I need help with some Ap Chemistry

Question

Nitroglycerin is a very powerful explosive which decomposes to four different gases when detonated.

2C3H5(NO3)3 (l) → 3N2 (g) + ½ O2 (g) + 6CO2 (g) + 5H2O (g)

(a) Calculate the ΔH° for this process given the following information.  

Species	ΔHf° (kJ/mol)
C3H5(NO3)3  (l)	-364
CO2 (g)	-393.5
H2O (g)	-241.8

(b) Calculate the amount of heat liberated when 18.0 g of nitroglycerin decomposes.

(c) What volume, in liters, of N2 (g) would be produced at 25.0°C and 1.00 atm of pressure in the above reaction if it involves the transfer of 500.0 kJ of heat?

(d) At 25°C the ΔS is known to be 156 J/K. Is this reaction thermodynamically favored at this temperature?  Justify your answer.

Answer 1

2C3H5(NO3)3 (l) → 3N2 (g) + ½ O2 (g) + 6CO2 (g) + 5H2O (g)

(a) ∆H_rxn = ∑n∆H_products - ∑n∆H_reactants

∑n∆H_products = 3x0 + 1/2x0 + 6x-393.5 + 5x-241.8 = -3570 kJ

∑n∆H_reactants = 2x-364 = -728 kJ

∆H_rxn = -3570 kJ - (-728 kJ)

∆H_rxn = -2842 kJ

(b) 18.0 g NG x 1 mol NG/227 g x -2842 kJ/2 mol NG = -225 kJ

(c) 25ºC and 1 atm = STP and at STP 1 mol of gas = 22.4 liters

500.0 kJ x 3 mol N2/-2842 kJ x 22.4 L/mol = 11.8 L N2 gas

(d) For this rx to be favored, the ∆G should be negative.

∆G = ∆H - T∆S

∆G = -2842 - T(0.156) note:changed units of ∆S to agree with units of ∆H

∆G = -2842 - (298x0.156) = -2842 - (- 46.5)

∆G = ~ -2780 kJ

The reaction would be thermodynamically favored at 25ºC because ∆G is negative at this temperature.