To solve this problem, we will use the ICE Initial, Change, Equilibrium table method. First, let's set up the ICE table:`` N2 g + 3H2 g 2NH3 g Initial: 0.1M 0.2M 0MChange: -x -3x +2xEquilibrium: 0.1-x 0.2-3x 2x``Now, we can write the expression for the equilibrium constant, Kc:Kc = [NH3]^2 / [N2] * [H2]^3 Given Kc = 6.0 x 10^2, we can substitute the equilibrium concentrations into the expression:6.0 x 10^2 = 2x ^2 / 0.1 - x * 0.2 - 3x ^3 Now, we need to solve for x. Since this is a complex equation, we can make an assumption that x is small compared to the initial concentrations of N2 and H2. This means that 0.1 - x 0.1 and 0.2 - 3x 0.2. This simplifies the equation to:6.0 x 10^2 = 2x ^2 / 0.1 * 0.2^3 Now, we can solve for x: 2x ^2 = 6.0 x 10^2 * 0.1 * 0.2^3 2x ^2 = 6.0 x 10^2 * 0.1 * 0.008 2x ^2 = 4.82x = sqrt 4.8 2x = 2.19x = 1.095Now that we have the value of x, we can find the concentration of NH3 at equilibrium:[NH3] = 2x = 2 * 1.095 = 2.19 MSo, the concentration of NH3 at equilibrium is 2.19 M.