Homework_7_Su15

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Name ________________________________________

Homework #7 Summer 2015

THERMODYNAMICS

(Assume R = 8.314 J mol -1 K -1 )

1. What is the first law of thermodynamics?

2. Consider an apple mass m = 0.10 kg held at a height h = 1 m above the ground. It has a potential energy PE (energy due to its position in space) due to the gravitational field of

PE = mgh

(where m is in kg and g is acceleration due to gravity = 10.0 ms -2 )

The apple is dropped and accelerates toward the ground, increasing its kinetic energy (energy due to motion)

KE = mv 2 /2 where v is the velocity (speed) in m/s. The total energy

E tot

= KE + PE

Use the first law of thermodynamics to calculate the following

Height h (m) PE (J) KE(J) E

TOT

V(m/s)

1

0.5

0.0*

* the moment before it hits the ground and stops

3. The apple comes to rest on the ground, where has its PE and KE gone?

4. In the process of falling what happened to the PE did it increase or decrease?

5. What is the PE of the apple at rest on the ground?

6. List the types of motion that the nuclei can have in the molecules of a gas or liquid

7. What are the source(s) of the potential energy in molecules?

8. Does the first law of thermodynamics predict the direction of spontaneous processes?

9. In a chemical reaction the enthalpy change is measured using calorimetry, what is the source of the enthalpy change, why is the internal energy changing?

10. Calculate the standard enthalpy change, ΔH o , for the formation of 1 mol of strontium carbonate (the material that gives the red color in fireworks) from its elements.

Sr (s)

C(graphite )

3

2

O

2

(g)

SrCO

3

(s)

The informatio n available is

(1) Sr (s)

1

2

O

2

(g)

SrO (s)

H

 

592 kJ

(2) SrO (s)

CO

2

(g)

SrCO

3

(s)

H

 

234 kJ

(3) C(graphite )

O

2

(g)

CO

2

(g)

H

 

394 kJ

11. Which of the following statements could be used to define the first law of thermodynamics? a.

Heat flows from a hot body to a cold body b.

Energy cannot be created or destroyed but can only change from one form to another c.

The absolute entropy of a crystal at 0K is zero d.

In a spontaneous process  S univ

< 0

12. Which of the following statements is consistent with the second law of thermodynamics? a.

Heat flows from a hot body to a cold body b.

Energy cannot be created or destroyed but can only change from one form to another c.

The absolute entropy of a crystal at 0K is zero d.

In a spontaneous process  S univ

< 0

13. Which of the following statements is a definition of the third law of thermodynamics? a.

Heat flows from a hot body to a cold body b.

Energy cannot be created or destroyed but can only change from one form to another c.

The absolute entropy of a crystal at 0K is zero d.

In a spontaneous process  S univ

< 0

14. Are Exothermic reactions spontaneous? a.

Always b.

Never c.

Sometimes

15. Are Endothermic reactions spontaneous? a.

Always b.

Never c.

Sometimes

16. Are reactions in which ΔS univ

> 0 spontaneous? a.

Always b.

Never c.

Sometimes

17. Are reactions in which ΔS univ

< 0 spontaneous? a.

Always b.

Never c.

Sometimes

18. Are reactions in which ΔG sys

> 0 spontaneous? a.

Always b.

Never c.

Sometimes

19. Are reactions in which ΔG sys

< 0 spontaneous? a.

Always b.

Never c.

Sometimes

20. Are reactions in which ΔS sys

> 0 spontaneous? a.

Always b.

Never c.

Sometimes

21. Which of the following statements is false? a.

Entropy can be considered a measure of energy dispersal per Kelvin b.

S = -k ln(W) c.

 G =  H-T  S d.

 S surr

=  H sys

/T e.

The entropy of the universe is always increasing

22. The spontaneous melting of ice at room temperature is a.

An enthalpy driven process b.

An entropy driven process c.

Neither a or b

23. The dissolution of NaCl in water NaCl(s)  Na + (aq) + Cl (aq) is a.

An enthalpy driven process b.

An entropy driven process c.

Neither a or b

24. For the process C

6

H

6

(l)  C

6

H

6

(g) a.

 S < 0 b.

 S > 0 c.

 S = 0 d.

Depends on the temperature e.

Impossible to say

25. For the process 4Al(s)+3O

2

(g)  2Al

2

O

3

(s) a.

 S < 0 b.

 S > 0 c.

 S = 0 d.

Depends on the temperature e.

Impossible to say

26. For the process 2O

2

(g)+N

2

(g)  2NO

2

(g) a.

 S < 0 b.

 S > 0 c.

 S = 0 d.

Depends on the temperature e.

Impossible to say

27. What is ΔS for the reaction at 25 o C

2SO

2

(g) + O

2

(g)  2SO

3

(g) given

SO

2

(g)

O

2

(g)

SO

3

(g)

S 0 (j/K/mol)

248.1

205.3

256.6 a.

-188.3 JK -1 mol -1 b.

+188.3 JK -1 mol -1 c.

-196.8 JK -1 mol -1 d.

+196.8 JK -1 mol -1 e.

none of the above

28. Which of the following gases at 1 atm and 298K will have the highest entropy? a. Ne b. Ar c. Kr d. Xe e. They will all be the same

29. Which of the following gases at 1 atm and 298K will have the smallest entropy? a. Ne b. Ar c. Kr d. Xe e. They will all be the same

30. Which allotrope of C will have the highest entropy? a. Diamond b. Graphite c. neither since Diamond and Graphite will have the same entropy

31. Consider 1 mol of H

2

O which will have the greatest entropy a. Ice b. liquid water c. water vapor d. They are all the same

32. Which will have the larger entropy? a. 1 mole of KClO

3

(aq) b. 1 mole of KClO

3

(s) c. neither

33. Which will have the larger entropy? a. 1 mole of Argon gas (molar mass 39.948 g/mol) b. 1 mole of NO gas (molar mass 30.006 g/mol) c. impossible to say

Borax (Na

2

[B

4

O

5

(OH)

4

].8H

2

O) is soluble in water

Na

2

[B

4

O

5

(OH)

4

].8H

2

O(s) ↔ 2 Na + (aq) + [B

4

O

5

(OH)

4

] 2- (aq) + 8 H

2

O (l)

The solubility product K sp

= [Na+] 2 [B

4

O

5

(OH)

42-

] is measured at temperatures ranging from 15 – 55 o C. A graph is made of ln(K sp

) vs. 1/T (see below) it is found to follow a straight line ln(K sp

) = -12406(1/T) + 37.88

Where T is in Kelvin ln(K sp

) vs 1/T for Borax

0

-1

0,003 0,0031 0,0032 0,0033 0,0034 0,0035

-2

-3

-4

-5

-6

1/T (K -1 )

Ряд1

Линейная (Ряд1) y = -12406x + 37,88

R² = 0,9368

34. What is the enthalpy  H o for this process? a.

-12.406 kJ/mol b.

+12.406 kJ/mol c.

103.14 kJ/mol d.

-103.14 kJ/mol e.

37.88 J/mol

35. What is the entropy change  S o for this process a.

37.88 J/(mol.K) b.

-37.88 J/(mol.K) c.

-12.406 J/(mol.K) d.

314.93 J/(mol.K) e.

-314.93 J/(mol.K)

36. Consider the reaction for the oxidation of SO

2

(g) at 25 o C and 1 atm

2SO

2

(g) + O

2

(g)  2SO

3

(g)

The relevant thermodynamic data for this process is given in the table below a.

Calculate ΔH o (6 marks) b.

Is the process exothermic or endothermic? (2 marks) c.

Calculate , ΔS o (6 marks) d.

Calculate ΔG o (6 marks) e.

Above what temperature would you expect this to be non-spontaneous? (6 marks) f.

Estimate the equilibrium constant and the position of the equilibrium for the above reaction at 400 o C (6 marks)

EXTRA CREDIT QUESTION FOR THOSE WHO HAVE TAKEN CALCULUS

Use calculus to prove that the Mechanical Energy E = KE + PE for the case of the falling apple is constant for every moment during the fall

EXTRA CREDIT QUESTION ON BOLTZMAN’S LAW OF ENTROPY

Consider the figure right. Let us assume there is 1 mole of Argon gas in vessel B at the beginning. a.

By treating each atom as either being in vessel A or B come up with W the number of possibilities where half the atoms are in vessel A and half in vessel B. (Assume the every possible arrangement of the atoms in the system is

equivalent) b.

Compare your answer to the number of possibilities where the all of the atoms end up in vessel B c.

Calculate  S for this process using Boltzmann’s Equation d.

Does your answer show that the process in the figure is spontaneous?

(Explain) e.

Prove that the configuration with the atoms split 50:50 between A and B the most probable configuration

Hint: If I have N particles and I want the probability that p are in vessel A, the number of permutations W p

are

W p

=

N !

p !

Where

N!

=

N x (N-1) x (N-2) x (N3)…. x 2 x 1

Also you may find it helpful to use Sterling’s approximation, which states that for large values of N log

( ) »

N log

( ) -

N

Have fun!

W p

=

N !

p !

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