Question

A: 2 NH3 (g) + 46 kJ <-> N2 (g) + 3 H2 (g)

For the reaction at equilibrium given above, give 2 ways that you can shift the equibrium to the left.

B: 2 NH_3(g) + 46 kJ <-> N_2(g) + 3 H_2(g)

For the reaction at equilbrium given above, calculate the numeric value of Kc if the concentrations at equilibrium are 0.50 M NH_3(g) , 0.44 M N_2(g), and 0.20 M H_2(g).

C: 2 NH_3(g) + 46 kJ <-> N_2(g) + 3 H_2(g)

If the equillibrium reaction above has an equilibrium constant of 2.6 at a particular temperature, determine the concentration of NH₃ (in Molarity), if the concentration of N₂ and H₂ are both 4.4 M. Show your work.

Answer 1

A)

2 NH3 (g) + 46 kJ <-> N2 (g) + 3 H2 (g)

a) As given it is an endothermic equation, in which heat is required for reaction hence we can shift the equilibrium by decreasing the temperature

b) we can shift the equilibrium to left by adding more product to the system

B: 2 NH_3(g) + 46 kJ <-> N_2(g) + 3 H_2(g)

Kc = [H₂]³ [N₂] / [NH₃]²

^Given: 0.50 M NH_3(g) , 0.44 M N_2(g), and 0.20 M H_2(g).

Kc = [0.2]³ [0.44] / [0.5]²

Kc =0.01408

3)

Kc = [H₂]³ [N₂] / [NH₃]²

given: N₂ and H₂ are both 4.4 M and equilibrium constant of 2.6

2.6 = [4.4]³ [4.4] / [NH₃]²

[NH₃]² = 144.16

[NH₃] = 12.01 M