The equilibrium constant (K) of the reaction below is K = 2.5 x 10-3, with initi

Question

The equilibrium constant (K) of the reaction below is K = 2.5 x 10-3, with initial concentrations as follows: [H2] = 1.0 x 10-2 M, [N2] = 4.0 M, and [NH3] = 1.0 x 10-4M. N2(g) + 3 H2 (g) --> 2 NH3 (g) Nitrogen + hydrogen --> ammonia 4. Write the equation for the equilibrium constant (Keq) of the reaction of nitrogen + hydrogen --> ammonia as shown above.

5. If the concentration of the reactant H2 was increased from 1.0 x 10-2 M to 2.5 x 10-1M, calculate the reaction quotient (Q) and determine which way the equilibrium position would shift.

6. If the concentration of the reactant H2 was decreased from 1.0 x 10-2 M to 2.7 x 10-4M, calculate the reaction quotient (Q) and determine which way the equilibrium position would shift.

7. If the concentration of the product NH3 was increased from 1.0 x 10-4 M to 5.6 x 10-3M, calculate the reaction quotient (Q) and determine which way the equilibrium position would shift.

Explanation / Answer

N2(g) + 3 H2 (g) --> 2 NH3 (g)

equilibrium constant :

Keq = [NH3]^2 / [N2] [H2]^3

1)

Qc= [NH3]^2/[N2][H2]^3

         = (1.0x10^-4)^2/ [2.5X10^-1]^3 X4

Qc = 1.6 x 10^-7

K= 2.5 x 10^-3            

Qc < Kc

reaction proceeds to right

2)

Qc= (1.0x10^-4)^2/ [2.7X10^-4]^3 X4

        = 127

Qc > Kc

reaction proceeds to left

3)

Qc= (5.6 x10^-3)^2/ [1X10^-2]^3 X4

          = 7.84

Qc > Kc

reaction proceeds to left

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.