CO(g) + H₂O(g) < = > CO₂(g) + H₂(g)
P(i) 1380-torr 1730-torr 0 0
ΔP -ΔP - ΔP +ΔP +ΔP
P(eq) 1380-ΔP 1730-ΔP ΔP<Equal>ΔP
Kp = P(CO₂)∙P(H₂)/P(CO)∙P(H₂O) = (ΔP)²/(1380-ΔP)(1730-ΔP) = 0.0611
Solve for ΔP using the Quadratic Formula ...
I got ΔP = 305-torr P(CO₂)∙ ... Verified by substituting into above Kp expression => Kp = 0.061075 ~ 0.0611.
Consider the following reaction:
CO(g)+H2O(g)⇌CO2(g)+H2(g)
Kp=0.0611 at 2000 K
A reaction mixture initially contains a CO partial pressure of 1380 torr and a H2O partial pressure of 1730 torr at 2000 K.
Calculate the equilibrium partial pressure of CO2.
2 answers
Formatting is tough here... The quagmire of symbols below the equation is the ICE table for the problem analysis.