Verified answer

find dGo for the reaction:

N2 (g) + 3 H2 (g) → 2 NH3 (g)

dG^o rxn = dG prod - dG reactants

dG^o rxn = dG (2 moles NH3 @ -16.67 kJ/mol) - zero

dG^o rxn = -33.34 kJ

dG^o rxn = - RTlnK

-33,340 J = - 8.314 J/mol-K)(256K) lnK

ln K = 15.66

doing n .... e^x

K = 6.3532 e6

your answer, rounded to 3 sig figs is

K = 6.35 X 10^6

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p.s.

the key to this problem is either know that ..., or to look up that

the standard dGf 's , the dG^o of formation for elements in their most natural state is Zero

dHf's & dG^o of formation are zero for elements in their mostr natural state

see any textbook or: http://bilbo.chm.uri.edu/CHM112/tables/thermtable....

so once again

dG^o rxn = dG^o formation prod - dG^o formation reactants

dG^o rxn = (2 mol NH3 @ -16.67 kJ/mol) - zero for N2 & H2

dG^o rxn = -33.34 kJ

which gave your final answer above as

K = 6.35 X 10^6

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p.p.s.

often times when someone sees this question they are reminde of times when they were asked to solve for dG at some other temp. they use the formula :

dG = dH -Tds

they then mistakenly use this dG @ this new temp, instead of using dG^o the standard G

the formula we use, uses dG^o , the standard G:

dG^o rxn = - RTlnK

this formula comes from the equation:

dG = dG^o rxn + RTlnK

which, knowing that at equilibrium dG = 0....

became simplified as

dG^o rxn = - RTlnK

it uses the dG^o rxn, something which has a value other than zero... & then calculates the K at a temp at which the dG was eliminated out of the equation as it is zero @ equilibrium

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your answer is

as calulated above , rounded to 3 sig figs is

K = 6.35 X 10^6

You have to convert all this equilibrium constants into equation H3PO4 -> H+ + H2PO4- K1 = 6.9 x 10-3 mol/L equation 1 H+ + PO43- -> HPO42- K2 = 2.1 x 1012 mol/L equation 2 H2PO4- -> HPO42- + H+ K3 = 6.2 x 10-8 mol/L equation 3 and if i'm right, K1 = [H+][H2PO4-] / [H3PO4 ] , the same for the others Then you have to rewrite the last with the three first : H3PO4 + 3 H2O -> 3H3O+ + PO43- equation 4 eq. 4 = 1 - 2 + 3 You write the good constants for this last equation, and then you look with the 1, 2 and 3. It's not very difficult, but you have to attentive with all this equations.

ΔG°= 2 ΔGf0(NH₃)-3 ΔGf0(H₂)-ΔGf0(N₂)

Now ΔG= ΔH-TΔS

For elements: ΔH=0 and the entropies are tabulated (e.g. Beilstein)

We get:

ΔGf0(NH₃)=-16.67 KJ/mol

ΔGf0(H₂)=-256K*131 J/molK=-33.54 KJ/mol K

ΔGf0(N₂)= 256K* 192 J/mol K=49.15 KJ/mol

-> ΔG°= (-2*16.67-3*33.54- 49.15)KJ/mol= -183.11 KJ/mol

ΔG°=-RT lnK-> ln K=- ΔG°/(RT)= 86.03 -> K= 2.31*10^37