To find the equilibrium concentrations of NOBr, NO, and Br2, we can set up an ICE Initial, Change, Equilibrium table:`` 2NOBr g <=> 2NO g + Br2 g Initial: 0.10 M 0 M 0 MChange: -2x M +2x M +x MEquilibrium:0.10-2x M 2x M x M``The equilibrium constant expression for this reaction is:Kc = [NO]^2[Br2] / [NOBr]^2We can plug in the equilibrium concentrations into the expression:1.5 x 10^-5 = 2x ^2 x / 0.10 - 2x ^2Now we need to solve for x: 1.5 x 10^-5 0.10 - 2x ^2 = 2x ^2 x It's a cubic equation, which can be difficult to solve directly. However, since Kc is very small, we can assume that x is also very small, and therefore 2x is also very small compared to 0.10. So, we can approximate: 1.5 x 10^-5 0.10 ^2 = 2x ^2 x Now we can solve for x: 1.5 x 10^-5 0.01 = 4x^31.5 x 10^-7 = 4x^3x^3 = 1.5 x 10^-7 / 4x^3 3.75 x 10^-8x 3.36 x 10^-3Now we can find the equilibrium concentrations:[NOBr] = 0.10 - 2x 0.10 - 2 3.36 x 10^-3 0.0933 M[NO] = 2x 2 3.36 x 10^-3 6.72 x 10^-3 M[Br2] = x 3.36 x 10^-3 MSo, the equilibrium concentrations are approximately [NOBr] = 0.0933 M, [NO] = 6.72 x 10^-3 M, and [Br2] = 3.36 x 10^-3 M.