.........2ClF(g) ==> F2(g)+Cl2(g)
I......0.5789.....0.5980..0.05037
C........+x..........-x......-x
E....0.5789+x...0.5980-x..0.05037-x
You know Eq for F2 is 0.5980-x = 0.5802
That allows you to calculate x and that allows you to calculate eq concns of each. Plug these into Kc expression and solve for Kc.
Post your work if you get stuck.
The initial concentration for the compounds involved in the reaction shown were determined to be [ClF(g)] = 0.5789 mol/L, [F2(g)] = 0.5980 mol/L, [Cl2(g)] = 0.05037 mol/L. Calculate the value of the equilibrium constant (Kc) at 2500 K if the equilibrium concentration of F2(g) was 0.5802 mol/L.
2ClF(g) = F2(g)+Cl2(g)
1 answer
1 answer