For the isomerization reaction:
Butane ⇌ isobutane
K_p equals 25 at 500°C.If the initial pressures of butane and isobutane are 20.atm and 0.0 atm,respectively,what are the pressures of the two gases at equilibrium?

Question

Quizplus · Accepted Answer

Using the equilibrium constant expression:
Kp = P(isobutane) / (P(butane))^3

We can write an ICE table:
Initial        Change          Equilibrium
P(butane)    -x                20 - x
P(isobutane)  +x                x

At equilibrium:
Kp = P(isobutane) / (P(butane))^3
25 = x / (20 - x)^3
x = 3.58 atm

Plugging x into the equilibrium expressions, we get:
P(butane) = 0.77 atm
P(isobutane) = 19.2 atm

Therefore, the best choice is A.

For the Isomerization Reaction