Consider the formation of hydrogen fluoride: H2(g) + F2(g) 2HF(g) If a 4.4 L nic

Question

Consider the formation of hydrogen fluoride: H2(g) + F2(g) 2HF(g) If a 4.4 L nickel reaction container (glass cannot be used because it reacts with HF) filled with 0.0067 M H2 is connected to a 2.9 L container filled with 0.035 M F2. The equilibrium constant, Kp, is 7.8 Times 1014 (Hint, this is a very large number, what does that imply?) Calculate the molar concentration of HF at equilibrium. A further hint is provided after the first attempt in the feedback. Suppose a 1.00 L nickel reaction container filled with 0.0057 M H2 is connected to a 4.00 L container filled with 0.069 M F2. Calculate the molar concentration of H2 at equilibrium. A hint is provided after the first attempt in the feedback.

Explanation / Answer

There are multiple questions here . i am allowed to answer only 1 at a time. I will answer 1st question for you.Please ask other as different question.

since Kp is very large number , it will not be a reversible reaction.

Number of moles of H2 = M*V = 0.0067*4.4 =0.02948 mol
Number of moles of F2 = M*V = 0.035*2.9=0.1015 mol

since 1 mol of H2 reacts with 1 mol of F2, H2 is limiting reahent.

we will use H2 for our further calculations.
Mol of HF formed = 2* mol of H2
= 2* 0.02948
= 0.05896 mol

Volume = 4.4 L

Concentration of HF = number of moles /volume
                   = 0.05896/4.4
                    =0.0134 M

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