DG0 = (2*DG0f,co2)-(2*DG0f,CO + 1*DG0f,O2)
= (2*-394.4)-(2*-137.2+1*0)
= -514.4 Kj
DG = DG0 + RTln Q
R = ideal gas constant = 8.314 j.k-1.mol-1
T = 298 k
Q = pCO2^2/pCO^2*pO2
= 0.025^2/(0.65^2*34)
DG = -514.4*10^3 +8.314*298ln(0.025^2/(0.65^2*34))
= -539.3 Kj
5.14. Determine A Gº and ArxnG for the following reaction at 25°C, using data in Appendix 2. The partial pre...
ΔΗ, (at 298K AG; (at 298 K in kJ/mol) Compound Compound AH (at 298K in kJ/mol in kJ/mol) AG (at 298 K. in kJ/mol 0 0 210.5 70.18 Ho (g) Hg (1) Hg,CI, (s) Ag (s) Agor(s AgCl(s) 0 265.37 106.76 -96.90 -109.80 0 -628.8 - 1582.3 -100.37 -127.01 0 -704 2 -1675.7 0 1 (g) AICI, (s) Al O (5) Ar (9) Au (s) BaSO's) 0 K(s) KBr(s) KCI (s) 0 -393.8 436.5 -567.3 -327.9 0 -1362.3 380.7 -408.5...
Using enthalpies of formation (Appendix C), calculate ΔH ° for the following reaction at 25°C. Also calculate ΔS ° for this reaction from standard entropies at 25°C. Use these values to calculate ΔG ° for the reaction at this temperature. COCl2(g) + H2O(l ) h CO2(g) + 2HCl(g). Appendix C Thermodynamic Quantities for Substances and Ions at 25°C Substance or lon AH; (kW/mol) 0 AG: (kJ/mol) 0 S° (J/mol-K) 20.87 ΔΗ, (kJ/mol) -946.3 - 33422 AG; {kj/mol) --859,3 -2793 (J/mol-K)...
Calculate the standard free-energy change and the equilibrium constant Kc for the following reaction at 25°C. See Appendix C for data. Fe(s) + Cu2+ (aq) = Fe2+ (aq) + Cu(s) Find equilibrium constan + K at 25°C Appendix C Thermodynamic Quantities for Substances and Ions at 25°C Substance or lon AH; (kJ/mol) AG; (kJ/mol (J/mol-K) 20.87 Substance or lon Ba(OH),(s) ΔΗ: (kJ/mol) -946.3 --3342.2 AG (kJ/mol) -859.3 -2793 (J/mol K) 107.1 427 Ba(OH), 8H,O(s) BaSO (8) - 1473.2 -1362.3 132.2...