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
Truong-Son N.
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 = ???)#

Related questions
Impact of this question
3136 views around the world
You can reuse this answer
Creative Commons License