Question

For the following reaction 2 NO(g) + Cl2(g) ⇋ 2 NOCl(g) The equilibrium concentrations of the gases at 1200°C are 0.026 M for NO, 0.596 M for Cl2, and 0.851 M for NOCl. Calculate the value of Kc.

Answer 1

Kc = 1.797 * 10³

Step-by-step explanation:

2 NO(g) + Cl₂(g) ⇋ 2 NOCl(g)

The equilibrium constant for the above reaction , can be written as the product of the concentration of product raised to the power of stoichiometric coefficients in a balanced equation of dissociation divided by the product of the concentration of reactant raised to the power of stoichiometric coefficients in the balanced equation of dissociation .

Hence ,

Kc = [ NOCl(g) ]² / [ NO(g) ] ² [Cl₂(g)]

From the question ,

[ NOCl(g) ] = 0.851 M

[ NO(g) ] = 0.026 M

[Cl₂(g) ] = 0.596 M

Now, putting it in the above equation ,

Kc = [ NOCl(g) ]² / [ NO(g) ] ² [Cl₂(g)]

Kc = ( 0.851 M )² / (0.026 M )² ( 0.596 M )

Kc = 1.81 * 10³