Question

1. Consider the reaction

2NH₃(g) + 3N₂O(g)------->4N₂(g) + 3H₂O(g)

Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.98 moles of NH₃(g) react at standard conditions.
delta S°_surroundings = _______J/K

2. Consider the reaction

2Fe(s) + 3Cl₂(g)------>2FeCl₃(s)

Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 2.25 moles of Fe(s) react at standard conditions.
delta S°_surroundings = ________ J/K

Answer 1

entropy of surrounding can be obtained from heat of reaction.
heat of reaction
dHrxn = sum of enthalpy formation of products-sum of enthalpy formation
of reactants
2NH3(g) + 3N2O(g)------->4N2(g) + 3H2O(g)

dHrxn = 4*dHf(N2) + 3*dHf(H2O) - 2*dHf(NH3) - 3dHf(n2O)
= 4*0 kj + 3*[-241.8] -2*[-46.1] - 3*[82.1]
= -879.5 Kj

this is for 2 mols of NH3 ,
when 1.98 mols NH3 is there dHrxn = [-879.5 kj/2mols]*1.98 mols
= - 870.705 Kj

dSsurr =-dH/T , where T= 298 K

= + 870.705 x10^3 J/298 K = 2921.83 j/K
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2Fe(s) + 3Cl2(g)------>2FeCl3(s)

dHrxn = 2* dHf(FeCl3) - 2*dHf(fe) - 3*dHf(Cl2)
= 2*-399.5 kj -0 kj- 0 kj = -799 KJ

this dHrxn is for 2 mols of Fe
for 2.25 mols of Fe
dHrxn = -799 KJ/2mols*2.25 mol = -898.875 KJ

dSsurr = -dHrxn/T = 898.875 x10^3 / 298 = 3016.35 J/K
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