Question

1. Consider the reaction

2NO(g) + O2(g)2NO2(g)

Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 2.26 moles of NO(g) react at standard conditions.

S°surroundings =

2. Consider the reaction:

NH₄Cl(aq)NH₃(g) + HCl(aq)

Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.61 moles of NH₄Cl(aq) react at standard conditions.

S°_surroundings =

Answer 1

2NO(g) + O2(g)---->2NO2(g)

to calculate dSsurr , we need    dHrxn

ΔHorxn =ΔHof(Products) - ΔHof(reactants)

dHf values can be obtained from thermodynamic tables

ΔHorxn = 2*ΔHof(NO2) -2* ΔHof(NO) - 1* ΔHof(o2)

ΔHorxn = 2*33.2 kj - 2* 90.3 KJ - 0.00 kj
= -114.2 KJ
this is for 2 mols NO. When 2.26 mol NO is used
ΔHorxn= -114.2 kj/2 *2.26 = -129.046 KJ

ΔSsurr = -ΔHorxn/T = 129046 J /298 K = 433. J/K

Answer
ΔSsurr = 433. J/K

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NH4Cl(aq)-->NH3(g) + HCl(aq)
ΔHorxn =ΔHof(Products) - ΔHof(reactants)

ΔHorxn = 1*ΔHof(NH3) - 1* ΔHof(HCl) - 1* ΔHof(NH4Cl)
= [-46.1 KJ] - [-167.2] - [-299.7 ] KJ
= 420.8 KJ

ΔHorxn for 1.61 mol NH4Cl
==> ΔHorxn =677.488 KJ

ΔSsurr = -ΔHorxn/T = -677488 J /298 K =- 2273. J/K

round to 3 sig.fig
Answer
ΔSsurr = -2270. J/K
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