.........2BrCl ==> Br2 + Cl2
I.......4.680.......0.....0
C........-2x........x.....x
E......4.680-x......x.....x
Kc = .... Substitute the E line into Kc expression and solve for (BrCl)
Then %BrCl = (BrCl)/4.680)*100 = ?
If the initial concentration of BrCl(g) is 4.680 mol/L, calculate the % of BrCl(g) left over after the reaction reaches equilibrium according to the balanced equation. The value of Kc at 500.0 K is 32.00. The initial concentration of the reaction products is 0 mol/L.
2BrCl(g) = Br2(g)+Cl2(g)
1 answer
1 answer