Introduction:
In this scenario, we have two moles of ammonia (NH3) introduced into a 500 mL vessel at a high temperature. The decomposition reaction of ammonia is given as 2NH3 → N2 + 3H2. At equilibrium, we find that only 1 mole of ammonia remains. We need to determine the equilibrium constant, denoted as K, for this reaction.

Equilibrium Constant (K):
The equilibrium constant (K) represents the ratio of the product concentrations to the reactant concentrations at equilibrium. For a general reaction aA + bB ↔ cC + dD, the equilibrium constant expression is given as:
K = ([C]^c x [D]^d) / ([A]^a x [B]^b)

Initial and Equilibrium Concentrations:
To determine the equilibrium constant (K) for the given reaction, we need to determine the concentrations of all species involved at equilibrium. Given that two moles of ammonia are initially introduced and only 1 mole is present at equilibrium, we can assume that the concentrations are halved.

Equilibrium Expression:
Based on the balanced equation, the equilibrium expression for the given reaction is:
K = ([N2] x [H2]^3) / [NH3]^2

Partial Pressure and Concentration Relationship:
Since the reaction is taking place in a vessel, we can assume that the gases behave ideally and follow the ideal gas law. The relationship between the concentration (c) and partial pressure (P) of a gas is given as:
P = nRT/V
Where P is the pressure, n is the number of moles, R is the ideal gas constant, T is the temperature in Kelvin, and V is the volume.

Calculating K:
To calculate the equilibrium constant (K), we need to determine the concentrations of nitrogen gas (N2), hydrogen gas (H2), and ammonia (NH3) at equilibrium. Since the volume of the vessel is given as 500 mL, we can convert it to liters by dividing by 1000.

Conclusion:
By determining the equilibrium concentrations of N2, H2, and NH3, and plugging the values into the equilibrium constant expression, we can calculate the equilibrium constant (K) for the given decomposition reaction of ammonia. The value of K will depend on the specific concentrations of the species at equilibrium and the temperature at which the reaction occurs.

you have given initial [NH3]. If [NH3] remaining = 1 mole/L, then 1mole/L reacted2NH3(g) N2(g)+3H2(g). So change in concentration of NH3 = -1 mole/L. This means that because 2 mole NH3 make 1 mole of N2, the change in [N2] will +0.5mol/L and that for H2 will be +1.5mol/L. and equilibrium concentration of NH3 will be 1 mole/LPlug into Keq= [N2][H2}^3/[NH3]^2 and find Keq=(1.5^3)(0.5) = 1.687 nearly equal to 1.7