We start with the balanced chemical equation for the equilibrium.

#"CO" + "H"_2"O" ⇌ "CO"_2 +"H"_2#

Then we write the #K_P# expression.

#K_P = (P_("CO"_2)P_("H"_2))/(P_("CO")P_("H"_2"O"))#

Next, we set up an ICE table:

Insert these values into the #K_P# expression:

#K_P = (P_("CO"_2)P_("H"_2))/(P_("CO")P_("H"_2"O")) = (x × x)/((1360-x)(1750-x)) = 0.0611#

#x^2/(2 380 000 – 3110x + x^2) = 0.0611#

#x^2 = 145 418 – 190.02x + 0.0611x^2#

Putting all terms on the left, we get

#x^2 - 0.0611x^2 + 190.02x -145 418 = 0#

#x^2(1-0.0611) + 190.02x -145 418 = 0#

#0.9389x^2 + 190.02x -145 418 = 0#

Use your calculator or an on-line quadratic equation solver such as the one at

http://www.math.com/students/calculators/source/quadratic.htm

#x = 305#

#P_("CO") = (1360 – x)" Torr" = (1360 - 305)" Torr" = "1055 Torr"#

#P_("H"_2"O") = (1750 – x)" Torr" = (1750 – 305)" Torr" = "1445 Torr"#

#P_(CO_2) = P_(H_2) = x" Torr" = "305 Torr"#

**Check: **#(305 × 305)/(1055 × 1445) = 0.0610# (Close enough!)

The video below shows how to use an ICE table to solve a #K_p# problem.