Consider the reaction
4NH3(g) +5O2(g) ⇌ 4NO(g)+6H2O(g) where ΔS° rxn= 180.5 J/K
Using standard thermodynamic data (in the Chemistry
References), calculate the entropy change of the surroundings
and the universe at 25°C.
4 NH3(g) + 5 O2(g) 4 NO(g) + 6H2O(g)
∆Hrxn = enthalpy of formation of products - enthalpy of formation of reactants
= 6×(-241.82) + 4×90.25 - 5× 0 - 4×(-46.11)
= -905.48kJ/mol
Heat released will be transferred to the surrounding .
∆Ssurroundings = -(∆Hrxn)/T = -(-905.48×1000J)/(273.15+25)K
= 3036.995 J/K = 3037.0 J/K. (Answer)
∆Ssys = 180.5J/K
∆Suniverse = 3036.995 + 180.5 = 3217.5 J/K (Answer)
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