For the isomerization reaction: butane ⇌ isobutane
K_p equals 25 at 500°C.If the initial pressures of butane and isobutane are 5.0.atm and 0.0 atm,respectively,what are the pressures of the two gases at equilibrium?

Question

Quizplus · Accepted Answer

The equilibrium constant, K, can be written as:

K = P(isobutane) / P(butane)

At equilibrium, the pressures can be expressed as:

P(butane) = 5.0 - x 
P(isobutane) = x

where x is the change in pressure from the initial state.

Substituting these into the equilibrium expression:

25 = x / (5.0 - x)

Solving for x:

x = 0.165 atm

Therefore, the pressures at equilibrium are:

P(butane) = 5.0 - 0.165 = 0.19 atm 
P(isobutane) = 0.165 atm

Therefore, the best choice is A, which gives the correct pressures for butane and isobutane at equilibrium.

For the Isomerization Reaction: Butane ⇌ Isobutane