Wize University Chemistry Textbook > Thermodynamics Part 2: (2nd Law-entropy, 3rd Law, Gibbs Free Energy, Spontaneity)

Entropy [detailed-more equations]

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Additional Equations for Entropy
This equation is specifically for a reversible process:
ΔS=qrevT\Delta S=\frac{q_{rev}}{T}ΔS=Tqrev​​
 For a reversible process 
ΔSuniv=ΔSsys+ΔSsurr=0\Delta S_{univ}=\Delta S_{sys} +\Delta S_{surr} =0ΔSuniv​=ΔSsys​+ΔSsurr​=0
 For an irreversible (spontaneous) process 
ΔSuniv=ΔSsys+ΔSsurr>0\Delta S_{univ} =\Delta S_{sys} +\Delta S_{surr} >0ΔSuniv​=ΔSsys​+ΔSsurr​>0

 Note: If no subscript is given on ΔS assume it is ΔSsys 
 Spontaneous (irreversible) process: A process which proceeds in one direction and may only proceed backwards if work 
 is done. eg. Piston, electric current, natural processes
loss of usable work
are spontaneous reactions always fast? No!
think about diamonds->graphite 
 Non-spontaneous process: A process which only proceeds by the addition of work. 
ex. of a non-spontaneous irreversible process is pushing a piston to force a gas into a smaller volume 
spontaneous is natural whereas non spontaneous is unnatural (needs human intervention)
 Reversible Process: A Process that is non-spontaneous in both directions 
 2nd Law: Applications The 6 cases 
 Case 1: Spontaneous Process in an isolated system 
q=0Why ?q=0 \hspace{30pt} \text{Why ?} q=0Why ?
ΔSsurr=0\Delta S_{surr} =0ΔSsurr​=0
ΔSuniv=ΔSsys+ΔSsurr>0\Delta S_{univ} =\Delta S_{sys} +\Delta S_{surr} >0ΔSuniv​=ΔSsys​+ΔSsurr​>0
ΔSsys>0\Delta S_{sys} >0ΔSsys​>0

Case 2: Isochoric or Isobaric Processes 
deltaS=(n)Cv or p ln⁡(T2(T1))deltaS=\left(n\right)Cv\ or\ p\ \ln\left(\frac{T2}{\left(T1\right)}\right)deltaS=(n)Cv or p ln((T1)T2​)

 Case 3: Isothermal Processes 
ΔU=0Why ?\Delta U=0 \hspace{30pt} \text{Why ?} ΔU=0Why ?
q=−wq=-wq=−w
ΔS=nRln⁡V2V1=nRln⁡p1p2\Delta S=nR \ln \frac{V_2}{V_1}=nR \ln \frac{p_1}{p_2}ΔS=nRlnV1​V2​​=nRlnp2​p1​​

 Case 4: Adiabatic Reversible Process 
q=0Why ?q=0 \hspace{30pt} \text{Why ?} q=0Why ?
ΔS=qrevT=0\Delta S=\frac{q_{rev}}{T} =0ΔS=Tqrev​​=0
For a reversible adiabatic process: ΔS_system = 0, ΔS_universe = 0.
For an irreversible adiabatic process: ΔS_system > 0, ΔS_surroundings = 0, and thus ΔS_universe > 0.
 Case 5: Reversible Phase Change (Phase change at the melting/boiling point) 
ΔS=ΔHT\Delta S=\frac{\Delta H}{T}ΔS=TΔH​
 Remember. If the phase Change is happening at a different temperature we can use Kirchoff’s Law to deal with that. 

 Case 6: Chemical Reactions 
 In the same way we can use standard heats of formation to calculate a reaction enthalpy at room temperature and 
 pressure, we can use So  
ΔrSo=∑iviSm,io(products)−∑jvjSm,jo(reactants)\Delta_rS^o =\sum_iv_iS^o_{m,i}(products) -\sum _j v_j S^o _{m,j} (reactants)Δr​So=∑i​vi​Sm,io​(products)−∑j​vj​Sm,jo​(reactants)

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a.  An open flask containing 5 moles of water is heated from 25oC to 30oC what is the change in entropy. Cp (H2O(l)) = 4.18 J g-1 K -1 . Give your answer in J/K 

 Isobaric Heating;  
deltaS=n(Cv)ln (T2/T1)
ΔS=Cv=ln⁡(T2T1)=4.18JgKln⁡(303K298K)=0.0696JgK\Delta S=C_v=\ln \bigg(\cfrac{T_2}{T_1}\bigg)=4.18\cfrac{J}{gK}\ln  \bigg(\cfrac{303K}{298K}\bigg)=0.0696\cfrac{J}{gK}ΔS=Cv​=ln(T1​T2​​)=4.18gKJ​ln(298K303K​)=0.0696gKJ​
ΔS=0.0696JgK×5mols×18gmol=6.26JK\Delta S=0.0696\cfrac{J}{gK}\times 5mols\times 18\cfrac{g}{mol}=6.26\cfrac{J}{K}ΔS=0.0696gKJ​×5mols×18molg​=6.26KJ​

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b.  A sealed 5L glass ampule full of F2(g) at 1atm at 25oC is found in a laboratory at UBC. We know that this type of glass ampule can easily withstand pressures up to 5atm. What would the change in entropy be if the F2(g) ampule was heated until the pressure reached 5atm? You can assume ideal gas behaviour for F2 and that the glass ampule does not change in volume. 
 Isochoric Heating:

ΔS=nCvln⁡(T2T1)=n32Rln⁡(T2T1)\Delta S=nC_v\ln \bigg(\cfrac{T_2}{T_1}\bigg)=n\cfrac{3}{2}R\ln \bigg(\cfrac{T_2}{T_1}\bigg)ΔS=nCv​ln(T1​T2​​)=n23​Rln(T1​T2​​)

 Ideal gas law for Initial moles

n=PVRT=(1atm)(5L)(0.08206LatmK−1mol−1(298K)=0.204moln=\cfrac{PV}{RT}=\cfrac{(1atm)(5L)}{(0.08206Latm K^{-1}mol^{-1}(298K)}=0.204moln=RTPV​=(0.08206LatmK−1mol−1(298K)(1atm)(5L)​=0.204mol

 Ideal gas law for final temperature

 T=PVnR=(5atm)(5L)(0.204mol)(0.08206LatmK−1mol−1=1493KT=\frac{PV}{nR}=\frac{(5atm)(5L)}{(0.204mol)(0.08206LatmK^{-1}mol^{-1}}=1493KT=nRPV​=(0.204mol)(0.08206LatmK−1mol−1(5atm)(5L)​=1493K

deltaS=nCvln⁡(T2(T1))=0.204moles(32R)ln⁡(1493K(298K))=4.1JKdeltaS=nCv\ln\left(\frac{T2}{\left(T1\right)}\right)=0.204moles\left(\frac{3}{2}R\right)\ln\left(\frac{1493K}{\left(298K\right)}\right)=\frac{4.1J}{K}deltaS=nCvln((T1)T2​)=0.204moles(23​R)ln((298K)1493K​)=K4.1J​

Extra Practice

Entropy Change of a Reaction

 Using the following data, calculate the standard entropy change of the reaction shown below. 
2Al(s)+3ZnO(s)→Al2O3(s)+Zn(s)2Al_{(s)}+3ZnO_{(s)}\to Al_2O_{3(s)}+Zn_{(s)}2Al(s)​+3ZnO(s)​→Al2​O3(s)​+Zn(s)​
So(Al(s))=28.3J/mol.KS^o(Al_{(s)})=28.3J/mol.KSo(Al(s)​)=28.3J/mol.K

Entropy

What is the entropy change (in J/K) for the formation of 3.00 g of steam from liquid water at 100oC, given that the enthalpy of vaporization for water is 40.1 kJ/mol?

Entropy

Which of the following reactions has a positive ΔSo? 
1.  CaCO3 (s) → CaO (s) + O2 (g)
2. N2O4 (g) → 2 NO2 (g)

Entropy

Use the standard molar entropy values in the data sheet to calculate the standard entropy change for the reaction:
CO2 (g) + 2 H2O (l) → CH4 (g) + 2 O2 (g)           ΔHo = +890 kJ  

Wize University Chemistry Textbook > Thermodynamics Part 2: (2nd Law-entropy, 3rd Law, Gibbs Free Energy, Spontaneity)

Entropy [detailed-more equations]

Popular Courses

Additional Equations for Entropy

2nd Law: Applications The 6 cases

$\Delta S_{surr} =0$

$\Delta S_{univ} =\Delta S_{sys} +\Delta S_{surr} >0$

$\Delta S_{sys} >0$

Isochoric Heating:

$\Delta S=nC_v\ln \bigg(\cfrac{T_2}{T_1}\bigg)=n\cfrac{3}{2}R\ln \bigg(\cfrac{T_2}{T_1}\bigg)$

Ideal gas law for Initial moles

$n=\cfrac{PV}{RT}=\cfrac{(1atm)(5L)}{(0.08206Latm K^{-1}mol^{-1}(298K)}=0.204mol$

Ideal gas law for final temperature

$T=\frac{PV}{nR}=\frac{(5atm)(5L)}{(0.204mol)(0.08206LatmK^{-1}mol^{-1}}=1493K$

$deltaS=nCv\ln\left(\frac{T2}{\left(T1\right)}\right)=0.204moles\left(\frac{3}{2}R\right)\ln\left(\frac{1493K}{\left(298K\right)}\right)=\frac{4.1J}{K}$

Extra Practice

Entropy Change of a Reaction

Entropy

Entropy

Entropy

Wize University Chemistry Textbook > Thermodynamics Part 2: (2nd Law-entropy, 3rd Law, Gibbs Free Energy, Spontaneity)

Entropy [detailed-more equations]

Popular Courses

Additional Equations for Entropy

2nd Law: Applications The 6 cases

ΔSsurr=0\Delta S_{surr} =0ΔSsurr​=0

ΔSuniv=ΔSsys+ΔSsurr>0\Delta S_{univ} =\Delta S_{sys} +\Delta S_{surr} >0ΔSuniv​=ΔSsys​+ΔSsurr​>0

ΔSsys>0\Delta S_{sys} >0ΔSsys​>0

Isochoric Heating:

ΔS=nCvln⁡(T2T1)=n32Rln⁡(T2T1)\Delta S=nC_v\ln \bigg(\cfrac{T_2}{T_1}\bigg)=n\cfrac{3}{2}R\ln \bigg(\cfrac{T_2}{T_1}\bigg)ΔS=nCv​ln(T1​T2​​)=n23​Rln(T1​T2​​)

Ideal gas law for Initial moles

n=PVRT=(1atm)(5L)(0.08206LatmK−1mol−1(298K)=0.204moln=\cfrac{PV}{RT}=\cfrac{(1atm)(5L)}{(0.08206Latm K^{-1}mol^{-1}(298K)}=0.204moln=RTPV​=(0.08206LatmK−1mol−1(298K)(1atm)(5L)​=0.204mol

Ideal gas law for final temperature

T=PVnR=(5atm)(5L)(0.204mol)(0.08206LatmK−1mol−1=1493KT=\frac{PV}{nR}=\frac{(5atm)(5L)}{(0.204mol)(0.08206LatmK^{-1}mol^{-1}}=1493KT=nRPV​=(0.204mol)(0.08206LatmK−1mol−1(5atm)(5L)​=1493K

deltaS=nCvln⁡(T2(T1))=0.204moles(32R)ln⁡(1493K(298K))=4.1JKdeltaS=nCv\ln\left(\frac{T2}{\left(T1\right)}\right)=0.204moles\left(\frac{3}{2}R\right)\ln\left(\frac{1493K}{\left(298K\right)}\right)=\frac{4.1J}{K}deltaS=nCvln((T1)T2​)=0.204moles(23​R)ln((298K)1493K​)=K4.1J​

Extra Practice

Entropy Change of a Reaction

Entropy

Entropy

Entropy

$\Delta S_{surr} =0$

$\Delta S_{univ} =\Delta S_{sys} +\Delta S_{surr} >0$

$\Delta S_{sys} >0$

$\Delta S=nC_v\ln \bigg(\cfrac{T_2}{T_1}\bigg)=n\cfrac{3}{2}R\ln \bigg(\cfrac{T_2}{T_1}\bigg)$

$n=\cfrac{PV}{RT}=\cfrac{(1atm)(5L)}{(0.08206Latm K^{-1}mol^{-1}(298K)}=0.204mol$

$T=\frac{PV}{nR}=\frac{(5atm)(5L)}{(0.204mol)(0.08206LatmK^{-1}mol^{-1}}=1493K$

$deltaS=nCv\ln\left(\frac{T2}{\left(T1\right)}\right)=0.204moles\left(\frac{3}{2}R\right)\ln\left(\frac{1493K}{\left(298K\right)}\right)=\frac{4.1J}{K}$