reaction free energy?

Question

Calculate the reaction free energy of

H2(g) + I2(g) 2HI(g)
when the concentrations are 0.026 mol L-1 (H2), 0.33 mol L-1 (I2), and 1.84 mol L-1 (HI), and the temperature is 700K. For this reaction Kc = 54 at 700 K.

a. +11.6 kJ mol-1
b. 0 kJ mol-1
c. -11.6 kJ mol-1
d. +2273 kJ mol-1

Answer 1

H2(g) + I2(g) ==> 2HI(g)
[H2] = 0.026 mol/L, [i2] = 0.33 mol/L, and [HI] = 1.84 mol/L. T = 700K and Kc = 54.
Why do you give BOTH chemical concentrations and reaction constant Kc? It is supposed to have:
[HI]^2/([H2][I2] = 1.84^2 /(0.026*0.33) = 394.6 =?= 54?
Based on [HI]^2/([H2][I2] = 1.84^2 /(0.026*0.33) = 394.6, we may have:
ΔG = -kT Ln{ [HI]^2/([H2][I2] }
= -8.314*700*Ln(394.6)
= -34790 (J)
=-34.8(KJ)
This answer can not find any match.
However, based on Kc = 54, we may have:
ΔG = -kT Ln(Kc)
= -8.314*700*Ln 54
= - 23.2(KJ)
Since that is for formation of 2Mol of HI, for one mol the result needs to be divided by 2. But then the free energy would not call for "this reaction (to form 2Mol of HI)". It should be called "Formation"
The answer is C.