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Calculate the free-energy change for ammonia synthesis at 25 ? C(298 K) given the following sets...

Calculate the free-energy change for ammonia synthesis at 25 ? C(298 K) given the following sets of partial pressures:

a.1.0 atm N2, 3.0 atm H2, 0.020 atm NH3

b. 0.010 atm N2, 0.030 atm H2, 2.0 atm NH3

Solutions

Expert Solution

a) N2 (g) + 3 H2 (g) <---------------> 2 NH3 (g)

Qp = P2NH3 / PN2 x P3H2

     = (0.020)2 / 1.0 x (3.0)^3

     = 1.48 x 10^-5

T = 25 + 273 =298 K

R = universal gas constant = 8.314 x 10^-3 kJ/K mol

standard Gibbs free energy G0 = ?

relation between these two

G0 = - RT ln Qp

          = - 2.303 RT log Qp

          = - 2.303 x 8.314 x 10^-3 x 298 x log (1.48 x 10^-5)

          = 27.55 kJ

standard Gibbs free energy G0 = 27.5 kJ

b)

Qp = P2NH3 / PN2 x P3H2

     = (2.0)2 / 0.010 x (0.030)^3

     = 1.48 x 10^7

G0 = - RT ln Qp

          = - 2.303 RT log Qp

          = - 2.303 x 8.314 x 10^-3 x 298 x log (1.48 x 10^7)

          = - 40.9 kJ

standard Gibbs free energy G0 = - 40.9 kJ

queen_honey_blossom answered 1 year ago
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