In a reactor, the gas phase reaction A ? B is at equilibrium with yA = 0.75 and yB = 0.25. An inert gas is added to the mixture, but no A or B is added or removed. Then the system is allowed to reach a new equilibrium at the same T and P as the original equilibrium. Both equilibrium states can be modeled as ideal solutions. Which of the following correctly describes the effect of adding the inert on the system?

Question

A. The reaction will shift to the right, forming more moles of B than in the original equilibrium.
B. The reaction will shift to the left, forming more moles of A than in the original equilibrium.
C. The numbers of moles of A and B will be the same as in the original equilibrium.
D. Answers A and B are both possible but the correct one cannot be determined from the given information.
E. The scenario is impossible because adding inert gas will inevitably increase the pressure.

Answer 1

A. Incorrect. The equilibrium expression for this ideal gas system is K = yB/yA. Adding an inert gas will lower both mole fractions, but will not change their ratio.
B. Incorrect. The equilibrium expression for this ideal gas system is K = yB/yA. Adding an inert gas will lower both mole fractions, but will not change their ratio.
C. Correct. The inert gas will change the mole fractions of A and B, but the change will cancel out when calculating their ratio. End result is that no change in the number of moles of either will occur. Note that if the “ideal solution” specification wasn’t present, then this would not necessarily be true, as adding an inert could influence the fugacity coefficients of A and B.
D. Incorrect. Only one of the three options is possible.
E. Incorrect. This is true if the inert gas is being added to a constant volume vessel, but there is nothing unrealistic about maintaining constant T and P while allowing V to expand as gas is added.