At 350 k, kc = 0.142 for the reaction 2 brcl(g) ⇀↽ br2(g) + cl2(g) an equilibrium mixture at this temperature contains equal concentrations of bromine and chlorine, 0.043 mol/l. what is the equilibrium concentration of brcl? answer in units of m.
0.114 mol/l The equilibrium equation will be: Kc = ([Br2][Cl2])/[BrCl]^2 The square factor for BrCl is due to the 2 coefficient on that side of the equation. Now solve for BrCl, substitute the known values and calculate. Kc = ([Br2][Cl2])/[BrCl]^2 [BrCl]^2 * Kc = ([Br2][Cl2]) [BrCl]^2 = ([Br2][Cl2])/Kc [BrCl] = sqrt(([Br2][Cl2])/Kc) [BrCl] = sqrt(0.043 mol/l * 0.043 mol/l / 0.142) [BrCl] = sqrt(0.001849 mol^2/l^2 / 0.142) [BrCl] = sqrt(0.013021127 mol^2/l^2) [BrCl] = 0.114110152 mol/l Rounding to 3 significant figures gives 0.114 mol/l
