The equilibrium constant, Kp = 3.4 for the isomerization reaction:cis-2-butene ⇌ trans-2-butene. If a flask initially contains 0.25 atm of cis-2-butene what is the equilibrium pressure (atm) for trans-2-butene?

The equilibrium pressure of trans-2-butene is 0.193 atm.

Explanation:

Kp = Ptrans-2-butene/Pcis-2-butene

Kp = 3.4

Initial pressure of cis-2-butene = 0.25 atm

Let the equilibrium pressure of trans-2-butene be y

From the equation of reaction,

Mole ratio of cis-2-butene to trans-2-butene is 1:1, therefore equilibrium pressure of cis-2-butene is (0.25 - y)

3.4 = y/(0.25 - y)

y = 3.4(0.25 - y)

y = 0.85 - 3.4y

y + 3.4y = 0.85

4.4y = 0.85

y = 0.85/4.4 = 0.193 atm

idk what the question is about