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.

Kc = 1.797 * 10³

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