(b)Calculate△s°(JK-mol') △H=SN,04e）-2SNO2g） =304.2-2(240.0) =-175.8 JK*mol (c)Calculate AG(kJ mol) △G=△H°-T△S =-57,240Jmo-(25+273)K(-175.8JK-mol =-4851.6Jmol =-4.85kmo (d)Calculate Kp at 50C K,-0.ma60 4851J mol1 (e)If the total pressure at 50C is 1 bar,calculate the partial pressure of each gas Since the total pressure is I bar,then pNo2 +pN2o4=1 ButK。=6.09=Po Thus,PN204=6.09(PNO2)2 PNOz PNo2+6.09(pNo2)2=1 6.09(pNo2)2+pNo2-1=0 Solving this quadratic,pNo2=0.331 atm PN204=1-PNO2=1bar-0.331 bar =0.669 bar 6 (b) Calculate ΔSo (J K-1 mol-1) ΔHo = So (N2O4(g)) - 2 So (NO2(g)) = 304.2 – 2(240.0) = -175.8 J K-1 mol-1 (c) Calculate ΔGo (kJ mol-1) ΔGo = ΔHo – TΔSo = -57,240 J mol-1 – (25 + 273)K(-175.8 J K-1 mol-1) = -4851.6 J mol-1 = -4.85 kJ mol-1 (d) Calculate Kp at 50o C o G J mol RT . J K mol ( )K Ke e p . − − − ⎛ ⎞⎛ −Δ ⎞ ⎜ ⎟⎜ ⎟ ⎟ ⎝ ⎠⎝ + ⎠ = = = 1 1 1 4851 8 314 50 273 6 09 (e) If the total pressure at 50o C is 1 bar, calculate the partial pressure of each gas. Since the total pressure is 1 bar, then pNO2 + pN2O4 = 1 But, N O p NO p K . p = = 2 4 2 2 6 09 Thus, pN2O4 = 6.09(pNO2) 2 pNO2 + 6.09(pNO2) 2 = 1 6.09(pNO2) 2 + pNO2 – 1 = 0 Solving this quadratic, pNO2 = 0.331 atm pN2O4 = 1 – pNO2 = 1bar - 0.331 bar = 0.669 bar 6