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
Where C+2H2O=CO2+2H2 has Kp = 3.99
and
H2+CO2=H2O+CO has Kp = .759
1 Answer
So you are asking... what is the
#"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) + 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 = ???)#