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200 300 500 400 ThermochemSolutions and suchAcids and BasesEquilibrium 100 200 100 500 400 300 200 100 500 400 300 200 100 500 400 300

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Page 1: 200 300 500 400 ThermochemSolutions and suchAcids and BasesEquilibrium 100 200 100 500 400 300 200 100 500 400 300 200 100 500 400 300
Page 2: 200 300 500 400 ThermochemSolutions and suchAcids and BasesEquilibrium 100 200 100 500 400 300 200 100 500 400 300 200 100 500 400 300

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Thermochem

Solutions and such

Acids and Bases

Equilibrium

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Page 3: 200 300 500 400 ThermochemSolutions and suchAcids and BasesEquilibrium 100 200 100 500 400 300 200 100 500 400 300 200 100 500 400 300

Thermochem 100

The total energy of the world is constant. Energy can neither be created

nor destroyed.

What is the law of conservation of energy?

Page 4: 200 300 500 400 ThermochemSolutions and suchAcids and BasesEquilibrium 100 200 100 500 400 300 200 100 500 400 300 200 100 500 400 300

Thermochem 200

What is the equation for a change in temperature?

ΔT = Tf - Ti

Page 5: 200 300 500 400 ThermochemSolutions and suchAcids and BasesEquilibrium 100 200 100 500 400 300 200 100 500 400 300 200 100 500 400 300

Thermochem 300

In the equation for calculation of heat transfer: q= mcΔT what does each letter

represent?Q = heat m= mass c = specific heat capacity ΔT = change in

temperature

Page 6: 200 300 500 400 ThermochemSolutions and suchAcids and BasesEquilibrium 100 200 100 500 400 300 200 100 500 400 300 200 100 500 400 300

Thermochem 400

What is the equivalence of 0ºC in Kelvins?

273.15K

Page 7: 200 300 500 400 ThermochemSolutions and suchAcids and BasesEquilibrium 100 200 100 500 400 300 200 100 500 400 300 200 100 500 400 300

Thermochem 500

How do you spell …

LOLCalculate the specific heat capacity of canola

oil. If the mass of the oil is 60g, the initial temperature is 35ºC and the final temperature

is 5.2ºC with a heat change of 4.0x103?

C= q/m∆T = -4.0x103J

(60.0g)(5.2ºC – 35.0ºC) = 2.24 J/gºC (pg 635)

Page 8: 200 300 500 400 ThermochemSolutions and suchAcids and BasesEquilibrium 100 200 100 500 400 300 200 100 500 400 300 200 100 500 400 300

Solutions and Such 100

Is ethanol (CH3CH20H) soluble in water?

Yes, infinitely soluble!

Page 9: 200 300 500 400 ThermochemSolutions and suchAcids and BasesEquilibrium 100 200 100 500 400 300 200 100 500 400 300 200 100 500 400 300

Solutions and Such 200

True or False, Polar compounds can dissolve in non-polar solvents. Explain.

False, polar compounds can only dissolve polar solvents and non-polar compounds can only dissolve in non-polar solvents!!

Page 10: 200 300 500 400 ThermochemSolutions and suchAcids and BasesEquilibrium 100 200 100 500 400 300 200 100 500 400 300 200 100 500 400 300

Solutions and Such 300

Define Miscibility and give an example of a misicble and immisicble solution.

Miscible – compounds that dissolve readily in each other in any proportion. (example: water

and alcohol)

Immiscible – liquids that do not readily dissolve in each other (example oil and water)

Page 11: 200 300 500 400 ThermochemSolutions and suchAcids and BasesEquilibrium 100 200 100 500 400 300 200 100 500 400 300 200 100 500 400 300

Solutions and Such 400

What is a dipole – dipole attraction?

The attraction between the opposite charge on two different

polar molecules is called a dipole – dipole attraction.

Page 12: 200 300 500 400 ThermochemSolutions and suchAcids and BasesEquilibrium 100 200 100 500 400 300 200 100 500 400 300 200 100 500 400 300

Solutions and Such 500

Calculate the volume of isoproply alcohol used to make a 500mL solution.

The volume/volume % is 70%.

Volume/volume percent = volume of solute x 100% volume of solution

Volume of solute = 70% x 500mL100%

= 350mL (pg 262)

Page 13: 200 300 500 400 ThermochemSolutions and suchAcids and BasesEquilibrium 100 200 100 500 400 300 200 100 500 400 300 200 100 500 400 300

Equilibrium 100

What is Le Châtelier’s Principle?

If a stress is applied to a system at equilibrium, the equilibrium will shift to

relieve the stress.

Page 14: 200 300 500 400 ThermochemSolutions and suchAcids and BasesEquilibrium 100 200 100 500 400 300 200 100 500 400 300 200 100 500 400 300

Equilibrium 200

Given the balanced formula:

2SO2(g) + O2 2SO3(g) + heat Increase SO2 concentration

Increase temperatureWhich way will the reaction shift?

Increase SO2 concentration = shift right

Increase temperature = shift left

Page 15: 200 300 500 400 ThermochemSolutions and suchAcids and BasesEquilibrium 100 200 100 500 400 300 200 100 500 400 300 200 100 500 400 300

Equilibrium 300

What is a catalyst, and how does it affect a reaction?

A catalyst is a substance that increases the rate of a chemical reaction without being consumed by the reaction. The catalyst is always left over.

Page 16: 200 300 500 400 ThermochemSolutions and suchAcids and BasesEquilibrium 100 200 100 500 400 300 200 100 500 400 300 200 100 500 400 300

Equilibrium 400

For the Reaction: CO(g)+NO2(g) CO2(g)+NO(g)

Given the ΔE= -226kJ and the Ea(fwd)= 134kJ

What is the Ea(rev)?

ΔE=Ea(rev)-Ea(fwd)

Ea(rev)=ΔE+Ea(fwd)

Ea(rev)=360Kj

Page 17: 200 300 500 400 ThermochemSolutions and suchAcids and BasesEquilibrium 100 200 100 500 400 300 200 100 500 400 300 200 100 500 400 300

Equilibrium 500

Step one: NO2(g) + NO2(g) NO3(g)+NO(g)

Step two: NO3(g)+CO(g) NO2(g)+CO2(g)

NO2(g)+NO2(g)+NO3(g)+CO(g) NO3(g)+NO(g)

+NO2(g)+CO2(g)

The Balanced Equation is: NO2(g)+CO(g) NO(g)+CO2(g)

The reaction Intermediate is NO3.

Given the following two step reaction:

Step one: NO2(g) + NO2(g) NO3(g)+NO(g)

Step two: NO3(g)+CO(g) NO2(g)+CO2(g)

What is the overall balanced equation for the reaction and what is the reaction intermediate.

Page 18: 200 300 500 400 ThermochemSolutions and suchAcids and BasesEquilibrium 100 200 100 500 400 300 200 100 500 400 300 200 100 500 400 300

Acids and Bases 100

What are the properties of an acid?

•pH between -1 and 7

•Sour taste

•Strong acids dissociate completely

•Conduct electricity

•Turns blue litmus paper red

•Have no characteristic feel

•Neutralize basic solutions, forming salt and water

Page 19: 200 300 500 400 ThermochemSolutions and suchAcids and BasesEquilibrium 100 200 100 500 400 300 200 100 500 400 300 200 100 500 400 300

Acids and Bases 200

Explain the Arrhenius, and the Brǿnsted Lowry Theories.

Arrhenius theory states: An acid is a substance that dissociates in water to produce one or more hydrogen ions, H+. A Base is a substance that dissociates in water to produce one or more Hydroxide ions, OH ֿ.

Brǿnsted Lowry theory states: An acid is a substance from which one proton (H+ ion) can be removed. A base is a substance that can

remove a proton (H+ ion) from an acid.

Page 20: 200 300 500 400 ThermochemSolutions and suchAcids and BasesEquilibrium 100 200 100 500 400 300 200 100 500 400 300 200 100 500 400 300

Acids and Bases 300

Identify the conjugate acid-base pairs of the following reaction:

HBr(g)+H2O(l) H3O+(aq)+Brˉ

HBr(g)+H2O(l) H3O+(aq)

+Brˉ

Conjugate acid-base pair

Conjugate acid-base pair

Page 21: 200 300 500 400 ThermochemSolutions and suchAcids and BasesEquilibrium 100 200 100 500 400 300 200 100 500 400 300 200 100 500 400 300

Acids and Bases 400

Calculate the pH of a solution with [H3O+]=3.8x10-3.

pH= -log(H3O+)

pH= -log(3.8x10-3)

pH= 2.42

Page 22: 200 300 500 400 ThermochemSolutions and suchAcids and BasesEquilibrium 100 200 100 500 400 300 200 100 500 400 300 200 100 500 400 300

Acids and Bases 500Given:

CH3CH2COOH(aq)+H2O(l) CH3CH2COOֿ(aq)+H3O(aq)

The initial [CH3CH2COOH]= 0.10 mol/L and pH= 2.96

Find Ka.

Concentration (mol/L) CH3CH2COOH(aq)+ H2O(l) CH3CH2COOֿ(aq)+ H3O(aq)

Initial 0.10 0 0

Change -x +x +x

Equilibrium 0.10 – x +x +x

Ka= (x)(x) (0.10-x)

[H3O+]= 10-2.96

[H3O+]= 1.1 x 10-

3

Ka= (1.1x10-3)(1.1x10-3) (0.10-1.1x10-3)

Ka= 1.2x10-5


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