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4. The initial concentration for the compounds involved in the reaction shown were determined to be [ClF(g)] = 0.5789 mol/L, [F...Asked by plzhelp
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)
2ClF(g) = F2(g)+Cl2(g)
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Answered by
DrBob222
.........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.
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.
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