At a certain temperature, this reaction establishes an equilibrium with the given equilibrium constant, kc. 3 a ( g ) + 2 b ( g ) − ⇀ ↽ − 4 c ( g ) k c = 2.93 × 10 29 if, at this temperature, 2.30 mol of a and 3.90 mol of b are placed in a 1.00 l container, what are the concentrations of a, b, and c at equilibrium?

The equilibrium concentration of A, B and C are 0.02 M, 2.38 M and 3.04 M respectively.

Explanation:

We are given:

Initial moles of A = 2.30 moles

Initial moles of B = 3.90 moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

So, concentration of A =

Concentration of B =

The given chemical equation follows:

Initial:          2.30       3.90

At eqllm:   2.30-3x   3.90-2x    4x

The expression of for above equation follows:

We are given:

Putting values in above expression, we get:

So, the equilibrium concentration of A =

The equilibrium concentration of B =

The equilibrium concentration of C =

Hence, the equilibrium concentration of A, B and C are 0.02 M, 2.38 M and 3.04 M respectively.

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