Question

Consider the formation of hydrogen fluoride: H2(g) + F2(g) ↔ 2HF(g) If a 1.8 L nickel reaction container (glass cannot be used because it reacts with HF) filled with 0.0098 M H2 is connected to a 2.8 L container filled with 0.030 M F2. The equilibrium constant, Kp, is 7.8 x 1014 (Hint, this is a very large number, what does that imply?) Calculate the molar concentration of HF at equilibrium.

Answer 1

1. First of all, the large number 7.8x10¹⁴ indicates that the reaction goes almost to completion.

Second, the total container volume is 4.6 L

You have, initially:
0.0098 M x1.8 L = 0.01764 mol H2; 0.01764 mol / 4.6 L = 0.00383 M H2
0.030 M x2.8 L = 0.084 mol F2; 0.084 mol / 4.6 L = 0.0182 M F2
0 moles HF

During reaction, you get the following changes:
(x represents the extent of reaction; V represents total volume)
-xV moles H2; -x M H2
-xV moles F2; -x M F2
+2xV moles HF; -2x M HF

And so at equilibrium you have:
(0.00383 - x) M H2
(0.0182 - x) M F2
(+2x) M HF
satisfying the equilibrium expression [HF]^2 / [H2][F2] = 7.8 x 10^14.

Now, you can go ahead and solve the quadratic equation, or use the approximation that the reaction must go essentially to completion: i.e. x = 0.00383 since H2 is the limiting reagent. Then we have at equilibrium "zero" H2 (not really but it's very small compared to the others) and:
0.0182 - 0.003833 = 0.01437 M F2
2(0.003833) = 0.00766 M HF