Chemistry, 08.04.2020 04:41 Natavia3402

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

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Answer from: shealwaysknows23

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.

Answer from: Quest

The correct answer is c. orbital shape. "the angular momentum quantum number l describes the shape of an orbital. it corresponds to each orbital type, i.e. (0,1,2,−1)=(s,p,d,f,g,"

Answer from: Quest

Answerδh > 0explanationc₆h₆ (s) → c₆h₆ (l)if you are changing it from a solid to a liquid, you are pretty much melting the solid. this requires heat energy. therefore, the reaction is endothermic (takes in energy); the system enthalpy increases as the reaction proceeds: δh > 0.

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