To solve this problem, we need to first write the balanced chemical equation for the reaction:N g + 3H g 2NH g Next, we need to determine the initial concentrations of the reactants and products. Since we are not given the volume of the reaction, we can assume a volume of 1 L for simplicity. This means that the initial concentrations are:[N] = 2.1 moles/L[H] = 0.8 moles/L[NH] = 0 moles/LNow, let's define the changes in the concentrations at equilibrium using a variable x:[N] = 2.1 - x moles/L[H] = 0.8 - 3x moles/L[NH] = 2x moles/LWe can now write the expression for the equilibrium constant, K:K = [NH] / [N] * [H] Plugging in the equilibrium concentrations and the given value of K 3.5 x 10^-4 : 3.5 x 10^-4 = 2x / 2.1 - x * 0.8 - 3x This is a complex equation to solve algebraically. However, we can make an assumption that x is small compared to the initial concentrations of N and H. This is reasonable because the equilibrium constant is very small, indicating that the reaction does not proceed very far to the right. Therefore, we can approximate: 3.5 x 10^-4 2x / 2.1 * 0.8 Now, we can solve for x: 2x = 3.5 x 10^-4 * 2.1 * 0.8 2x = 3.5 x 10^-4 * 2.1 * 0.512 2x = 3.78 x 10^-42x = sqrt 3.78 x 10^-4 2x 0.01945x 0.009725Now we can find the equilibrium concentration of ammonia:[NH] = 2x 2 * 0.009725 0.01945 moles/LSo, the equilibrium concentration of ammonia is approximately 0.01945 moles/L.