For the isomerization reaction:
butane ⇌ isobutane
Kp 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

For the isomerization reaction: 
butane ⇌ isobutane
Kp 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?

anonymous · Answer

$$K _{p}= {products \over reactants} = 25 $$ Also the initial pressures add to 5.0 It's just math from here.$$butane + isobutane = 5.0$$$${isobutane \over butane} = 25$$ Therefore $${isobutane \over 5.0 - isobutane} = 25$$$$isobutane = 125 - 25 isobutane$$$$isobutane = 4.81atm$$$$butane = 5.0atm -4.81atm = .19atm$$ Does this make work? The two still sum to 5.0 and if you divide 4.81 by .19 you get approximately 25 so yes it does work.