Question

Determine the value of Kc for the following reaction if the equilibrium concentrations are as follows: [H2]eq = 0.14 M, [Cl2]eq = 0.39 M, [HCl]eq = 1.6 M.H2(g) + Cl2(g) ⇌ 2 HCl(g)8.7 × 10-22.1 × 10-229473.4 × 10-2

Answer 1

Answer:

Kc = 47

Explanation:

From the given equation ,

H₂(g) + Cl₂(g) ⇌ 2 HCl(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 = [HCl(g) ]² / [ Cl₂(g) ] [ H₂(g) ]

From the question ,  

[HCl(g) ] = 1.6 M

[ Cl₂(g) ] =  0.39 M

[ H₂(g) ] = 0.14 M

Now, putting it in the above equation ,  

Kc = [HCl(g) ]² / [ Cl₂(g) ] [ H₂(g) ]

Kc = [ 1.6 M ]² / [ 0.39 M ][ 0.14 M ]

Kc = 47