(a)Calculate(in kJ/mol)for the reaction. H=2HNOe）-HN2e）-H(O2e =180.0-0-0 =180.0 kJ/mol (b)Calculate S(in JKmol)for the reaction. S=2(NO()-S(N2()-S(O2(g) =2(211)-192-205 =+25 JK!mol (c)Calculate G(in kJ/mol)for the reaction at 500C G=H°.TS° =180,000J/mol-(500+273)K(25JKmo =+160675J/mol =+160.7kJ/mol (d)Calculate the value of Kp at 500C. G°=H°-298(S) =180,000J/mol-298K(25 JK!mol) =+172550Jmol -172550Jmo1 8.314JK-mo-1(500+273)K =2.19×10-12 (e)If po2=0.01 atm and pN2=0.7 atm,calculate pNo (in atm)at equilibrium at 500C. K品 Thus,PNO(KPN P =(2.19x102(0.7x0.01)12=1.24x10atm(a) Calculate ° (in kJ/mol) for the reaction. H° = 2 Hf°(NO(g)) - Hf°(N2(g)) - Hf°(O2(g)) = 180.0 – 0 – 0 = 180.0 kJ/mol (b) Calculate S° (in J K-1 mol-1) for the reaction. S° = 2 S°(NO(g)) - S°(N2(g)) - S°(O2(g)) = 2(211) – 192 – 205 = +25 J K-1 mol-1 (c) Calculate G (in kJ/mol) for the reaction at 500o C . G = H° - T S° = 180,000 J/mol – (500+273) K (25 J K-1 mol-1) = +160675 J/mol = +160.7 kJ/mol (d) Calculate the value of Kp at 500°C. Go = H° - 298( S°) = 180,000 J/mol – 298 K (25 J K-1 mol-1) = +172550 J/mol Kp = o G J mol exp exp . RT . J K mol ( )K − − − − ⎡⎤⎡ −Δ − ⎤ = = ⎢⎥⎢ ⎥ ⎣⎦⎣ + ⎦ 1 12 1 1 172550 2 19 10 8 314 500 273 × (e) If pO2 = 0.01 atm and pN2 = 0.7 atm, calculate pNO (in atm) at equilibrium at 500°C. ON 22 NO 2 p pp p K = Thus, pNO = (Kp pN2 pO2) 1/2 = (2.19 x 10-12 (0.7 x 0.01))1/2 = 1.24 x 10-7 atm