Question

9) The reaction between ammonia and oxygen gas is an eqilibrium process, shown as an equation below. The reaction K. = 2.2 x 109 at 25°C. 4 NH, u + 30, 2N,+ 6H,0 + heat Would this reaction be considered product-favored or reactant-favored at equilibrium? a) (1) b) Write out the equilibrium expression for this reaction: (3) K = c) Suppose some of the nitrogen gas was removed from the reaction after equilibrium is reached. Would this shift the reaction to the left, or to the right? (2) d) Suppose the equilibrium system was heated. How would this change the concentration of the oxygen gas? (2) (1) e) Give the definition of the equilibrium point for a reversible reaction:

Answer 1

a) We can see that K_eq is >>1 , there fore it means the reaction is nearly completed and hence it is product favoured.

b) K_eq = [H₂O]⁶ * [N₂]² / [O₂]³ * [NH₃]⁴

c) If some of the nitrogen gas is removed from the reacrion, there will be decrease in K_eq and hence formation of nitrogen gas will take place and hence the reaction will move towards the right.