[kimi85]

The reaction below has Kc = 0.0425. When the reaction is initiated and allowed to come to equilibrium in a 40.0-L flask, 1.60 mol CH4 and 1.65 mol H2 are present. If an additional 0.10 mol H2 is added to the flask, what is the difference between Qc and Kc, if any, and will the number of grams of Mo increase, decrease, or remain the same?

Mo(s) + CH4(g) <==> Mo2C(s) + 2 H2(g)

Thank you very much.

[sdekivit]

when H₂ is added to the existing equilibrium, Qc is not equal to Kc anymore, but the system will adapt in such a way that Qc will become equal to Kc in time.

So the question is actiually: what happens to the equilibrium when H₂ is added --> will it shift to the left, right or remain unchanged ?

[AWK]

I understand quotient Qc is before equilibrium is reached. Only when Qc .EQ. Kc reaction remains unchanged

[kimi85]

here are the choices:
a.The difference is 0.0479; Mo increases because Qc > Kc.
b. The difference is 0.0423; Mo decreases because Qc < Kc.
 c.The difference is 0.0054; Mo is unchanged because it is a solid.
 d. The difference is 0.0054; Mo increases because Qc > Kc.

[vladi307]

I think that the d answer is correct. Qc>Kc because we put an additional amount of product into the equilibrium system. Mass of Mo must be increased because the reaction turns to the left (Le Chatelier principle).

[sdekivit]

d is indeed correct

Qc will be 0.0479 and thus Qc > Kc. Therefore Qc must decrease so more Mo(s) will be formed.

(Keq = [H₂]² / [CH₄] --> to let K decrease [CH₄] must increase and thus more Mo(s) will also be formed)

[kimi85]

I got it. Thank you very much. Yea, the correct answer is d.