For questions consider the chemical reaction for the formation of ammonia: N_2(g

Question

For questions consider the chemical reaction for the formation of ammonia: N_2(g) + 3H_2(g) 2NH_3(g) Delta G degree = -34 kJ What is the equilibrium constant (K) for the formation of ammonia at 298 K? 9.1 times 10^5 7.7 times 10^3 1.3 times 10^-3 1.7 times 10^-6 If you start with 0.60 atm N_2(g), 0.44 atm H_2(g), and 0.10 atm NH_3 (g) at 298K, then the reaction proceeds to the because right Delta G_SYS = -38 kJ left Delta G_SYS = -38 kJ right Delta G_SYS = +38 kJ left Delta G_SYS = +38 kJ left Delta G_SYS = 0 kJ

Explanation / Answer

21

find K

so

dG° = -RT*ln(Kc)

so

Kc = exp(-dG°/(RT))

Kc = exp(34000/(8.314*298))

Kc = 911742.414

Q22

Calculate Q

Q = NH3^2 / N2 * H2^3

Q = (0.1^2 )/ ((0.6)(0.44^3))

Q = 0.19565

since Qp is in pressure, change to concentration

Qp = Qc*(RT)^dn

Qp*(RT)^dn = Qc

Qc = 0.19565 * (0.082*298)^2

Qc = 116.82

since Qc << Kc, then this is not favoured

then this goes to the left

dGsystem must be possitive

and cnat be 0

so..

+38 and left is the answer (D)

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