Consider the formation of hydrogen fluoride: H2(g) + F2(g) 2HF(g) If a 2.4 L nic
ID: 482301 • Letter: C
Question
Consider the formation of hydrogen fluoride:
H2(g) + F2(g) 2HF(g)
If a 2.4 L nickel reaction container (glass cannot be used because it reacts with HF) filled with 0.0054 M H2 is connected to a 3.8 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.
BONUS Suppose a 2.00 L nickel reaction container filled with 0.0072 M H2 is connected to a 5.00 L container filled with 0.158 M F2. Calculate the molar concentration of H2 at equilibrium.
Explanation / Answer
Given that 2.4 L nickel reaction container filled with 0.0054 M H2 is connected to a 3.8 L container filled with
0.030 M F2.
Total volume = 2.4 L + 3.8 L = 6.2 L
Moles of H2 = Molarity x volume = 0.0054 M x 2.4 L = 0.01296 mol
[H2] = 0.01296 mol/ 6.2 L = 2.09 x 10-3 M = 0.00209 M
Moles of F2 = Molarity x volume = 0.03 M x 3.8 L = 0.0114 mol
[F2] = 0.0114 mol/ 6.2 L = 0.01838 M
Hence,
H2(g) + F2(g) 2HF(g)
Initial 0.00209 mol 0.01838 mol 0
At equilibrium 0.00209 - X 0.01838 - X 2X
Then,
No of moles of H2 are less than F2. So, H2 is the limiting reagent.
So,
at equilibrium, concentratin of H2 will be zero.
Then, 0.00209 - X = 0
X = 0.00209 M
Therefore,
molar concentration of HF at equilibrium = 2X = 2 x 0.00209 M = 0.00418 M