Suppose a 3.00 L nickel reaction container filled with 0.0072 M H2 is connected

Question

Suppose a 3.00 L nickel reaction container filled with 0.0072 M H2 is connected to a 3.00 L container filled with 0.247 M F2. Calculate the molar concentration of H2 at equilibrium.

-Hint, find the equilibrium concentration of HF and F2 (as in the above problem), then use the equilibrium expression to find the hydrogen concentration.

-When you do the limiting reactant calculation you essentially use up all the limiting reactant. So, hydrogen concentration is approximately zero. But, it can't be identically equal to zero or the Keq expression can not be correct. You will find how small it really is.

Explanation / Answer

Consider the formation of hydrogen fluoride:

H2(g) + F2(g) 2HF(g)

the total container volume is 6.0 L

You have, initially:
0.0072 M * 3.0 L = 0.0216 mol H2;         0.0216 mol / 6.0 L = 0.0036 M H2
0.247 M * 3.0 L = 0.741 mol F2;            0.741 mol / 6.0 L = 0.1235 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:
(0.0036 - x) M H2
(0.1235 - x) M F2
(+2x) M HF
satisfying the equilibrium expression [HF]2/[H2][F2] = 7.8 x 1014       Assuming K = 7.8 x 1014

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 = 0.0036 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:
0.1235 - 0.0036 = 0.1199 M F2
2(0.0036) = 0.0072 M HF

The actual value of [H2] is calculated as follows:
[HF]2/[H2][F2] = 7.8x1014
(0.0072)2/[H2](0.1199) = 7.8x1014
[H2] = 5.54x10-19 M

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