Initial,atm 0.75 0 0 Change,atm -2x +2x +x Equilibrium,atm 0.75-2x 2x Atequilibrium. Pso Po=K PNo,2 (0.75-X7=448x103 (2x)2x Here,Kp is so small,we can make the approximation that x<<0.75,so the expression simplifies to 075=4.48x10-9 (2x)2x 4x3=(0.75)2(4.48×10-13) x3=(0.75)2(4.48×10-13)/4 x3=6.3×10-14 x=3.98×10-5 Thus,pNo2=0.75-2x=0.75 atm pw0=2x=7.96x103atm Po2=x=3.98 x 105 atm Check:(7.96x102(3.98x10)/(0.75-2(3.98x102=4.48x1013=K C2.Given the following information,calculate the ionization energy of lithium(in kJmol): Lattice energy of LiF(s) 1050 kJ mol! Electron affinity of fluorine -328 kJ mol △H°(Lifs) -617 kJ mol Enthalpy of sublimation of lithium metal 161 kJ mol Initial, atm 0.75 0 0 Change, atm -2x +2x +x Equilibrium, atm 0.75 – 2x 2x x At equilibrium, NO O p NO p p K p ( x) x . ( . x) − = = × − 2 2 2 2 2 13 2 2 4 48 10 0 75 Here, Kp is so small, we can make the approximation that x << 0.75, so the expression simplifies to ( x) x . (. ) x (. )( . ) x ( . ) ( . )/ x . x . − − − − − = × = × = × = × = × 2 13 2 32 1 32 13 3 14 5 2 4 48 10 0 75 4 0 75 4 48 10 0 75 4 48 10 4 6 3 10 3 98 10 3 Thus, pNO2 = 0.75 – 2x = 0.75 atm pNO = 2x = 7.96 x 10-5 atm pO2 = x = 3.98 x 10-5 atm Check: (7.96 x 10-5) 2 (3.98 x 10-5) / (0.75 – 2(3.98 x 10-6))2 = 4.48 x 10-13 = Kp C2. Given the following information, calculate the ionization energy of lithium (in kJ mol-1): Lattice energy of LiF(s) 1050 kJ mol-1 Electron affinity of fluorine − 328 kJ mol-1 ΔHf o (LiF(s)) − 617 kJ mol-1 Enthalpy of sublimation of lithium metal 161 kJ mol-1 4