Part a
N2(g)+3H2(g)=2NH3(g)
By using values of enthalpy and entropy from table
∆H°rxn=2×∆Hf(NH3)−3×∆Hf(H2)−∆Hf(N2)
∆H°rxn=2×−45.9−3×0−0
∆H°rxn=-91.8 kJ/mol
∆S°rxn=2×S°(NH3)−S°(N2)−3×S°(H2)
∆S°rxn=2×199−191.5−3×130.6
∆S°rxn=−197.3J/K
By using relationship
∆H°−T∆S°=−RTLnKp
Kp=1
−91.8×10³+T×197.3=0
T=465.28K
Part b)
AtT=400°C=673K
∆H°-T∆S°=-RTLnKp
-91.8×10³-673×-197.3=-8.314×673×lnKp
lnKp=−7.3245
Kp=6.592×10−4
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