The equilibrium constant,K_p,equals 3.40 for the isomerization reaction:
Cis-2-butene ⇌ trans-2-butene.
If a flask initially contains 0.250 atm of cis-2-butene and 0.145 atm of trans-2-butene,what is the equilibrium pressure of each gas?

Question

Quizplus · Accepted Answer

We can use the equilibrium expression for this reaction: K_p=(Ptrans-2-butene)/(Pcis-2-butene) We can rearrange this expression to solve for Ptrans-2-butene: Ptrans-2-butene=K_p*Pcis-2-butene At equilibrium, the ratio of Ptrans-2-butene to Pcis-2-butene should be equal to the equilibrium constant, K_p: K_p=(Ptrans-2-butene)/(Pcis-2-butene) Let x be the change in pressure from the initial pressure for both cis-2-butene and trans-2-butene. Then the equilibrium pressures will be: Pcis-2-butene=0.250 - x Ptrans-2-butene=0.145 + x Substituting into the equilibrium expression: K_p = (0.145 + x)/(0.250 - x) Solving for x: x = 0.0787 atm Substituting x into the expressions for Pcis-2-butene and Ptrans-2-butene: Pcis-2-butene = 0.250 - 0.0787 = 0.171 atm Ptrans-2-butene = 0.145 + 0.0787 = 0.223 atm Therefore, the equilibrium pressure of cis-2-butene is 0.171 atm and the equilibrium pressure of trans-2-butene is 0.223 atm. The closest answer choice is D, which gives the correct values to two significant figures.

The Equilibrium Constant,Kp,equals 3.40 for the Isomerization Reaction

The Equilibrium Constant,K_p,equals 3.40 for the Isomerization Reaction