Asked by OVA9000
For the following equilibrium; Br2 (g) + Cl2 (g) ⇄ 2BrCl (g)
the Kc = 7.00 at 127oC. If 0.450 mol of Br2 (g) and 0.450 mol of Cl2 (g) are introduced into a 3.00 L container at 400 K, what will the equilibrium concentration of BrCl(g) be?
the Kc = 7.00 at 127oC. If 0.450 mol of Br2 (g) and 0.450 mol of Cl2 (g) are introduced into a 3.00 L container at 400 K, what will the equilibrium concentration of BrCl(g) be?
Answers
Answered by
DrBob222
I assume you want the concns at 127 C.
M = mols/L = 0.450/3 = 0.150M
............Br2 + Cl2 ==> 2BrCl
I.........0.150..0.150......0
C............x.....-x......2x
E......0.150-x...0.150-x....2x
Substitute into Kc expression and solve for x and 2x.
M = mols/L = 0.450/3 = 0.150M
............Br2 + Cl2 ==> 2BrCl
I.........0.150..0.150......0
C............x.....-x......2x
E......0.150-x...0.150-x....2x
Substitute into Kc expression and solve for x and 2x.