14.09.2020

Lesson 9: Enthalpy, Entropy, and Free Energy

. 4

Faq

Chemistry
Step-by-step answer
P Answered by Master

T_m=157.22=-115.8^oC\\T_b=350.34K=77.34^oC

Explanation:

Hello,

In this case, given the enthalpies and entropies of boh fusion and boiling, we can compute both the fusion and boiling points as shown below, considering ethanol's molar mass is 46 g/mol:

T_m=\frac{\Delta _fH}{\Delta _fS} =\frac{108.9\frac{J}{g}*\frac{46g}{1mol} }{31.6\frac{J}{mol*K} } \\\\T_m=157.22=-115.8^oC\\\\T_b=\frac{\Delta _bH}{\Delta _bS} =\frac{837\frac{J}{g}*\frac{46g}{1mol} }{109.9\frac{J}{mol*K} } \\\\T_b=350.34K=77.34^oC

Best regards.

Chemistry
Step-by-step answer
P Answered by Specialist

T_m=157.22=-115.8^oC\\T_b=350.34K=77.34^oC

Explanation:

Hello,

In this case, given the enthalpies and entropies of boh fusion and boiling, we can compute both the fusion and boiling points as shown below, considering ethanol's molar mass is 46 g/mol:

T_m=\frac{\Delta _fH}{\Delta _fS} =\frac{108.9\frac{J}{g}*\frac{46g}{1mol} }{31.6\frac{J}{mol*K} } \\\\T_m=157.22=-115.8^oC\\\\T_b=\frac{\Delta _bH}{\Delta _bS} =\frac{837\frac{J}{g}*\frac{46g}{1mol} }{109.9\frac{J}{mol*K} } \\\\T_b=350.34K=77.34^oC

Best regards.

Engineering
Step-by-step answer
P Answered by Specialist

A) 251.8 kj/kg

B) 0.9150 kj/kg-k

C) 155.4 kj/kg

F) 1.50

G) 3.95 kw

H) 2.6 kw

Explanation:

Given conditions :

air conditioner : R -134a

compressor efficiency (nc) = 90%.

T1 = 9⁰c,  T3 = 39⁰c, mass flow rate = 0.027 kg/s

A) Specific enthalpy at the compressor inlet

at T = 9⁰c the saturated vapor (x) = 1

from the R-134a property table

h1 = 251.8 kj/kg

B ) specific entropy ( kj/kg-k) at the compressor inlet

at T = 9⁰c the saturated vapor (x) = 1

s = 0.9150 kj/kg-k ( from the R-134a property table )

C) specific enthalpy at the compressor exit

at T3 = 39⁰c , s2 = s1

has = 165.12 kj/kg

h2 = 155.4 kj/kg

attached below is the remaining solution to some of the problems


An air-conditioner which uses R-134a operates on the ideal vapor compression refrigeration cycle wit
Engineering
Step-by-step answer
P Answered by Specialist

A) 251.8 kj/kg

B) 0.9150 kj/kg-k

C) 155.4 kj/kg

F) 1.50

G) 3.95 kw

H) 2.6 kw

Explanation:

Given conditions :

air conditioner : R -134a

compressor efficiency (nc) = 90%.

T1 = 9⁰c,  T3 = 39⁰c, mass flow rate = 0.027 kg/s

A) Specific enthalpy at the compressor inlet

at T = 9⁰c the saturated vapor (x) = 1

from the R-134a property table

h1 = 251.8 kj/kg

B ) specific entropy ( kj/kg-k) at the compressor inlet

at T = 9⁰c the saturated vapor (x) = 1

s = 0.9150 kj/kg-k ( from the R-134a property table )

C) specific enthalpy at the compressor exit

at T3 = 39⁰c , s2 = s1

has = 165.12 kj/kg

h2 = 155.4 kj/kg

attached below is the remaining solution to some of the problems


An air-conditioner which uses R-134a operates on the ideal vapor compression refrigeration cycle wit
Chemistry
Step-by-step answer
P Answered by Specialist

Explanation:

First, we will calculate the entropies as follow.

 \Delta S_{2} = C_{p_{2}} ln \frac{T_{2}}{T_{1}} J/K mol

As,   T_{1} = 298 K,         T_{2} = 373.8 K

Putting the given values into the above formula we get,

     \Delta S_{2} = C_{p_{2}} ln \frac{T_{2}}{T_{1}} J/K mol

                 = 81.6 ln (\frac{373.8}{298})

                 = 18.5 J/K mol

Now,

       \Delta S_{3} = \frac{\Delta H_{vap}}{T_{b.p}}

                  = \frac{35270 J}{373.8}

                  = 94.2 J/mol K

Also,

    \Delta S_{4} = C_{p_{4}} ln \frac{T_{2}}{T_{1}}

                = 43.89 \times ln (\frac{800}{373.8})

                = 33.4 J/K mol

Now, we will calculate the entropy of one mole of methanol vapor at 800 K as follows.

   \Delta S_{T} = \Delta S^{o}_{1} + \Delta S_{2} + \Delta S_{3} + \Delta S_{4}

                 = (126.8 + 18.5 + 94.2 + 33.4) J/K mol

                 = 272.9 J/K mol

Thus, we can conclude that the entropy of one mole of methanol vapor at 800 K is 272.9 J/K mol.

Chemistry
Step-by-step answer
P Answered by Specialist

Explanation:

First, we will calculate the entropies as follow.

 \Delta S_{2} = C_{p_{2}} ln \frac{T_{2}}{T_{1}} J/K mol

As,   T_{1} = 298 K,         T_{2} = 373.8 K

Putting the given values into the above formula we get,

     \Delta S_{2} = C_{p_{2}} ln \frac{T_{2}}{T_{1}} J/K mol

                 = 81.6 ln (\frac{373.8}{298})

                 = 18.5 J/K mol

Now,

       \Delta S_{3} = \frac{\Delta H_{vap}}{T_{b.p}}

                  = \frac{35270 J}{373.8}

                  = 94.2 J/mol K

Also,

    \Delta S_{4} = C_{p_{4}} ln \frac{T_{2}}{T_{1}}

                = 43.89 \times ln (\frac{800}{373.8})

                = 33.4 J/K mol

Now, we will calculate the entropy of one mole of methanol vapor at 800 K as follows.

   \Delta S_{T} = \Delta S^{o}_{1} + \Delta S_{2} + \Delta S_{3} + \Delta S_{4}

                 = (126.8 + 18.5 + 94.2 + 33.4) J/K mol

                 = 272.9 J/K mol

Thus, we can conclude that the entropy of one mole of methanol vapor at 800 K is 272.9 J/K mol.

Chemistry
Step-by-step answer
P Answered by Master

a. ΔG : It seems that the proper definition is not given, anyway, it is the free energy change.

b. ΔS : 3. Change in entropy.

c. Keq : 6. Equilibrium constant.

d. ΔH : 7. Change in enthalpy.

e. ΔG° : 9. Standard free energy change.

f. J : 8. Joule.

Explanation:

Hello,

In this case, we can match the symbol with the proper definition as shown below:

a. ΔG : It seems that the proper definition is not given, anyway, it is the free energy change and it uses G since it is better referred to the Gibbs free energy.

b. ΔS : 3. Change in entropy.

c. Keq : 6. Equilibrium constant.

d. ΔH : 7. Change in enthalpy.

e. ΔG° : 9. Standard free energy change.

f. J : 8. Joule.

Best regards.

Chemistry
Step-by-step answer
P Answered by PhD

ΔS = -0.272 J/K

Explanation:

Step 1: Data given

Its normal freezing point is –38.9 °C

molar enthalpy of fusion is ∆Hfusion = 2.29 kJ/mol

Mass of Hg = 5.590 grams

Step 2:

ΔG = ΔH - TΔS

At the normal freezing point, or any phase change in general,  

ΔG =0

0  =  Δ

Hfus

Tfus  Δ

S

fus

Δ

S

fus = Δ

Hfus

/Tfus

Δ

S

fus = 2290 J/mol / 234.25 K

Δ

S

fus = 9.776 J/mol*K

Since fusion is  from solid to liquid. Freezing is the opposite process, so the entropy change of freezing is -9.776 J/mol*K

Step 3: Calculate moles Hg

Moles Hg = 5.590 grams / 200.59 g/mol

Moles Hg = 0.02787 moles

Step 4: Calculate the entropy change of the system:

Δ

S = -9.776 J/mol*K * 0.02787 moles

ΔS = -0.272 J/K

Chemistry
Step-by-step answer
P Answered by PhD

c. The statement is true

Explanation:

Endothermic reactions are defined as the reactions in which energy of the product is greater than the energy of the reactants. The total energy is absorbed in the form of heat and  for the reaction comes out to be positive.

Exothermic reactions are defined as the reactions in which energy of the product is lesser than the energy of the reactants. The total energy is released in the form of heat and  for the reaction comes out to be negative.

As the given value of \Delta H^0=-9kJ/mol, the heat is released and is exothermic.Thus the statement that the reaction is exothermic is true.

Chemistry
Step-by-step answer
P Answered by PhD

ΔS = -0.272 J/K

Explanation:

Step 1: Data given

Its normal freezing point is –38.9 °C

molar enthalpy of fusion is ∆Hfusion = 2.29 kJ/mol

Mass of Hg = 5.590 grams

Step 2:

ΔG = ΔH - TΔS

At the normal freezing point, or any phase change in general,  

ΔG =0

0  =  Δ

Hfus

Tfus  Δ

S

fus

Δ

S

fus = Δ

Hfus

/Tfus

Δ

S

fus = 2290 J/mol / 234.25 K

Δ

S

fus = 9.776 J/mol*K

Since fusion is  from solid to liquid. Freezing is the opposite process, so the entropy change of freezing is -9.776 J/mol*K

Step 3: Calculate moles Hg

Moles Hg = 5.590 grams / 200.59 g/mol

Moles Hg = 0.02787 moles

Step 4: Calculate the entropy change of the system:

Δ

S = -9.776 J/mol*K * 0.02787 moles

ΔS = -0.272 J/K

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