Problem 3

Dissociatin of molecular Bromine
Br_2 == 2 Br

Constants

data = {a→37.32 J/(K mol), b→0.5 10^(-3) J/(K^2mol), c→ - 1.26 (J K)/mol, R→8.315 J/(K mol)} ;

R = 8.315 J/(K mol) ;

Specific heats for a diatomic and monotomic Bromine

cp1 = a + b T + c/T^2 ;

cp2 = 5/2.R ;

cpBr2[T_] := (a + b T + c/T^2)/.data

cpBr[T_] := cp2/.data

Values at T= 298 K

ΔH298 = 192861 J ;

ΔG298 = 161682 J ;

ΔS298 = - (ΔG298 - ΔH298)/(298.15 K)

(104.575 J)/K

cpBr2[x]

(37.32 J)/(K mol) + (1.26 J K)/(mol x^2) + (0.0005 J x)/(K^2 mol)

At T= 500, 1000  K we get fo ΔH

ΔH500 = ΔH298 + ∫_500^298mol cpBr2[K x] K x + ∫_298^5002 mol cpBr[K x] Kx

193680. J

ΔH1000 = ΔH298 + ∫_1000^298mol cpBr2[K x] K x + ∫_298^10002 mol cpBr[K x] Kx

195620. J

Entropy at T = 500 K

ΔS500 = ΔS298 + ∫_500^298 (mol cpBr2[K x] K )/(K x) x + ∫_298^500 (2 mol cpBr[K x] K)/(K x) x

(106.676 J)/K

At T = 1000

ΔS1000 = ΔS298 + ∫_1000^298 (mol cpBr2[K x] K)/(K x) x + ∫_298^1000 (2 mol cpBr[K x] K)/(K x) x

(109.375 J)/K

Gibbs energy at 500 K and 1000 K

ΔG500 = ΔH500 - 500 K ΔS500

ΔG1000 = ΔH1000 - 1000 K ΔS1000

140342. J

86245. J

Equilibrium constants

K500 = Exp[-ΔG500/(R mol 500. K)]

2.18665*10^-15

K1000 = Exp[-ΔG1000/(R  mol 1000. K)]

0.0000312899

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