Nitrogen and water react to form nitrogen monoxide and hydrogen, like this: n2(g) + 2h2o(g) → 2no(g) +2h2(g) also, a chemist finds that at a certain temperature the equilibrium mixture of nitrogen, water, nitrogen monoxide, and hydrogen has the following composition: compound concentration at equilibrium h20 no h2 0.25 m 1.3 m 0.33 m 1.2 m calculate the value of the equilibrium constant kc for this reaction. round your answer to 2 significant digits

Kc for this reaction is 0.43

Explanation:

This is the equilibrium:

N₂(g) + 2H₂O(g) → 2NO(g) +2H₂(g)

And we have all the concentration at equilibrium:

N₂: 0.25M

H₂ : 1.3M

NO: 0.33M

H₂: 1.2M

They are ok, because they are in MOLARITY. (mol/L)

Let's make the expression for Kc

Kc = ( [NO]² . [H₂]² ) / ([N₂] . [H₂O]²)

Kc = (0.33² . 1.2²) / (0.25 . 1.2²)

Kc = 0.4356

In two significant digits. 0.43

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole