Consider the formation of hydrogen fluoride: H_2(g) + F_2(g) xharr 2HF(g) if a 3
ID: 534364 • Letter: C
Question
Consider the formation of hydrogen fluoride: H_2(g) + F_2(g) xharr 2HF(g) if a 3.0 L nickel reaction container (glass cannot be used because it reacts with HF) filled with 0.0055 M H_2 is connected to a 3.4 L container filled with 0.033 M F_2. The equilibrium constant. K_p, is 7.8 times 1014 . Calculate the molar concentration of HF at equilibrium. A further hint is provided after the first attempt in the feedback. Answer: _________ BONUS Suppose a 2.00 L nickel reaction container filled with 0.0072 M H_2 is connected to a 5.00 L container filled with 0.287 M F_2. Calculate the molar concentration of H_2 at equilibrium.Explanation / Answer
Solution:
Calculation of number of moles of H2 and F2:
Volume of H2 = 3.0 L
Molarity of H2 = 0.0055M
Therefore, number of moles of H2 = Molarity x volume of H2 = 0.0055 M x 3.0 L = 0.0165 mol
Similarly, number of moles of F2 = 0.033 M x 3.4 L = 0.1122 mol
Identification of limiting reagent:
From the above reaction, it is clear that, 1 mol of H2 reacts with 1 mol of F2 for complete reaction.
Thus, 0.0165 mol of H2 will react with 0.0165 ol of F2 for complete reaction.
Thus, H2 is the limiting reagent in the reaction.
Calculation of molar concentration of HF:
The value of Kp is 7.8 x 1014
The greater value of Kp implies that the reaction is about to complete.
Now, 1 mole of H2 forms 2 moles of HF after complete reaction.
Therefore, 0.0165 mol of H2 will form (2 x 0.0165 = 0.033) mol of HF after complete reaction.
Total volume of the container = 3.0 L + 3.4 L = 6.4 L
Therefore, molar concentration of HF = number of moles / volume in litres = 0.033 mol / 6.4 L = 0.005 M
Thus, the molar concentration of HF at equilibrium is 0.005 M.