The following questions are based on this reaction: N_2 (g) + 3 H_2 (g) 2 NH_3 (

Question

The following questions are based on this reaction: N_2 (g) + 3 H_2 (g) 2 NH_3 (g) Use the provided table to determine Delta G_rxn. For the above reaction at 298 K. Assuming Delta H_rxn. And Delta S_rxn do not change significantly with temperature, determine the temperature at which this reaction becomes spontaneous. Is the reaction spontaneous above this temperature, or below this temperature? Calculate the equilibrium constant (K_o) for this reaction at 298 K. Calculate Delta G when [N_2] = 0.010 M, [H_2] = 0.020 M, and [NH_3] = 0.50 M at 298 K. In which direction will the reaction proceed to reach equilibrium?

Explanation / Answer

a)

Grxn = Gproducts - Greactants = H - TS

HRxn = Hprod - Hfinal = 2*(-46.11) - (0)= -92.22 kJ/mol = -92220 J/mol

Srxn = Sprod - Sreact = (2*192.45) - (191.61 + 3*130.68) = -198.75 J/molK

Grxn = -92220 + 298*(-198.75 ) = -151447.5 J/mol = -151.44 kJ/mol

b)

for this to be in spotnanous

dG < 0 always

so

dG = H-Ts < 0

-92220 - T*-198.75 <0

198.75 *T <92220

T < 92220 /198.75

T < 464 K

temperature must be lower than 464 K

c)

dG = -RT*ln(K)

-151.44*10^3 = -8.314*298*ln(K)

K = exp((151.44*10^3)/(8.314*298))

K = 3.5153*10^26

d)

dG = dG° + RT*ln(Q)

dG = -151.44*1000 + 8.314*298*ln((0.5^2) / ((0.02^3)(0.01)))

dG = -114.388 J/mol = -114388.047 kJ/mol

this is spontanous, i.e. it favours products in equliibrium

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