For the isomerization reaction:
Butane ⇌ isobutane
K_p equals 25 at 500°C.If the initial pressures of butane and isobutane are 10.atm and 0.0 atm,respectively,what are the pressures of the two gases at equilibrium?

Question

Quizplus · Accepted Answer

From the given equilibrium constant and initial pressures, we can set up an ICE table:

Butane ⇌ isobutane 
I: 10 atm      0.0 atm 
C: -x           +x 
E: 10-x         x

Using the equilibrium constant expression: 
Kc = [isobutane] / [butane] 
25 = x / (10-x) 
25(10-x) = x 
250 - 25x = x 
26x = 250 
x = 9.62

Therefore, the equilibrium pressures are: 
P(butane) = 10 - x = 0.38 atm 
P(isobutane) = x = 9.62 atm

So, the correct answer is option A.

For the Isomerization Reaction