Chapter 4:

Electrolytes:

    Electrolytes conduct electricity when dissolved because they produce ions in aqueous solutions

        Strong Electrolytes: lots of ions

        Weak Electrolytes: few ions

        Non-electrolytes: molecular substance

            Non-ionic. Covalently bonded molecular substance

                Non-metals

Acids:

    Strong acid = strong electrolyte

        Ion that acids make is H+  (protons)

    Weak acids = weak electrolyte

        Mostly stay together as molecules

            Few H+ ions produced

Precipitation reactions

    Double Displacement Reactions leading to insoluble product.

    Start using states of matter in equations

        (aq) dissolved, (s) solid, (l) liquid, (g) gas

    Memorize or at least very familiar with solubility rules.

        Solubility Rules

Displacement Reactions

    DDR's often Redox reactions

        AB + CD ==> AD + CB

    SDR's always Redox reactions

        A + BC ==> AC + B if metal

        A + BC ==> AB + C if non-metal

Net Ionic Equations

  This is a big change from Reg Chem. All equations that can be put in     Net Ionic format MUST be put in Net Ionic format.

Know solubility rules.

Write complete balanced equation, with states of matter

Rewrite Equation

    Ionize, dissociate anything soluble

    Cancel spectator ions.

    Write net ionic equation

Acid/Base Reactions

    Learn Strong Acids: If not strong must be weak

    Learn Strong Bases: Group I and Soluble Group II hydroxides

    Weak base: ammonium hydroxide, amines.

    Neutralization: Acid + Base ==> Salt + Water

        Net ionic equation is always H⁺(aq) + OH^-(aq)==> H₂O(l)

        Exception: if base is insoluble it is not present as ions so can't        cancel spectators

Redox (Review your notes from Reg. Chem.)

Redox

Chapter 5 Thermodynamics

Energy and work and heat are equivalent

First Law of Thermodynamics Energy is conserved

    Exothermic reactions release stored heat to environment

        Delta H system is -

        Delta H environment is  +

Enthalpy is state function (stored heat)

Enthalpies of Reaction (both physical changes and chemical changes)

    kJ or kJ/mol  Check to see if the reaction produces many moles or just 1

    A reaction that burns 4 moles of propane will produce more energy than one that burns 2 moles.

        CH₄ (g) + 2 O₂ (g) --> CO₂ (g) + 2 H₂O (g)                deltaH = -802 kJ ( per mole of CH4 implied)

        2CH₄ (g) + 4 O₂ (g) --> 2CO₂ (g) + 4 H₂O (g)             deltaH = -1604 kJ ( per  2mole of CH4 implied)

Example: How many joules are released by the burning of 28.0 grams methane in air, given that delta H_combustion methane is -802 kJ/mol

    Balance equation: CH₄ (g) + 2 O₂ (g) --> CO₂ (g) + 2 H₂O (g)                                     deltaH = -802 Kj/mol

Hess' Law.  Since Enthalpy is a state function it is independent of path. So we can add subtract or multiply intermediate reactions and if we can get them to add up to the equation of interest we can calculate the overall enthalpy change by doing the same algebraic step to the enthalpies of reaction. Flip the sign if the equation is reversed, multiply by 3 if reaction tripled etc.

Example: Calculate the enthalpy of reaction for: 2S (s) + 3O₂ (g) --> 2 SO₃ (g)

    Given:1.)   S (s) + O₂ (g) --> SO₂(g)                        deltaH= -296.9kJ

             2.)   2SO₃(g) --> 2SO₂(g)  + O₂(g)                              + 196.6 kJ

              Want 2S as a reactant, multiply reaction #1  by 2  remember multiply enthalpy also

                    2S (s) + 2O₂ (g) --> 2SO₂(g)                        deltaH=  (2)-296.9kJ

                   2SO₂(g)  + O₂(g) -->  2SO₃(g)                                   (-1) 196.6 kJ

                    2S (s) + 3O₂ (g) --> 2 SO₃ (g)                                        -790.4 kJ

Enthalpies of formation

Enthalpy required to make 1 mole of production from its constituent elements in their stable form at 25 °C and 1 atm. Units are kJ    (per 1 mole is implied by definition)

Enthalpy of  formation of any element in its stable form at 25 ºC and 1 atm is 0 kJ.

Calorimetry

Key concept: Heat lost by the hot object's cooling will be gained by the cool object's warming.

Heat capacity: J/gºC  how much energy is required to raise the temperature of a substance.

Calorimeter constant J/°C  how many Joules it takes to change a calorimeter's temperature 1°C

(Must keep the calorimeter parameters constant: use exact same mass of water each trial.

Heat capacity of water 4.184 J/gºC

Heat of fusion of water 333.5 J/g   energy required to melt (+) or liberated when freezing water.

Heat of vaporization of water 2257 J/g  (+) vaporizing, (-) condensing.

Key relationship deltaH = m Cp delta T

Example easy:

25.0 grams of water at 0 c  is added to a calorimeter at 22.5 C if the ending temperature is 16 C What is the calorimeter constant. 

Heat gained by cold object is lost by warm object or reaction

Heat gained by cold

25.0 grams x 4.184 J/g ºC x (16 ºC - 0 ºC) = + 1673.6 J.

Calorimeter constant is Energy per degree of temp change

-1673.6J/ -6.5 Cº = 257.5 J/ºC

Now running a reaction in the exact same calorimeter, If the calorimeter is seen to rise 21.0 ºC how many calories are released by the chemical reaction?

257.5 J/ºC  x 21 ºC = 5407.5 J  since gain by calorimeter = loss by chemicals of reaction

- 5407.5 J.

Example- more thorough and complicated

Modern Concept of the Atom/Atomic Structure (Chapter 6 Brown, LeMay and Others)

Photoelectric effect

Specific threshold frequency (not amplitude!!!! in sharp contrast to mechanical waves)

Above threshold, KE electron increases linearly as v light increases.

KE electron = ½ mv² = hv - hv0

Where hv is the energy of the incident photon

hv0 is the energy required to remove the electron from the metal.

Somewhat related

E=mc²

So

m = E/c²

but we say that photons have no rest mass.

When hydrogen is excited by passing an electric arc through it. Only 4 visible spectra are emitted. Evidence that energy in the atom is quantized λ’s 410 nm, 434 nm, 486nm and 656. All hydrogen always the same. Only some wavelength possible implies that only some energy levels are present.  So electrons are jumping between 4  orbitals or 4 n-numbers.

λ = hc/ ΔE