In the presence of a catalyst, n-butane (CH3CH2CH2CH3, abbreviated "A") and iso-butane (CH3CH( CH3)CH3, abbreviated " B ") can reach chemical equilibrium in an isomerization reaction: A(g)B(g) (a) Calculate the equilibrium constant of this reaction at 500K. State any assumption(s). (b) Assuming n-butane and iso-butane form an ideal gas mixture, calculate the equilibrium mole fraction of iso-butane, yB, if the reaction is run at 500K,70 bar. (c) At 450K, the equilibrium mixture from Part (b) is pressurized isothermally until the first drop of liquid appears. Calculate the pressure at which this happens, and the mole fraction of iso-butane, xB, in the first drop of liquid. Hence, calculate the equilibrium constant of the reaction A(l)B(l) at 500K, knowing that chemical equilibrium is also attained in the liquid phase. State any assumption(s). (d) Consider the gas phase reaction at 500K and 70 bar again. At such high pressure, the ideal gas assumption may be invalid, but we can still assume they form an ideal solution (since they are chemically similar). Pure n-butane gas and pure iso-butane gas are both wellmodeled by the virial equation of state, resulting in the following expression for the fugacity coefficient: i=exp(RTBiP) where BA=216cm3/mol for n-butane and BB=195cm3/mol for iso-butane. Redo Parts (b) and (c), this time with the ideal solution assumption for the gas phase