Assume that the change in concentration of N2O4 is small enough to be neglected in the following problem. Calculate the equilibrium concentration of both species in 1.00 L of a solution prepared from 0.129 mol of N2O4 with chloroform as the solvent.

The question is incomplete, here is the complete question:

Assume that the change in concentration of N₂O₄ is small enough to be neglected in the following problem. Calculate the equilibrium concentration of both species in 1.00 L of a solution prepared from 0.129 mol of N₂O₄ with chloroform as the solvent.

N₂O₄ ⇆ 2NO₂ ;   Kc = 1.07×10⁻⁵  in chloroform

The equilibrium concentration of and are 0.1233 M and 0.0114 M respectively

Explanation:

We are given:

Initial moles of = 0.129 moles

Volume of the solution = 1.00 L

The molarity is calculated by using the equation:

Initial concentration of

For the given chemical equation:

Initial:          0.129

At eqllm:     0.129-x        2x

The expression of for above equation follows:

We are given:

Putting values in above expression, we get:

Neglecting the value of x = -0.006 because concentration cannot be negative

So, equilibrium concentration of

Equilibrium concentration of

Hence, the equilibrium concentration of and are 0.1233 M and 0.0114 M respectively

hope this !

explanation:

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