Question

2.50 mol NOCl was placed in a 2.50 L reaction vessel at 400ºC. After equilibrium was established, it was found that 28% of the NOCl had dissociated according to the equation 2NOCl(g) 2NO(g) + Cl 2(g). Calculate the equilibrium constant, K c, for the reaction.

Answer 1

Answer:

Explanation:

2NOCl(g)    ⇄      2NO(g) + Cl 2(g)

C ( 1 - .28 )            .28 C         .14 C

Kc = [ NO ]²  x [ Cl₂ ] / [ NOCl ]²

= (.28 C )² x .14 C / C² ( 1 - .28 )²

= .021173 x C

C = concentration of reactant

= 2.5 / 2/5 = 1 M

Kc = .021173 x 1

= 211.73 x 10⁻⁴ M .