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)C(s)+CO2(g)⇌2CO(g)
if
"C"(s) + 2"H"_2"O"(g) rightleftharpoons "CO"_2(g) + 2"H"_2(g)C(s)+2H2O(g)⇌CO2(g)+2H2(g) ,K_(p1) = 3.99Kp1=3.99
"H"_2(g) + "CO"_2(g) rightleftharpoons "H"_2"O"(g) + "CO"(g)H2(g)+CO2(g)⇌H2O(g)+CO(g) ,K_(p2) = 0.759Kp2=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)Kp→K−1p , i.e.1/K_p1Kp . - Scaled reactions have
K_p -> K_p^cKp→Kcp wherecc is the constant everything is scaled by. - Added reactions have
K_p = K_(p1)K_(p2)cdotsKp=Kp1Kp2⋯
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 = ???)