Using the conditions given, calculate equilibrium constants designated for A-F.

Question

Using the conditions given, calculate equilibrium constants designated for A-F. Rank from the highest value of equilibrium constant, to the lowest. The temperature is 500C. A. C2H4(g) + F2(g) <=> C2H4F2(g) initial conc: [C2H4] = 0.10 M; [F2] = 0.10 M equilibrium conc: [C2H4F2] = 0.020 M

B. C2H4(g) + F2(g) <=> C2H4F2(g) initial conc: [C2H4F2] = 0.25 M equilibrium conc: [F2] = 0.165 M

C. 2 BrNO(g) <=> 2 NO(g) + Br2(g) initial conc: [BrNO] = 1.0 M equilibrium conc: [BrNO] = 0.10 M

D. 2 BrNO(g) <=> 2 NO(g) + Br2(g) initial conc: [NO] = 0.15; [Br2] = 0.19M equilibrium conc: [BrNO] = 0.010M

E. 2 KClO3(s) <=> 2 KCl(s) + 3 O2(g) initial conditions: 1.0 mol of KClO3 in a 1.0 L flask equilibrium conditions: 0.0020 mol O2 created

F. 2 KClO3(s) <=> 2 KCl(s) + 3 O2(g) initial conditions: 0.01 mol of KClO3 in a 2.0 L flask equilibrium conditions: 0.0040 mol O2 created

Highest 1_______ 2________ 3________ 4________ 5________ 6________ Lowest

I REALLY need to know how to work these, so PLEASE show me the work.

Explanation / Answer

A. C2H4(g) + F2(g) <=> C2H4F2(g)

Kc = [C2H4F2] / [C2H4][F2]

Kc = 0.020 / (0.08 * 0.08)

Kc = 3.125

B. C2H4(g) + F2(g) <=> C2H4F2(g

Kc = (0.25 - 0.165) / (0.165 * 0.165)

Kc = 3.12

C. 2 BrNO(g) <=> 2 NO(g) + Br2(g)

Kc = [NO]2[Br2] / [BrNO]2

Kc = (0.90)2(0.45) / (0.10)2

Kc = 36.45

D. 2 BrNO(g) <=> 2 NO(g) + Br2(g)

Kc = (0.14)2(0.185) / (0.010)2

Kc = 36.26

C > D > A > B

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