Question

70. Use data from Appendix IIB to calculate the equilibrium is constants at 25°C for each reaction. -1.0 kJ/mol. a, 2 NO2(g)一一N204(g) b. Br2(g) + Cl2(g)--2 BrCl(g) for BrCl(g)

Answer 1

70 a .

$2NO_2(g) \rightleftharpoons N_2O_4(g)$

$\Delta G^0_{f, \: rxn} = \Delta G^0_{f, \: N_2O_4(g)} - 2\Delta G^0_{f, \: NO_2(g)} \\ \:\\ \: \Delta G^0_{f, \: rxn} = 99.8 - 2 \times 51.3 \\ \:\\ \: \Delta G^0_{f, \: rxn} = -2.8 \: kJ/mol$

$\Delta G^0_{f, \: rxn} = -RTlnK \\ \:\\ \: -2.8 \: kJ/mol \times 1000 \: J/kJ = - 8.314 \: J/mol/K \times 298.15 \times ln K \\ \:\\ \: lnK=-1.1297 \\ \:\\ \: K= 0.323$

b.

$Br_2(g) + Cl_2(g) \rightleftharpoons 2BrCl(g)$

$\Delta G^0_{f, \: rxn} =2 \times \Delta G^0_{f, \: BrCl(g)} - \Delta G^0_{f, \: Br_2(g)} - \Delta G^0_{f, \: Cl_2(g)} \\ \:\\ \: \Delta G^0_{f, \: rxn} = 2 \times (-1.0) - 0.0-3.1 \\ \:\\ \: \Delta G^0_{f, \: rxn} = -5.1 \: kJ/mol$

$\Delta G^0_{f, \: rxn} = -RTlnK \\ \:\\ \: -5.1 \: kJ/mol \times 1000 \: J/kJ = - 8.314 \: J/mol/K \times 298.15 \times ln K \\ \:\\ \: lnK=2.06 \\ \:\\ \: K= 7.83$