For the isomerization reaction cis-2-butene ⇌ trans-2-butene with an equilibrium constant Kp of 3.40, if a flask initially contains 0.250 atm of cis-2-butene and 0.165 atm of trans-2-butene, what is the equilibrium pressure of cis-2-butene?

• A. 0.150 atm
• B. 0.200 atm
• C. 0.100 atm
• D. 0.250 atm

Answer: (A)

Verified step by step guidance

1. Identify the initial pressures of the reactants and products: cis-2-butene = 0.250 atm and trans-2-butene = 0.165 atm.
2. Write the expression for the equilibrium constant Kp for the reaction: Kp = (P_trans-2-butene) / (P_cis-2-butene).
3. Set up an ICE (Initial, Change, Equilibrium) table to track the changes in pressure as the system reaches equilibrium. Let the change in pressure of cis-2-butene be -x and the change in pressure of trans-2-butene be +x.
4. Substitute the equilibrium pressures from the ICE table into the Kp expression: Kp = (0.165 + x) / (0.250 - x).
5. Solve the equation for x using the given Kp value of 3.40, and then calculate the equilibrium pressure of cis-2-butene as 0.250 - x.