Question

The equilibrium constant for the reaction N2(g) +3 H2(g) =2 NH3(g) at 715 k is 6.0× 10-2 . If in a particular reaction there are 0.25 mol l-1 of NH3 present, calculate the concentration of N2 at equilibrium​

Answer 1

Answer:

given = N2(g) +3 H2(g) =2 NH3(g)

            temprature = 715 k

            eq.constant = 6.0× 10-2

            [NH3] = 0.25mol⁻¹

to find = [N2]

solution  =

 Kc = [NH3]²/[N2][H2]³

 [H2] = 0.25M

 [NH3] = 0.06M

Kc= 6 x 10⁻²

6 x 10⁻² = 0.06 x 0.06  /  (0.025)³ x [N2]

[N2] = 0.06 X 0.06 / 0.06 X 0.25 X 0.25 X 0.25

[N2] =  0.0036 / 15625

[N2] = 0.084M

the concentration of N2 at equilibrium ia 0.084 M.