Multiple Choice
For the isomerization reaction butane ↔ isobutane, Kp equals 25 at 500°C. If the initial pressures of butane and isobutane are 10.0 atm and 0.0 atm, respectively, what are the pressures of the two gases at equilibrium?
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16. Chemical Equilibrium
Intro to Chemical Equilibrium
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The equilibrium constant, Kp, equals 3.40 at 25°C for the isomerization reaction: cis-2-butene ⇌ trans-2-butene. If a flask initially contains 1.00 atm of each gas, in what direction will the system shift to reach equilibrium?
A
The system will shift to the right, favoring the formation of trans-2-butene.
B
The system will shift to the left, favoring the formation of cis-2-butene.
C
The system is already at equilibrium and will not shift in either direction.
D
The system will shift in both directions equally.
Verified step by step guidance1
Identify the initial conditions: The flask contains 1.00 atm of cis-2-butene and 1.00 atm of trans-2-butene.
Write the expression for the equilibrium constant Kp for the reaction: Kp = (P_trans-2-butene) / (P_cis-2-butene).
Substitute the initial pressures into the Kp expression: Kp_initial = (1.00 atm) / (1.00 atm) = 1.00.
Compare the initial Kp value with the given equilibrium constant Kp = 3.40. Since Kp_initial < Kp, the reaction will shift to the right to increase the concentration of trans-2-butene.
Conclude that the system will shift to the right, favoring the formation of trans-2-butene, to reach equilibrium.
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