Consider the following reaction: N_2(9) + 3H_2(g) rightarrow 2NH_3(g) Report all

Question

Consider the following reaction: N_2(9) + 3H_2(g) rightarrow 2NH_3(g) Report all answers to three significant figures! a) Determine the standard free energy of the reaction at 298K? Delta G degree = -32 b) Determine the equilibrium constant at 298K? K = 4.07e5 c) Determine the reaction quotient when the partial pressures of the gases are the following. N_2 = 10.90 atm; H_2 = 12.80 atm; NH_3 = 3.70 atm Q = d) Determine the value of Delta G when the reaction mixture is the one described in Part c. Delta G = kJ

Explanation / Answer

a)

dG = Gprod - Greactant

dG = 2*NH3 - (N2 + 3H2)

dG = 2*(-16.4) - (0 + 3*0) = -32.8 kJ/mol

b)

For K:

dG = dG° + RT*ln(K)

so

dG° = -RT*ln(K)

-32.8 *1000 = -8.314*298*ln(K)

K = exp(32.8 *1000/(8.314*298))

K = 561724.9701

K = 5.6*10^5

c)

Q = [NH3]^2 / [N2][H2]^3

Q = (3.7^2) / ((10.90)(12.80^3))

Qp = 0.000598

change to Qc

Qp = Qc*(RT)^dn

Qc = Qp*(RT)^-dn

Qc = (0.000598)((0.082*298)^-(2-4))

Qc = 0.3570

d)

so

dG = dG° - RT*ln(Q)

dG = -32800 - 8.314*298*ln(0.3570)

dG =-30248.052 J

dG = -30.24 kJ/mol

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