5. An important step in the industrial production of hydrogen is the reaction of carbon monoxide with water:
CO(g) + H2O(g) <==> CO2(g) + H2(g)
(a) Use the law of mass action to write the equilibrium expression for this reaction.
By using the law of mass action, We can express the equilibrium expression as:
K = [CO2][H2]
(b) At 500 C, the equilibrium constant for this reaction is 3.9. Suppose that the equilibrium partial pressures of CO and H2O are both 0.10 atm and that of CO2 is 0.70 atm. Calculate the equilibrium partial pressure of H2(g).
We can substitute the information provided during the question into the equation that we have already derived.
3.9 = (0.70 atm)[H2]
(0.10 atm)(0.10 atm)
[H2] = 0.56 atm
19. Suppose that K1 and K2 are the respective equilibrium constants for the two reactions:
XeF6(g) + H2O(g) <==>
XeOF4(g) + 2 HF(g)
XeO4(g) + XeF6(g) <==>
XeOF4(g) + XeO3F2(g)
Give the equilibrium constant for the reaction:
XeO4(g) + 2 HF(g) <==>
XeO3F2(g) + H2O(g)
To find K3 we must first add the first two equations to each other. (Note that I reversed the second equation.)
XeF6(g) + H2O(g) <==> XeOF4(g) + 2 HF(g)
XeOF4(g) + XeO3F2(g) <==> XeO4(g) + XeF6(g)
XeO3F2(g) + H2O(g) <==> XeO4 + 2 HF(g)
This equation is the third equation. We can speculate that K3 is just the ration of K1 over K2.
K3 = K1/K2