Question

n-butane and isobutane are in equilibrium at 25 degrees centigrade:
  
n-butane (g)                           isobutane (g)       
  
A one liter flask with the two species at equilibrium contains 0.170 mole isobutane and 0.0680 mole n-butane. Then the concentration of n-butane is increase by 0.200 moles. (0.200 mole n-butane is added.) What is the new equilibrium concentration for isobutane? Assume value for K _c remains constant throughout process. Hint--this is a hard problem. At equilibrium (which you are given) you can find K _C. Now, the equilibrium changes. You must use ICE to find new values.

		9.22 x 10-2M
		1.57 x 10-1M
		9.25 x 10-3M
		3.50 x 10-2M
		0.312 M
		7.67 x 10-5M
		0.190 M
		3.12 x 10-2 M

Answer 1

n-buteene o isobutane (9) 0.068 ateam conch 0.174 kca (isobutant > Kc =/ 0.17 ) ) ķ=2.5 (0.068) (n-butane] Now in this 0.2 moles of n-butane is colded h-butame =) 0:2 +0.068 =) 0:260 moles In butane] = 0.268 M here Recetunts are being adcled means reaction will shift in fesward direction. n-butane o isobutane (8) initical (2) 0.268M 0.17 chengecc) -2 +2 at egm (E) 0.268-2 017+2 k= ( Isobutene] 2:5= 0:17+ve 0.268-2 [n-butare] 2:5 (0-268-0) = (0:17 +x) =) 0:67 - 2-52 = 017 + z = 0.1428 isobutene = 0.17 +0.142 cIsobutane] = 0.312M La