Chemistry 360 Name _____________KEY________________ Dr. Jean M. Standard Fall 2016

Exam 2 Solutions

1.) (14 points) Determine

€

∂H∂ P

#

$ %

&

' ( T

for a gas obeying the equation of state

Z  = PVmRT

  =   1   +   BP   +   CP2 ,

where B and C are constants.

Since a partial derivative of H is requested, it is helpful to start with the fundamental equation for H,

dH   =  T dS  +  V dP. Take the partial derivative with respect to V (at constant T) of both sides of the equation,

∂H∂ P

!

"#

$

%&T

 =   T ∂ S∂ P

!

"#

$

%&T

 +  V .

The partial derivative on the right side of the equation cannot be determined from the equation of state. However, one of the Maxwell relations can be used,

∂ S∂ P!

"#

$

%&T

  =  − ∂V∂T!

"#

$

%&P

.

Substituting yields

∂H∂ P

!

"#

$

%&T

  =   −T ∂V∂T

!

"#

$

%&P

 +  V .

Now, the partial derivative on the right side can be calculated from the equation of state. Since the equation of

state is Z  = PVmRT

  =   1   +   BP   +   CP2 , solving for Vm gives

PVmRT

  =   1   +   BP   +   CP2

Vm   =  RTP

1   +   BP   +   CP2( )

or    Vm   =  RTP

  +   BRT   +   CRTP .

Now, converting to V instead of

€

Vm leads to the expression

V  =  nRTP

  +   nBRT   +   nCRTP .

2 1. Continued

Taking the partial derivative,

∂V∂T!

"#

$

%&P

  =   nRP   +   nBR  + nCRP .

Substituting into the expression for the partial derivative of H gives the result,

∂H∂ P!

"#

$

%&T

   =   −T ∂V∂T!

"#

$

%&p

+  V

=   −T nRP   +   nBR  + nCRP

!

"#

$

%&  +  V

=   − nRTP   +   nBRT   + nCRTP

!

"#

$

%&   +  V

=   −V  +  V

∂H∂ P!

"#

$

%&T

  =  0.

3

2.) (14 points) The standard molar enthalpy of formation of Fe2O3 (s) is

€

ΔH f! = −824.2 kJ/mol , and the

standard molar enthalpy of formation of SO2 (g) is

€

ΔH f! = −296.8 kJ/mol (both at 298 K). Use this

information, along with the standard molar enthalpy change of the following reaction at 298 K,

2 FeS2 (s)  +   112 O2 (g)     →    Fe2O3 (s) + 4 SO2 (g)     ΔHR

! = −1655 kJ/mol,

to determine the standard molar enthalpy change of the reaction shown below at 298 K:

€

Fe (s) + 2 S (s) → FeS2 (s) . [Note: The standard state of iron is Fe(s) and the standard state of sulfur is S(s) at 298 K.]

The formation reaction for Fe2O3 is

2 Fe s( ) + 32 O2 g( ) → Fe2O3 s( ) .

The formation reaction for SO2 is

S s( ) + O2 g( ) → SO2 g( ) .

The desired reaction can be constructed from the reactions given in the problem in the following way:

€

2 S s( ) + O2 g( ) → SO2 g( ) [ ] 1

2 2 Fe s( ) + 32 O2 g( ) → Fe2O3 s( )[ ]

12 Fe2O3 s( ) + 4 SO2 (g) → 2 FeS2 (s) + 11

2 O2 (g) [ ]

Fe s( ) + 2 S s( ) → FeS2 s( ) From the combination of reactions above, the enthalpy of reaction can be calculated from Hess' Law,

€

ΔH f! FeS2( ) = 2ΔH f

! SO2( ) + 12 ΔH f

! Fe2O3( ) − 12 ΔHr

! ,

where

€

ΔHr! corresponds to the standard molar enthalpy of the reaction between FeS2 and O2 given in the

problem. Substituting,

€

ΔH f! FeS2( ) = 2ΔH f

! SO2( ) + 12 ΔH f

! Fe2O3( ) − 12 ΔHr

! ,

= 2 −296.8 kJ/mol( ) + 12 −824.2 kJ/mol( ) − 1

2 −1655kJ/mol( )ΔH f

! FeS2( ) = −178.2 kJ/mol.

4

3.) (14 points) Explain why we are able to obtain absolute molar entropies of substances; that is, what is the theoretical foundation that allows the determination of absolute molar entropies? Discuss any trends that may be observed in the magnitudes of the absolute molar entropies of the following substances and the reasoning behind your answer: He (g), O2 (g), CH4 (g), C6H6 (g).

The opportunity for an absolute entropy scale is provided by the Third Law of Thermodynamics, which sets the absolute entropy of a perfect crystalline substance as 0 at 0 K. The absolute entropies tabulated above are all for gases of varying structural complexity. As the substances vary from a monatomic gas (He) to a diatomic gas (O2) to polyatomic gases (CH4 and C6H6), the entropy is expected to increase. This is a result of the increase in the number of degrees of freedom of the substances. As we saw in our discussion of the Equipartition Theorem, the number of degrees of freedom increases as the complexity of the molecule increases. For the complex molecules with more degrees of freedom, there are more ways of distributing the available energy among the degrees of freedom, which increases the disorder in the system. Thus, more complex substances have higher entropies than less complex substances.

5 4.) (15 points) True/false, short answer, multiple choice.

a.) True or False : In the statistical definition of entropy, S  = kB lnW , the quantity W corresponds to the energy.

b.) True or False : The Debye equation gives the form of the constant pressure molar heat capacity of a

solid at low temperature and has the form Cp,m!    T( ) = aT 2 .

c.) Short answer _____Adiabatic (or bomb)____ calorimetry is carried out at constant volume, whereas _____Solution______ calorimetry is carried out at constant pressure.

d.) Short answer

For two chemical reactions that combine to give a third reaction (the overall reaction), the enthalpy change of the overall reaction (3) is given as

€

ΔH3 = ΔH1 + ΔH2 , which is an example of the application of _________Hess' Law________ .

e.) Multiple Choice: Which of the following corresponds to the correct relationship?

1) T = ∂G∂ P!

"#

$

%&S

.

2) T = − ∂U∂V"

#$

%

&'S

.

3) T = − ∂ A∂V"

#$

%

&'S

.

4) T = ∂H∂ S

!

"#

$

%&P

.

6

5.) (14 points) Using the values in the table below reported at 298 K, determine ΔHR! at 700 K for the

hydrogenation of acetylene to ethane,

C2H2 g( ) +  2 H2 g( )     →    C2H6 g( ) .

Assume that

€

Cp,m is independent of temperature for all species in the reaction.

€

ΔH f!

(kJ/mol)

€

Cp,m

(Jmol–1K–1) C2H2 (g) 226.73 43.93 C2H6 (g) –84.68 52.63 H2 (g) 0 28.82

The first step is to determine the enthalpy of reaction at 298 K:

ΔHR

!  =  ΔH f! C2H6( )    −   ΔH f

! C2H2( )   −   2ΔH f! H2( )

=  −84.68 kJ/mol   −   226.73 kJ/mol( )   −   2 0( )ΔHR

!  =  −311.41 kJ/mol.

The temperature dependence of the molar enthalpy of reaction is given by

ΔHR! T2( )  =  ΔHR

! T1( )    +    ΔCp,m ⋅ T2 −T1( ).

This equation assumes that the molar heat capacities are independent of temperature. In the equation,

€

ΔCp,m is the difference in molar heat capacities between products and reactants with the appropriate signed stoichiometric coefficients included,

€

ΔCp,m = ν iCp,m i( )i∑ .

For this reaction, the heat capacity difference is

ΔCp,m  =    ν iCp,m i( )i∑

=   Cp,m C2H6( )   −   Cp,m C2H2( )   −   2Cp,m O2( ) =   52.63   −   43.93   −   2 28.82( ) Jmol−1K−1

ΔCp,m  =   −48.94 Jmol−1K−1.

Substituting, the molar enthalpy of reaction at 700 K is

ΔHR! 700 K( )  =  ΔHR

! 298 K( )   +   ΔCp,m ⋅ 700− 298 K( )

=  −311.41 kJ/mol   +   −48.94 Jmol−1K−1( ) 700− 298 K( ) 1 kJ1000 J⎛

⎝⎜

⎞

⎠⎟

=  −311.41   −19.67 kJ/mol

ΔHR! 700 K( )  =  −331.08 kJ/mol .

7 6.) (14 points) Calculate the entropy change

€

ΔS for the following transformation involving one mole of H2O,

€

H2O (s, 200 K) → H2O (ℓ, 300 K) .

You may need to use the following data:

€

T fus = 273 K ;

€

ΔH fus! = 6.01 kJ mol−1

€

Tvap = 373 K;

€

ΔHvap! = 40.65 kJ mol−1

€

Cp,m s( ) = 36.2 J mol−1K−1 ;

€

Cp,m ℓ( ) = 75.3 J mol−1K−1;

€

Cp,m g( ) = 33.6 J mol−1K−1

The process may be broken up into three steps: (1) heating the solid from 200 to 273 K; (2) the S-L phase transition at 273 K; and (3) heating the liquid from 273 to 300 K. The molar entropy change can then be written as a sum of these three steps,

ΔSm    =   Cp,m s( )TT1

Tfus∫ dT    +    ΔH fus

!

Tfus     +    

Cp,m ℓ( )TTfus

T2∫ dT .

Assuming that the heat capacities are independent of temperature (they are listed in the problem as constants), the expression becomes

ΔSm    =   Cp,m s( )TT1

Tfus∫ dT     +    ΔH fus

!

Tfus    +   

Cp,m ℓ( )TTfus

T2∫ dT

=   Cp,m s( ) 1TT1

Tfus∫ dT +    ΔH fus

!

Tfus    +   Cp,m ℓ( ) Tfus

T2∫ 1TdT

ΔSm    =   Cp,m s( ) lnTfusT1

⎛

⎝⎜

⎞

⎠⎟  +    

ΔH fus!

Tfus     +    Cp,m ℓ( ) ln T2

Tfus

⎛

⎝⎜⎜

⎞

⎠⎟⎟ .

Substituting,

ΔSm   =   Cp,m s( ) lnTfusT1

⎛

⎝⎜

⎞

⎠⎟ +

ΔH fus!

Tfus + Cp,m ℓ( ) ln T2

Tfus

⎛

⎝⎜⎜

⎞

⎠⎟⎟

= 36.2 J mol−1K−1( ) ln 273K200K⎛

⎝⎜

⎞

⎠⎟ +

6010 J/mol( )273K

+ 75.3J mol−1K−1( ) ln 300K273K⎛

⎝⎜

⎞

⎠⎟

= 11.3 + 22.0 + 7.1 J mol−1K−1

ΔSm = 40.4 J mol−1K−1.

So, for one mole of H2O, the entropy change is

€

ΔS = 40.4 J K−1 .

8 7.) (15 points) True/false, short answer, multiple choice.

a.) True or False: For the reaction 2 CO (g) + O2 (g) → 2 CO2 (g), the entropy change is expected to be negative.

b.) True or False : The standard state is defined to be the pure form of an element at 1 bar and 25°C.

c.) Short answer

The ___Second Law of Thermodynamics____ states that entropy increases for a spontaneous process in

an isolated system.

d.) Short answer

The _______Helmholtz Energy A_________ is defined as U −TS .

e.) Multiple Choice: At 298 K, ΔGR! = –158 kJ/mol for a certain chemical reaction. In addition, this reaction

has ΔHR! > 0 . Which of the following is a reasonable expectation for ΔGR

! at 500 K?

1) ΔGR

! = –195 kJ/mol .

2) ΔGR

! = –158 kJ/mol . 3) ΔGR

! = –141 kJ/mol .

4) ΔGR

! = +158 kJ/mol .