Question

Estimate the value of the equilibrium constant at 635 K for each of the following reactions. ?G?f for BrCl(g) is ?1.0 kJ/mol. The standard molar entropy, S?, for BrCl(g) is 240.0 J/mol?K.

2NO2(g)?N2O4(g). ?H?f for N2O4(g) is 9.16 kJ/mol.

Br2(g)+Cl2(g)?2BrCl(g) .?H?f for BrCl(g) is 14.6 kJ/mol.

Answer 1

From text book

compound	AGp(kJ/mol)
NO2	51.3
N2O4	99.8

2NO2(g)<===> N2O4(g).

$\Delta G_r^{\circ} =\sum n\times \Delta G^{\circ}(products)-\sum n\times \Delta G^{\circ}(reactants)$

▲Go-99.8-2 × 51.3

$\Delta G_r^{\circ} =-2.8 \ kJ/mol$

Formula to find k

$\Delta G_r^{\circ} =-RTlnK$

$K =e^{\frac{-\Delta G^{\circ}}{RT}}$

$K =e^{\frac{-(-2.8\times 10^3 J/mol)}{635 K\times 8.314 J/K/mol}}$

K = 1.70

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Br2(g)+Cl2(g)<===>2BrCl(g)

BrCl(g) is 14.6 kJ/mol.

compound	$\Delta H_f^{\circ}(kJ/mol)$	AGp(kJ/mol)	$S_f^{\circ}(J/mol/K)$
BrCl(g)	14.6	-1.0	240.0
Br2(g)	30.9	3.1	245.5
Cl2(g)	0	0	223.1

$\Delta G^{\circ}= 2(-1.0)-3.1 =-5.1\ kJ$

At 635 K

$\Delta G^{\circ}= -RTlnK$

$K=e^{\frac{-\Delta G}{RT}}$

$K=e^{\frac{-(-5.1\times 10^{-3})}{8.314\times 635}}$

K = 2.63