Chem II Ch. 16, 17, 18, 19

0.0(0)
studied byStudied by 0 people
GameKnowt Play
learnLearn
examPractice Test
spaced repetitionSpaced Repetition
heart puzzleMatch
flashcardsFlashcards
Card Sorting

1/79

encourage image

There's no tags or description

Looks like no tags are added yet.

Study Analytics
Name
Mastery
Learn
Test
Matching
Spaced

No study sessions yet.

80 Terms

1
New cards

Qsp < Ksp

No precipitate

2
New cards

Qsp = Ksp

No precipitate

3
New cards

Qsp > Ksp

Yes precipitate

4
New cards

Common Ion Effect

Influence (usually a decrease in solubility) an ion already in solution has on the solubility of an ionic compound

  • Usually decreases solubility if the acid shares a common ion with the salt

5
New cards

Solving For Molar Solubility w/ Common Ion Effect

  1. Identify the common ion between the two compounds

    • the given concentration of the other compound (not the one you find Ksp for) will be the concentration of the common ion

  2. Write rxn of main compound breaking down into its ions

  3. Make RICE table (main compound = solid) & put concentration of common ion

    • if main compound has OH, always assume the OH came from H2O ( [OH] from H2O = 1 × 10-7 M)

  4. Find Ksp and plug in equilibrium expression to find s

6
New cards

Metal hydroxides & compounds with strong or weak base anions have …

Higher solubility in acidic solutions

7
New cards

Mg/Ca/Sr/Ba with (OH)2 have …

Lower solubility in basic solutions because of the common ion effect

8
New cards

Most other metal hydroxides have …

Increased solubility in basic solutions due to the formation of a complex ion

9
New cards

Salts that contain a basic anion are …

More soluble in strong acids and the anion will react with H+ (aq) to form its weak conjugate acid

10
New cards

Steps for Expecting Solubility to be More, Less, or Same as in Pure Water Solution

  1. Write what ions each salt breaks into when dissolved in H2O

    a) If the salt shares a common ion with the acid (common ion effect), solubility will be less than in pure water solution

    b) If above doesn’t apply, continue to Step 2

  2. Write eqn of salt + strong acid

  • if it forms a weak conjugate acid, its solubility is more than in pure water solution

11
New cards

Selective Precipitation Definition

Using differences in solubility to separate mixtures of ions

12
New cards

Steps for Determining Concentration and Whether There is Precipitation

  1. Ignoring neutral ions, write out what precipitate is formed when combining solutions

    • Ex: NaCO3 + MgNO3: MgCO3 (s) → Mg2+ (aq) + CO32- (aq)

  2. Write out Ksp eqn and find its table value

    • You need to compare Q with Ksp, but need to find Q first

  3. Calculate concentration of both ions, plug them into Q, and compare with Ksp

  • Qsp > Ksp = a precipitate forms

  • Qsp < Ksp = no precipitate forms

  1. Create RICE Table

  2. Solve for concentration of each ions as normal

13
New cards

1st Law of Thermodynamics

1) the energy of the universe is conserved (constant)

2) the energy of an isolated system is conserved

3) the change in internal energy for a closed system is equal to the heat and work exchanged

14
New cards

Entropy, S - Information Theory Perspective

Allows us to calculate the absolute entropy of a system

15
New cards

2nd Law of Thermodynamics

A spontaneous process in an isolated system always increases the entropy of the system

16
New cards

Entropy, S General Definition

A measure of how much you don’t know about a macrostate

17
New cards

Factors that Increase Entropy

  • Increased Volume of gas at constant temp

  • More Matter on Product Side of Rxn

  • Increased Temperature

  • Ssolid < Sliquid < Sgas

18
New cards

3rd Law of Thermodynamics

The entropy of a perfect crystal at 0 K is zero

19
New cards

Entropy - Thermodynamic Perspective

The change in entropy is the reversible heat exchanged at temperature, T

20
New cards

Formula for Finding △S°universe

knowt flashcard image
21
New cards

If △S°universe is positive …

Then is product favored aka thermodynamically favored to occur

22
New cards

If △S°universe is negative …

Then is not product favored

23
New cards

At low temp …

△H°rxn has the most effect on sign of △S°universe

24
New cards

At high temp …

△S°rxn has the most effect on sign of △S°universe

25
New cards

Reversible

1) forward & backward change follows the same path

2) equations of state are always obeyed

3) the reversible path as infinitesimally small steps so that the system & surroundiings are always in equilibrium

26
New cards

Formula for Finding △S°rxn

knowt flashcard image
27
New cards

Factors in Determining Entropy

1) Phase

  • Ssolid < Sliquid < Sgas

2) Molar Mass

  • Heavier = higher entropy

3) Molecular Complexity

  • more complex = higher entropy

28
New cards

Formula for Finding △H°rxn

knowt flashcard image
29
New cards

Formula for Finding △Ssurroundings

knowt flashcard image
30
New cards

If Endothermic Rxn, Entropy of Surroundings …

Decreases

31
New cards

If Exothermic Rxn, Entropy of Surroundings …

Increases

32
New cards

When △Suniv > 0 and △G°rxn < 0 …

Spontaneous, K > 1; product favored

33
New cards

When △Suniv < 0 and △G°rxn > 0 …

Non spontaneous, K < 1; reactant favored

34
New cards

When △S°rxn = + and △H°rxn = + …

Spontaneous at High Temp

35
New cards

When △S°rxn = - and △H°rxn = - …

Spontaneous at Low Temp

36
New cards

When △S°rxn = + and △H°rxn = - …

Spontaneous at All Temp

37
New cards

When △S°rxn = - and △H°rxn = + …

Spontaneous at No Temp

38
New cards

Formula for Finding Tswitch

knowt flashcard image
39
New cards

Formula for Finding △G°rxn

knowt flashcard image
40
New cards

△G°rxn Definition

Change in free energy to create 1 mol of compound from its elements in their standard state

41
New cards

If Reverse a Rxn …

Flip sign of △G°rxn

42
New cards

If multiply a rxn by a factor …

Multiply △G°rxn by the factor

43
New cards

If add two rxns …

Add △G°rxn together

44
New cards

Formula for Finding △G

R = 8.314 × 10-3

<p>R = 8.314 × 10<sup>-3</sup></p>
45
New cards

The better the agent …

the worse the conjugate

46
New cards

Balancing Redox Rxns in Pure H2O

  1. Look at the main chemical rxn and identify what is oxidized and reduced

  2. Write out the oxidation half rxn and write out how many electrons is added and on which side of rxn

  3. Write out the reduction half rxn and do same thing

  4. Combine them back

47
New cards

Balancing Redox Rxns in Acidic Solutions

  1. Balance the electrons of the non-oxygen / central atoms

  2. Balance O atoms with H2O by adding however H2O on the other side to satisfy the amount of needed O atoms

  3. Balance H atoms with H+ by looking adding however H+ atoms on other side to satisfy the amount of needed H atoms

  4. Balance charge of e- by adding necessary electrons on either side

48
New cards

Balancing Redox Rxns in Basic Conditions

  1. Balance central atoms

  2. Balance O atoms with OH-

  3. Balance H atoms with H+

  4. Add OH- to both sides to turn H+ and OH- into H2O

  5. Balance charge with e-

49
New cards

Oxidation is always on …

The left

50
New cards

Reduction is always on …

The right

51
New cards

Reference Cell

Standard hydrogen electrode (SHE)

  • H2 (g, 1 atm)

52
New cards

Formula for Finding E°cell

cathode - E°anode

53
New cards

If E°cell > 0 and △G < 0

Spontaneous in the direction written, Q < K

54
New cards

If E°cell < 0 and △G > 0

Non spontaneous, Q < K, backward rxn

55
New cards

Candidates for Reduction

The more positive reduction potential

56
New cards

Candidates for What Gets Oxidized

The more negative reduction potential

57
New cards

If Atomic # > amu

Have too many neutrons, use β- - decay to get rid of a neutron

58
New cards

If Atomic # < amu

Don’t have enough neutrons, so use β+ (positron emission) or electron capture to make a neutron

59
New cards

If both heavy AND too few neutrons

Use α - decay

60
New cards

S° > 0

Have more mol of gas on product side

Product Favored

61
New cards

S° < 0

Have more mol of gas on reactant side

Reactant favored

62
New cards

s → l

l → g

g → s

△S > 0, spontaneous

63
New cards

l → s

g→ l

△S < 0, nonspontaneous

64
New cards

Formula for Finding Total Charge Passed (in coulombs)

Q = I * t

  • I = the current in amperes (C/s)

  • t = time in seconds

65
New cards

Formula for Converting Charge to Moles of Electrons

mol e- = Q/F

  • F = faraday’s constant

66
New cards

Steps for Finding Amount of Grams Able to be Electroplated

  1. Use Q = I*t to find total charge passed (answer will be in C units)

  2. Convert charge to mole e- by using mol e- = Q/F

  3. Write down half reaction including electrons added

  4. Divide answer from Step 2 by the amount of electrons

  5. Convert mol to grams

67
New cards

If Asked What Will Reduce H+ Spontaneously

Look at table with standard reduction potentials

  • metals above hydrogen

68
New cards

If Asked to Find Change in Gibbs free energy

Use △G = -nFEcell

  • n = number of electrons transferred

  • F = faraday’s constant

  • Ecell = Ecathode - Eanode

69
New cards

Good Candidates for being a Reducing Agent

More negative standard reduction potential

70
New cards

Good Candidates for being an Oxidizing Agent

Less negative standard reduction potential

71
New cards

Use Nernst Equation When …

The electrochemical cell is not at standard conditions

  • not at 1 M, 1 atm, or 298 K

72
New cards

Equation for Q in lnQ

[products / [reactants]

73
New cards

Form of electrochemical cell notation

reactant | product || reactant | product

74
New cards

The salt bridge completes the circuit by allowing __ to flow towards the anode cell and __ to flow towards the cathode cell

Anions, cations

75
New cards

In Chemical Rxn, what can be oxidized and reduced/ oxidizing and reducing agents?

Only species on the reactant side

76
New cards

For alpha decay …

Mass # decreases by 4, atomic # decreases by 2

77
New cards

For beta- decay …

Mass # stays the same, atomic # increases

78
New cards

For positron emission (beta+) and electron capture …

Mass # stays the same, atomic # decreases

79
New cards

in Ebinding = -△mc2 / A

△m = actual mass - sum of subatomic particle mass

80
New cards

Steps for Solving for Ebinding per nucleon

  1. Determine the number of protons and neutrons

  • M - A = N

  1. Multiply the number of protons and neutrons to the mass numbers given accordingly

  2. Find △m (actual mass - sum of subatomic particle mass) and convert to kg/mol by adding x 10-3 to the end of the number

  3. Plug numbers into equation where A = atomic number