How do I calculate Kp for C+CO2=2CO given the following reactions? C+2H2O=CO2+2H2 and H2+CO2=H2O+CO.

Where C+2H2O=CO2+2H2 has Kp = 3.99
and
H2+CO2=H2O+CO has Kp = .759

1 Answer
Jan 24, 2018

So you are asking... what is the K_p of

"C"(s) + "CO"_2(g) rightleftharpoons 2"CO"(g)

if

"C"(s) + 2"H"_2"O"(g) rightleftharpoons "CO"_2(g) + 2"H"_2(g), K_(p1) = 3.99

"H"_2(g) + "CO"_2(g) rightleftharpoons "H"_2"O"(g) + "CO"(g), K_(p2) = 0.759

Well, it's Hess's Law. Add up the reactions to cancel out the steps and get the overall reaction.

  • Reversed reactions have K_p -> K_p^(-1), i.e. 1/K_p.
  • Scaled reactions have K_p -> K_p^c where c is the constant everything is scaled by.
  • Added reactions have K_p = K_(p1)K_(p2)cdots

To do this, you want "C"(s) on the reactants side, so keep the first step as it is. Double step 2 so that 2"CO"(g) is a product. Therefore:

"C"(s) + cancel(2"H"_2"O"(g)) rightleftharpoons "CO"_2(g) + cancel(2"H"_2(g)), " "K_(p1) -> K_(p1)
ul(2(cancel("H"_2(g)) + "CO"_2(g) rightleftharpoons cancel("H"_2"O"(g)) + "CO"(g))), " "K_(p2) -> K_(p2)^2
"C"(s) + "CO"_2(g) rightleftharpoons 2"CO"(g)

This means

color(blue)(K_p = K_(p1)cdot K_(p2)^2 = ???)