We are asked to find the partial pressure of ethene given the equilibrium constant and the partial pressures of the other species. The reaction and equilibrium constant are shown below:
C₂H₄ (g) + H₂O (g) → C₂H₅OH (g) Kc = 9 x 10³
Since we are given the equilibrium constant, Kc, which is regarding the concentration of species. However, we want the value of Kp which is the equilibrium constant regarding partial pressures. To convert these values we use the following formula:
Kp = Kc · (RT)ⁿ
R = 0.08206 Latm/molK
T = 600 K
Kc = 9 x 10³
Δn = sum of stoichiometric coefficients of products - sum of stoichiometric coefficients of reactants = -1
Kp = Kc/RT = (9 x 10³)/(0.08206)(600)
Kp = 183
Kp = [Pethanol]/[Pwater][Pethene] = 183
183 = (200)/(400)(Pethene)
Pethene = (200)/(400)(183)
Pethene = 0.00273 atm
The partial pressure of ethene was found to be 0.00273 atm. This appears to be a very small number, however, it agrees with the scenario as we were told that the equilibrium constant was a very large number which suggests that the equilibrium falls far to the right. Based on this partial pressure of ethene, it appears that the majority of the ethene has reacted to form ethanol.
