Question

Estimate the value of the equilibrium constant at 625 K for each of the following reactions. ΔG∘f for BrCl(g) is −1.0 kJ/mol. The standard molar entropy, S∘, for BrCl(g) is 240.0 J/mol⋅K

1- 2NO2(g)⇌N2O4(g). ΔH∘f for N2O4(g) is 9.16 kJ/mol.

2- Br2(g)+Cl2(g)⇌2BrCl(g) .ΔH∘f for BrCl(g) is 14.6 kJ/mol.

Answer 1

1) Equilibrium equation: 2N0,(8) = 1,0. (8) AG. m = {1xAC(8,0) -2xACE(No.»]kJ/mol = ((1x99.8)-(2x51.3 )) kJ/mol =-2.8 kJ/mol Gas constant, R= 8.314 JK-mol-1 Temperature, T=25ºC +273 = 298 K Standard free energy change of the reaction, AG' = - RT In K. AGO -2.8x10 J/mol In K.q=- RT 8.314 JK-mol-2 x 298 K Kage = 0+1.13 = 3.10 ------------ Equilibrium equation: 2 NO,(8) = 1,04(8) AH. n = [1xAH:(8,00) – 2xAHNO,) ] kJ/mol = ((1x9.16)-(2x33.18 ))kJmol = -57.2 kJ/mol Temperature, T = 625 K According to VantHoff equation, K at 625K Kế Hº (1 11 In 20 á | In sang nở - R 62 --1208 this = 220–568x100 Kos = (3.10)(5.68x10°) = 1.76x10%

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2) Reaction: Bry (8)+ C1,(8) ► 2 BrCI(8) Standard free energy change of the reaction, com = mx4G(products) - 1x4G (reactarius) A CHE = [(2xApexey +) - (1xCp6) +1%AC%2(] = |(2x-1.0 kJmol-1) -((1X3.1 kJmol-")) +{1x0 kJm01-))) = -5.1 kJ/mol Temperature, T= 298 K Gas constant, R=8.314 JK-m01-1 Standard free energy change of the reaction, 4Gº = - RT in K In ka 400 RT -5.1x10 J/mol =2.06 8.314 JK-mol-? x 298 K) Equilibrium constant, K= 22.06 = 7.83 ----------- -------------- ------------ Equilibrium equation: Bry (8) + Cl2(g) → 2 BrCI(8) A n = [(2xAH'Exx)+) - 1x H120) + 1%AC%C40)] = (2x14.6 kJmol-1) - {1x30.91 kJm01-4) +{1x0 kJmol-) ) ) = -1.71 kJ/mol Temperature, T=625 K According to VantHoff equation, K at 625K In 783) = 8.314 IK-mol-1(298 K 625 K) -1.71x10* Jmol ( 1 )--0.361 konse = 231=0.697 K675 = (7.83) (0.697) = 5.46