Question

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

Answer 1

Given question is incomplete. The complete question is as follows.

n-butane and isobutane are in equilibrium at 25 degrees centigrade.

n-butane(g)
`\rightleftharpoons` 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.

Step-by-step explanation:

As the given reaction is as follows.

n-butane(g)
`\rightleftharpoons` isobutane(g)

Initial: x 0

Equilbm: x - 0.17 0.17

It is given that equilibrium concentration of n-butane is 0.068 M.

So, x - 0.17 = 0.0680

x = 0.238

Therefore, the initial concentration is 0.238.

Here,

`K_(c) = ([isobutane])/([n-butane])`

=
`(0.170)/(0.068)`

= 2.5

When 0.2 mol of n-butane is added. Hence, total moles will be as follows.

Total mole = (0.238 + 0.2)

= 0.438 mol

Hence, for 1 L volume the ICE table will be as follows.

n-butane(g)
`\rightleftharpoons` isobutane(g)

Initial: 0.438 0

Equilbm: 0.438 - x x

`K_(c) = (x)/(0.438 - x)`

2.5 =
`(x)/(0.438 - x)`

1.095 - 2.5x = x

x =
`(1.095)/(3.5)`

= 0.3128 M

Thus, we can conclude that the equilibrium concentration is 0.3128 M.