i am the law!

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King Edward VIII Queen Elizabeth I King Henry VII King Richard III King Ralph $100 $100 $100 $100 $100 $200 $200 $200 $200 $200 $300 $300 $300 $300 $300 $400 $400 $400 $400 $400 $500 $500 $500 $500 $500

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I Am The Law!. Provide definitions of the 2 nd Law of Thermodynamics: Firstly, in words Secondly, as an equation. Spontaneous or Not?. Which of the following processes would be spontaneous ? Ice melting at -5  C and 1 atm Dissolution of sugar in hot tea - PowerPoint PPT Presentation

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Page 1: I Am The Law!

King Edward VIII

Queen Elizabeth I

King Henry VII

King Richard III

King Ralph

$100 $100 $100 $100 $100

$200 $200 $200 $200 $200

$300 $300 $300 $300 $300

$400 $400 $400 $400 $400

$500 $500 $500 $500 $500

Page 2: I Am The Law!

I Am The Law!

• Provide definitions of the 2nd Law of Thermodynamics:

• Firstly, in words• Secondly, as an equation

Page 3: I Am The Law!

Spontaneous or Not?• Which of the following processes

would be spontaneous?

• Ice melting at -5 C and 1 atm• Dissolution of sugar in hot tea• Alignment of iron filings in a

magnetic field

Page 4: I Am The Law!

S

Which of the following reactions has a S of greater than zero?

• C2H4(g) + H2(g) C2H6(g)

• Be(OH)2(s) BeO(s)A + H2(g)

• N2O4(g) 2NO2(g)

• Ag+(aq) + Cl-(aq) AgCl(s)

Page 5: I Am The Law!

Ksp

If aqueous solutions of 2.5x10-3 M Ca(NO3)2 and 1.0x10-3 M KOH were mixed together, would a precipitate form?

Ksp (Ca(OH)2) = 6.5x10-6

Page 6: I Am The Law!

Spontaneity again…• Which of the following would be

spontaneous at all temperatures?

H = -25 kJ/mol; S = -27 J/K.mol H = +33 kJ/mol; S = -310 J/K.mol S = +122 kJ/mol; H = +12.2

J/K.mol S = +72 kJ/mol; H = -54.8 J/K.mol

Bonus: Which would be spontaneous only at high temperatures?

Page 7: I Am The Law!

Gibbs Free Energy Change

• What does it mean when a reaction has a G < 0 ?

• What does it mean when a reaction has a G > 0 ?

• What does it mean when a reaction has a G = 0 ?

Page 8: I Am The Law!

It’s just so common…

What is the common ion effect?

Give one example of this phenomenon.

Page 9: I Am The Law!

Michael Buffer(s)

Which of the following would not be a good acid/base pair for use as a buffer?

• HCl and NaCl• HC2H3O2 and NaC2H3O2

• HF and NaCl• NH3 and NH4Br• HC2H3O2 and KNO2

Bonus: What is Michael Buffer’s famous catchphrase?

Page 10: I Am The Law!

Solubility

• Which of the following compounds is the least water-soluble?

• BaCO3 (Ksp = 5.0 x 10-9)

• PbCl2 (Ksp = 1.7 x 10-5)

• Cu(OH)2 (Ksp = 4.8 x 10-20)

• CaF2 (Ksp = 3.9 x 10-11)

• AgI (Ksp = 8.3 x 10-17)

Page 11: I Am The Law!

Titrations• What is the meaning of the equivalence

point during the titration of an acid with a base?

• Where is the equivalence point (in terms of pH) when a strong acid and strong base are titrated?

• What about for when a weak acid is titrated with a strong base?

Page 12: I Am The Law!

pH plotsSketch the shape of the curve you would expect if you were to titrate a solution of ammonia with hydrochloric acid. (if pH was on the y-axis and HCl(mL) was on the x-axis)

Page 13: I Am The Law!

Buffers

What exactly are buffers, what are they made up of, and how do they work?

Give one real-life example of an important buffered solution.

Page 14: I Am The Law!

Buffer pH

• What is the pH of a solution containing 2.5 M HF and 3.2 M KF?

• One way to calculate this is through the use of an ICE table. What is the other, quicker method? What other information do you need to calculate the pH this way?

Page 15: I Am The Law!

And More Buffer pH

• What would be the pH of the solution from the last question, if

5 g of sodium hydroxide were dissolved in it?

Page 16: I Am The Law!

Solubility from Ksp• Solid silver chromate (Ag2CrO4) is added

to a sample of water and stirred. Analysis of the solution shows that in the solution, the silver ion concentration is 1.3x10-4 M. Assuming that the silver chromate dissociates completely in water, calculate its Ksp.

Page 17: I Am The Law!

Calculating S

Predict the sign of S for this reaction: C2H5OH(l) + 3O2(g) CO2(g) + 3H2O(g)

Then, Using data from Appendix C in your textbook, calculate a value for S. Do your signs agree?

Page 18: I Am The Law!

Titration again…

• 25 mL of hydrochloric acid is titrated with 0.5 M NaOH.

• The solution turns pink and stays pink after the addition of 37.5 mL NaOH.

• What is the significance of the color change?

• What is the molarity of the HCl?

Page 19: I Am The Law!

Calculating G• Using the enthalpy and entropy data in

Appendix C (textbook), calculate the standard free energy change for the following reaction:

• 2NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(g)

Page 20: I Am The Law!

• Consider the following solubility equilibria:

• Zn(OH)2(s) Zn2+ (aq) + 2OH-(aq)• Cu(OH)2(s) Cu2+ (aq) + 2OH-(aq)

• Which of the following reagents, when added to a solid mixture containing both Zn(OH)2(s) and Cu(OH)2(s), would be most effective at dissolving both solids?

(a) Cu(NO3)2(aq) (b) Zn(NO3)2(aq) (c) NaOH(aq) (d) CH3CO2H (acetic acid) (e) HCl

Solubility Equilibria

Page 21: I Am The Law!

Relative entropy values

• Consider the structures of diamond, graphite and buckminsterfullerene (all are different forms of elemental carbon), below:

buckminsterfullereneS (diamond) = 2.43 J/K.molS (graphite) = 5.69 J/K.mol

Give a possible value for the S for buckminsterfullerene…