To solve this problem, we need to use the balanced chemical equation for the reaction and the equilibrium constant expression. The balanced chemical equation for the reaction between NOBr, NO, and Br2 is:2 NOBr 2 NO + Br2Now, let's set up the equilibrium constant expression. For this reaction, the equilibrium constant Kc is given by:Kc = [NO]^2 [Br2] / [NOBr]^2We are given that the initial concentration of each species is 0.10 mol. Let's assume that the volume of the container is 1 L, so the initial concentrations are also 0.10 M. At equilibrium, the concentration of NOBr is given as 0.075 M. Let x be the change in concentration of NOBr. Since 2 moles of NO are produced for every 2 moles of NOBr consumed, the change in concentration of NO is also x. Similarly, since 1 mole of Br2 is produced for every 2 moles of NOBr consumed, the change in concentration of Br2 is x/2.Now, we can set up the equilibrium concentrations for each species:[NOBr] = 0.10 - x[NO] = 0.10 + x[Br2] = 0.10 + x/2We are given the equilibrium concentration of NOBr as 0.075 M, so we can solve for x:0.075 = 0.10 - xx = 0.10 - 0.075x = 0.025Now we can find the equilibrium concentrations of NO and Br2:[NO] = 0.10 + x = 0.10 + 0.025 = 0.125 M[Br2] = 0.10 + x/2 = 0.10 + 0.025/2 = 0.10 + 0.0125 = 0.1125 MSo, the equilibrium concentrations of NO and Br2 are 0.125 M and 0.1125 M, respectively.