Question

2. (10 pts) Determine , and when 100.0 g of hydrogen reacts with sufficient amount of nitrogen to form ammonia at standard thermodynamic conditions. Determine Δ H, ΔU , ΔA , ΔG ,ΔS , q and w. The enthalpies of formation and entropies are: ΔHf = { H2= 0, N2 =0, NH3 =-46.1} Sْ = { H2 =130.68 , N2 = 191.61 , NH3 = 192.45 }

Answer 1

3H2(g) + N2(g) ----> 2NH3(g)

DHrxn = (2*DHfNH3)- (DHfN2+3*DHfH2)

      = (2*-46.1) - ( 0 + 2*0)

      = -92.2 kj

DSrxn = (2*192.45)-(191.61+3*130.68)

      = -198.75 j/k.mol

No of moles of H2 = 100/2 = 50 mole

from equation. 3 mole H2= 2 mole NH3 = 1 mole N2

3 mole H2 = -92.2 kj

50 MOLE H2 = -92.2*50/3 = -1536.67 kj

DHrxn of 100 grams H2 = -1536.67 kj

dsrxn 100 grams H2 = -198.75*50/3 = -3312.5 j/k.mol.

DHrxn = DU+DnRT

DUrxn = DHrxn - DnRT

Dn = nproducts - nreactant

   = 2-4 = -2

      = -1536.67- (-2*(8.314/1000)* 273.15)

DUrxn = -1532.13 kj

DG = DH-TDS

   = -1536.67 - (273.15*(-3312.5/1000))

DGrxn   = -631.86 kj

DArxn = DU-TDS

    
   = -1532.13 - (273.15*(-3312.5/1000))

   = -627.32 kj

FROM FIRST LAW OF THERMODYNAMICS

w = DnRT

   = (-2*(8.314/1000)* 273.15)

   = -4.542 kj

q = DU+w

q = -1532.13-4.542 = -1536.672 kj