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.

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,

          $K_{c} = \frac{[isobutane]}{[n-butane]}$

                    = $\frac{0.170}{0.068}$

                    = 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

                   $K_{c} = \frac{x}{0.438 - x}$

                      2.5 = $\frac{x}{0.438 - x}$

                    1.095 - 2.5x = x

                             x = $\frac{1.095}{3.5}$

                                = 0.3128 M

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