For the equilibrium Br2(g) + Cl2(g) equilibrium reaction arrow 2 BrCl(g) at 400.
ID: 729207 • Letter: F
Question
For the equilibriumBr2(g) + Cl2(g) equilibrium reaction arrow 2 BrCl(g)
at 400. K, Kc = 7.0. If 0.29 mol of Br2 and 0.51 mol of Cl2 are introduced into a 1.5-L container at 400. K, what will be the equilibrium concentrations of Br2, Cl2 and BrCl?
Explanation / Answer
0.29 mol of Br2 and 0.51 mol of Cl2 are introduced........ so we can say dat based on that assumption. 0.29 moles/1.5 L = 0.1933 M for (Br2) and 0.51 moles/1.5= 0.34 M for (Cl2) initially. Set up and ICE chart. Br2 + Cl2 ==> 2BrCl initial change...(0.193-x)M (0.34-x)M 0 equil..... (0.193-x)M (0.34-x)m 2x 7= (2x)^2/ (.193-x)(.34-x) Substitute the equilibrium values into the Kc expression and solve for x and from dat we can calculate d value of equilibrium conc. for Br2, Cl2.,BrCl.