[petrichorr]

http://wallace.chem.ox.ac.uk/teaching/1.3%20Thermodynamics%20II.pdf
Question 2B/C
So far i have calculated K using ∆rH, ∆rS, ∆rG* at 298k as (∆rH=-92,400JMol-1, ∆rS=-199JKMol-1, and ∆rG= -33,000JMol-1) and have got a value of K as 746.5, however I am skeptical about it as K seems quite large, when calculating K i used the formula LnK=-∆rG/RT using my values from part A of the question and the temperature as the one where equilibrium is reached (600K), I'm unsure whether is alter my value of ∆rG* to take into account the equilibrium temperature or if that is already accounted for by LnK=-∆rG*/RT, the next problem i encounter is setting up the question to answer part C, I have worked out a formula for K in terms of pressure and the molefractions as K after cancelling down and simplifying as K= (Xnh3)2/((Xn2) +(Xh2)3 + P2) and worked out the molefractions of N2 and H2 as 0.1625 and 0.4875 respectively using the molefraction given of Nh3=0.35 as a guide. However when rearranging to find the total pressure i find that the value of P is very small (0.09) so there must either be a mistake in part B in working out K (which seems most likely) or an error in my assumptions for C, can anyone please point out/guide me as to where I'm going wrong, any help would be greatly appreciated!

[mjc123]

NO that is WRONG!! You cannot use lnK =-ΔG°/RT to calculate K at one temperature using a value of ΔG° at a different temperature. That is because the temperature variation is not only contained in the factor 1/RT, but is inherent in ΔG°, which is NOT a constant. You must use ΔG° at the temperature of interest. You know that ΔG = ΔH - TΔS; however it is not enough to simply use this equation with T = 600 and your calculated values of ΔH and ΔS because over this wide range of temperature ΔH and ΔS are not constant. You need to work out their values at 600K using the heat capacity data.
Your mole fractions are correct, but your expression for K looks wrong, unless those plus signs are meant to be multiplication signs.