The balanced chemical equation for the reaction between nitrogen gas N2 and hydrogen gas H2 to form ammonia NH3 is:N2 g + 3H2 g 2NH3 g Given the initial moles of N2 and H2, we can calculate the initial concentrations:[N2] = 0.50 moles / 1.00 L = 0.50 M[H2] = 0.50 moles / 1.00 L = 0.50 M[NH3] = 0 moles / 1.00 L = 0 M since no NH3 is present initially Let x be the change in concentration of N2 and H2 at equilibrium. Then, the change in concentration of NH3 will be 2x. The equilibrium concentrations will be:[N2] = 0.50 - x[H2] = 0.50 - 3x[NH3] = 2xNow, we can use the given Kc value to set up the equilibrium expression:Kc = [NH3]^2 / [N2] * [H2]^3 = 0.129Substitute the equilibrium concentrations into the expression:0.129 = 2x ^2 / 0.50 - x * 0.50 - 3x ^3 This is a complex cubic equation that can be solved either by numerical methods or by using a specialized calculator or software. Solving for x, we get:x 0.042Now, we can find the equilibrium concentrations:[N2] = 0.50 - x 0.50 - 0.042 0.458 M[H2] = 0.50 - 3x 0.50 - 3 0.042 0.374 M[NH3] = 2x 2 0.042 0.084 MSo, the equilibrium concentrations are approximately:[N2] 0.458 M[H2] 0.374 M[NH3] 0.084 M