Question

Need help with #6,7&9 to make sure I did it right. Will rate high if it is correct! Thank you.

6. Use given data to calculate the standard free- energy change for the reaction Pa(g) + 6 Cl2(g) → 4 PC13(g) at 298 K. (ii) What is AGº for the reverse of this reaction? PA(g) Cl2(g) PC13(g) 24.4 -299.6 kJ/mol AGE 9. Calculate AG at 298 K for a mixture of 2.0 atm N2, 3.0 atm H2, and 0.50 atm NH3 being used in the Haber process: N2(g) + 3 H2(g) ++2 NH3(g) AG° = -33.3 kJ for this reaction. a) -1222.8 kJ/mol, +1222.8 kJ/mol b) -1344.8 kJ/mol, +1344.8 kJ/mol c) -446.8 kJ/mol, +446.8 kJ/mol d) 153.2 kJ/mol, -153.2 kJ/mol e) -1222.8 kJ/mol, -1222.8 kJ/mol a) -44.9 kJ b) -13.4 kJ c) -46.6 kJ d) 44.9 kJ e) 46.6 kJ 7. In Section 5.7 we used Hess's law to calculate AH for the combustion of propane gas at 298 K: 3 CO2(g) + 4 H2O(l) → C3H3(g) + 5 O2(a) AH = +2220 kj Without using data from Appendix C, predict whether AGº for this reaction is more positive or less positive than ΔΗ°. a) Less positive b) More positive

Answer 1

tu (g) + 6C ca 4 Paz eg) A4 = 544) Product - Al Roactant = 4x (-299-6) - 610) - 94.4 = - 1222-8 kJ / moles 19%eovnse = -05 1 = 1222.8 kJ/male 31029) + 4H2011) ► cz Hg 8) + 5029) AHO = +2220 kJ AG° for this reaction ů less positive because . combustion reactions are spontaneous reactions. N₂ cg J + 3 H₂ cg) 2 NH3 cg). A9 - - 33.3 kJ = - 33-34103 J AG = oGº + 7.303 RT longo L (3) 3 x2 -33.3810° - 13319.84 - - 46.6KJ