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

a

explanation:

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.