \textcolor{b l u e}{2}$

$\textcolor{b l u e}{2} {\text{H"_text(2(g]) + color(blue)(2)"CO"_text(2(g]) rightleftharpoons color(blue)(2)"H"_2"O"_text((g]) + color(blue)(2)"CO}}_{\textrm{\left(g\right]}}$

The equilibrium constant for this reaction is equal to

${K}_{p 2}^{'} = \left({\left({\text{H"_2"O")^color(blue)(2) * ("CO")^color(blue)(2))/(("H"_2)^color(blue)(2) * ("CO}}_{2}\right)}^{\textcolor{b l u e}{2}}\right)$

But since the original reaction had

${K}_{2 p} = \left(\left({\text{H"_2"O") * ("CO"))/(("H"_2) * ("CO}}_{2}\right)\right)$

you can say that

${K}_{2 p}^{'} = {\left[\left(\left({\text{H"_2"O") * ("CO"))/(("H"_2) * ("CO}}_{2}\right)\right)\right]}^{\textcolor{b l u e}{2}} = {K}_{\textrm{p 2}}^{\textcolor{b l u e}{2}}$

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)#

${\text{C"_text((s]) + 2"CO"_text(2(g]) rightleftharpoons "CO"_text(2(g]) + 2"CO}}_{\textrm{\left(g\right]}}$

This is equivalent to

${\text{C"_text((s]) + "CO"_text(2(g]) rightleftharpoons 2"CO}}_{\textrm{\left(g\right]}} \to$ the target reaction

The equilibrium constant will be equal to

${K}_{p} = {K}_{p 1} \times {K}_{p 2}^{'}$

${K}_{p} = {K}_{p 1} \times {K}_{p 2}^{\textcolor{b l u e}{2}}$

${K}_{p} = 3.65 \times {0.633}^{2} = \textcolor{g r e e n}{1.46}$

The answer is rounded to three sig figs.