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

First, we can set up an ICE table to represent the reaction:
Initial:   10 atm        0 atm
Change:   -x              +x
Equilibrium: 10-x      x
Using the equilibrium constant expression, we can set up the equation:
25 = (x)/(10-x)
Solving for x, we get x = 0.38 atm. 
Substituting x back into our ICE table, we get the equilibrium pressures to be:
P(butane) = 10 - x = 9.62 atm, and 
P(isobutane) = x = 0.38 atm. 
Thus, the answer is A.

For the Isomerization Reaction: Butane ⇌ Isobutane