I2(g) + Cl2(g) < ==== > 2 ICl(g) An equilibrium mixture of the above contains 0.0100 atm I2(g), 0.0100 atm Cl2(g), and 0.0900 atm ICl(g) at 298 K. Calculate the Delta G at standard condition.

Question

This was a multiple choice question. I don't know what I was doing wrong. I tried using the formula:

delta G= delta G° + RT ln(Q)

I got delta G° = -30440 J/mol by using the G_f table 2(-5.52)-19.4+0 then convert to J

Q= (0.0900atm)² ÷ (0.0100atm) (0.0100atm)

After plugging everything in I got -19.55Kj/mol but that wasn't one of the choices but after guessing I found out the correct answer was -10.9Kj/mol. I need help on how to correctly do this problem.

Thanks

Answer 1

It says that the system is at equilibrium and gives the equilibrium concentrations, so no need to find Q. You simply find K (equilibrium constant). It will be the same value as you calculated for Q, but has a different meaning.

K = [ICl]² / [I₂][Cl₂] = (0.09)² / (0.010(0.01) = 81

Since the system is @ equilibrium, ∆G = 0 and so your equation of ∆G = ∆Gº + RT ln Q becomes

∆Gº = -RT ln K

∆Gº = - (0.008314 kJ/Kmol)(298K) ln 81

∆Gº = -2.477 x 4.39

∆Gº = - 10.89 kJ/mol