To solve this problem, we need to use the reaction for the dissociation of NOCl and the equilibrium constant expression. The reaction is:2 NOCl g 2 NO g + Cl2 g Let's denote the initial concentration of NOCl as [NOCl] = 0.10 M, and the initial concentrations of NO and Cl2 as [NO] = [Cl2] = 0 M. When the volume is changed to 1 L, the initial concentration of NOCl will change. Let's denote the new initial concentration as [NOCl'].Since the number of moles of NOCl remains constant, we can use the formula:n = C VWhere n is the number of moles, C is the concentration, and V is the volume. For the initial state:n = [NOCl] V = 0.10 M 2 L = 0.20 molesFor the new state:n' = [NOCl'] V' = [NOCl'] 1 LSince the number of moles remains constant:n = n'0.20 moles = [NOCl'] 1 L[NOCl'] = 0.20 MNow, let's denote the change in concentration at equilibrium as x. The equilibrium concentrations will be:[NOCl] = 0.20 M - 2x[NO] = 0 M + 2x[Cl2] = 0 M + xThe equilibrium constant expression is:Kc = [NO]^2 [Cl2] / [NOCl]^2 Substitute the equilibrium concentrations and the given Kc value:0.054 = 2x ^2 x / 0.20 - 2x ^2 Now we need to solve this equation for x. It's a cubic equation, and solving it, we get:x 0.0367Now we can find the new equilibrium concentration of NOCl:[NOCl] = 0.20 M - 2x 0.20 M - 2 0.0367 0.1266 MSo the new equilibrium concentration of NOCl is approximately 0.1266 M.