November+13th+-+pg.338


 * __Page 338.__

7.**

NH3(g) + HNO3(l) à NH4NO3(s)

a) **Reactants**

H°f(NH3) = -45.9 kJ H°f(HNO3) = -174.1 kJ


 * Products**

H°f(NH4NO3) = -365.6 kJ

H = H°f (products) - H°f (reactants) = -365.6 – [-45.9 + (-174.1)] = -365.6 – (-220) = -365.6 + 220 = **-145.6 kJ**

b)

c) 50T = 50000kg = 50000000g of NH4NO3

1 mol of NH4NO3 = 80.06g X mols = 50000000g = 624531.6 mols

-145.6 __kJ__ = __x kJ__ Mol 624531.6 mol = **- 9.1 x 107

8.** Enthalpy for the formation of anthracite (C52H16O) = -396.4 kJ

Anthracite Coal à 100kg = 100000g

2 C52H16O(s) + 111O2(g) à 104CO2(g) + 16H2O

Change in H = Sum of Hºf(products) – Sum of Hºf (reactants) = [(104)(-393.5) + (16)(-241.8)] – [(2)(_396.4) + (111)(0)] = (-40924 – 3868.8) – (-792.8 + 0) = (-44792.8 + 792.8) = -44, 000 kJ

1 mol anthracite = 656.68g x mols = 100000g = 152.28mol

1 mol  =  -22000kJ 152.28 = x kJ x = -3350160kJ = -3.4 x 106kJ
 * = -3.34 x 103MJ**