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UA / Chemistry / CH 118 / Which property of liquid referes to resistance to flow?

Which property of liquid referes to resistance to flow?

Which property of liquid referes to resistance to flow?

Description

School: University of Alabama - Tuscaloosa
Department: Chemistry
Course: Honors General Chemistry
Term: Fall 2015
Tags: General Chemistry
Cost: 50
Name: CH 118 Final Studyguide
Description: This study guide covers all chapters, 11-20. It has definitions, examples, etc.
Uploaded: 04/25/2018
5 Pages 35 Views 3 Unlocks
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CH 118-001 Final Studyguide


Which property of liquid referes to resistance to flow?



1

I.

Chapter 11 – Liquids, Solids, and Intermolecular Forces

Solid

Doesn't Flow

Extremely Slow Diffusion Diffusion

Liquid

Flows

Retain Shape Virtually

incompressible Assumes Container | Virtually Shape

Incompressible Assumes Container | Compressible Shape

ZOSO

Gas

Flows

Rapid Diffusion

re

a. Intermolecular Forces (AKA Between Molecules)


Who does cohesive force bind molecules together?



i. Ion-Dipole

1. Strongest Intermolecular Force 2. Ionic compound is mixed with a polar compound

a. Important when dissolving an ionic compound in water ii. Dipole-Dipole We also discuss several other topics like What are the bases of depreciation calculation?

1. Exists in all molecules that are polar

a. Polar molecules have electron-rich and electron-deficient


How is vapor pressure measured?



regions b. Positive end of one dipole attracts the negative end of We also discuss several other topics like What are the resources of fossil fuel?

another iii. Hydrogen Bonding (a super dipole-dipole force)

1. Hydrogen bonded to small electronegative atoms (F,0,N)

a. H has partial positive charge, while FON have partial

negative charge 2. Makes boiling point a lot higher iv. London Dispersion Forces (AKA Induced Dipole)

1. Present in all molecules 2. Result of fluctuations in the electron distribution within molecules

or atoms Stuea. Generally increases with molar mass

b. Larger electron cloud = greater dispersion force

i. Electrons held less tightly and can polarize more If you want to learn more check out What is the function justification?

easily c. Longer chains have a higher boiling point than spherical Don't forget about the age old question of What are the dimensions of emotional states?

chains b. Properties of Liquids

i. Viscosity – Resistance to flow

1. Water is not viscous 2. Measured in terms of time required to flow through a thin tube

3. Higher temp = lower viscosity ii. Surface Tension

1. Energy required to increase the surface area of a liquid by a unit

amount

a. Surface tension is higher on the surface of water because We also discuss several other topics like What is ‘water smart’ products and why is it said to be beneficial?
We also discuss several other topics like How is angular momentum defined?

there aren't water molecules on all sides 2. Cohesive force – binds like molecules together

CH 118-001 Final Studyguide

3. Adhesive force – binds molecules to a surface

a. Causes meniscus c. Changes of State

Sublimation

Melting

Vaporizing

Solid

Liquid

Gas

Freezing

Condensing

Deposition

i. Heat of fusion - enthalpy change associated with melting

1. Ex: AfH of water is 6.01 kJ/mol ii. Heat of vaporization - enthalpy change associated with vaporization

1. Ex: AHvap of water is 40.67 kJ/mol

2. Always positive because it is endothermic d. Vapor Pressure - The pressure of a gas in dynamic equilibrium with its liquid

1. Dynamic Equilibrium - things are happening at a molecular level,

but the overall system is not changing over time 2. Measured using manometer 3. Boiling point – vapor pressure equals the applied pressure

a. At latm, it is the normal boiling point e. Phase Diagram

vapor

Boiling Point

с

Lo

Н

Temp

Liquid

Melting point .

сон

Solid

Heat

CH 118-001 Final Studyguide

i. Left to right is exothermic ii. Triple point - where solid, liquid, and gas exist in equilibrium iii. Critical point – liquid and gas have the same density and are

indistinguishable f. Clausius-Clapyeron Equation

AHvap

i. Incea) = -slipar en la

1

In(P2) In(P1)

11.2

ii. Example 11.5 Methanol has a normal boiling point of 64.6°C and a heat of vaporization of 35.2 kJ/mol. what is the vapor pressure of methanol at 12.00CP

Ti-64.6°C + 273.15

- 337.8K T2 =12.0°C + 273.15

= 285. 21.

in P- AHMAD ( 7.)

in Pz : -35.2 X108 Amor

Pi 8.314 Fimon (285.24 - 12/37.84) in Pa = -2.31

-2.31 e 760 torr (1ookup)

P2 =(P, Je 2.31

P2= 75.4 torr A

CH 118-001 Final Studyguide

4.

II.

Chapter 12 - Crystalline Solids and Modern Materials a. Crystalline vs Amorphous i. Crystalline

1. Highly structured and exhibit long-range order ii. Amorphous

1. No order iii. Crystalline solids can be made amorphous by melting and cooling rapidly

1. AKA quartz to glasse b. Unit Cell – repeating identical structure

i. Packing Efficiency

1. %E = Watom * 100%

Vcell 2. Higher packing efficiency = higher melting point

m

3. Example 12.2. Calculate the packing efficiency of a simple Cubic unit cell.

Packing efficiency= Vatom x100%

Vcen

Study soup

Varom= 4/3 rr3

Vcell = 13 = (3r) 3 = 8,3 Packing efficiency: singl x100:1.

-52.361. A

8

ii. Coordination Number

1. Number of atoms an atom is in direct contact with 2. Higher packing efficiency generally means higher coordination

number iii. Edge Length

1. How long a side is

2. AKA volume of a unit cell iv. Crystal Lattice

1. 3D array of unit cells 2. Simple cubic

a. 1 atom per cell b. Coordination number = 6 c. Packing Efficiency = 52% d. 1 = 2r

CH 118-001 Final Studyguide

V3

e. AKA stack 2 pennies 3. Body-centered cubic

a. 2 atoms per cell b. Coordination number = 8 c. Packing Efficiency=68% d. 1 = 45

e. Ex: Lic1 4. Face-centered cubic UP

a. 4 atoms per cell b. Coordination number = 12 c. Packing efficiency = 74% d. 1 = 2V(2r)

e. Ex: NaCl

5. Example 12.3 A body-centered cubic unit cell has a volume of 4.32x10-23 cm3 Find the radius of the atom inom,

v=p3 l=35

l= 41 1/3

b= 3/4.32x10-23 cm3 = 3.5088x10cm

1.5193x10 cm

© Are 13 (3.5088 X10-8 cm)

1.5193x108 cmx 0.00m x P

= 152pm A

x

cm

10m

c. X-Ray Diffraction

i. Determines arrangement of atoms and measures distance between them ii. Beam is shot through the crystal and diffracts to give a pattern on a

photographic plate Constructive Interference

1. Same wavelength + same frequency = double magnitude iv. Destructive Interference

CH 118-001 Final Studyguide

1. Cancel out V. Bragg's Law

1. Allows us to find the distance between atomic layers 2. n2 = 2dsino 3. Example 12.1

SOUP

led

when an X-ray beam of 2=154 is incident on the

surface of an iron crystal, it produces a max reflection

at an angle of @= 32,6° Assuming n=l, calculate the separation between layers of iron atoms in the crystal,

d = na

2 sina d= 154pm

Zsin (32.60)

d=143 pm

d. Solids

Type

Unit Particle Forces Atoms/Molecules Intermolecular

Molecular

Examples Dry ice Sugar

Covalent (network)

Atoms connected Covalent bonds in covalent network

Diamond Quartz Graphite

Properties *Poor thermal and electrical conductors *Fairly soft *Low to somewhat high melting point *Very high melting point

*Very hard *poor conduction *Brittle *Hard * High melting point *Poor conductors *Soft to hard *low melting point to high *excellent conductors

Ionic

lons

Electrostatic attractions

NaCl and other soft salts

Metallic

Metal atoms

Metallic bonding

All metallic elements

Stud

e. Band Theory

CH 118-001 Final Studyguide

i. Valence band - Bonding molecular orbitals and contain the N valence

electrons ii. Conduction band – Antibonding and are completely empty iii. Conductor

1. No energy gap between valence and conduction bands iv. Semiconductor

1. Small energy gap between valence and conduction bands V. Insulator

1. Large energy gap between valence and conduction bands

CH 118-001 Final Studyguide

III. Chapter 13 – Solutions

a. Solution - homogenous mixture

i. Solvent – usually the component in the greatest amount

ii. Solute b. Concentration

i. Molarity (M)= moles solute

1. Molamy (1) L solution

mass of component in solution ii. Weight Percentage = -

on * 100%

total mass solution 1. For dilute solutions use ppm

mass of component in solution a.

ution * 106 " total mass solution 2. Even more dilute, use ppb (10%)

3. Example: made from 8.0g Nach and 100g H2O

dy Soup

Solution

WT1 = 89 Naci

- X1001 1089 Naci+H2O 27.4.1.

9 2.60'. H2O

iii. Mole Fraction (X)

1 mol component

total moles iv. Molality (m)

1 moles solute * kg solvent

2. In dilute solutions, Mæm c. Solution process

i. Solution - solvent molecules surround solute

1. If solvent is water, this is called hydration

2. Every time bonds are broken or formed, there is a AE ii. 3 attractive interactions

1. Solute - solute 2. Solvent - solvent 3. Solute – solvent

4. AH sol = AH (endo) + AH2(endo) + AH3(exo) d. Solubility

i, Dynamic equilibrium – solution that is in equilibrium with undissolved

solute

1. Saturated solution

a. Solute and solvent dissolve to form solution

b. Solution crystallizes to form solute and solvent 2. Supersaturated solution - more solute in solution than needed

ON

CH 118-001 Final Studyguide

a. Very unstable e. Factors affecting solubility

i. Like dissolves like ii. Miscible – pairs of liquids that dissolve in all proportions iii. Higher pressure means a higher amount of gas in liquid (increases

concentration of solution)

1. Henry's Law

a. Cg=kPg where K is Henry's Law Constant

b. Example: Solubility of Nz in H2O (25°c) and 0.18 atm is 5.3x10-4M.

Cg = kPa K=Cg = 5.3610-4 moll

0.78 atm

=6.8810-4 mol/l-arma

iv. Higher temp for gas in liquid = lower solubility f. Colligative Properties

i. Depend on the concentration, but not the nature, of the solute ii. Vapor Pressure

1. The vapor pressure of the solution is lower than the vapor pressure

of the pure solvent 2. Raoult’s Law

a. Psolution = XsolventP Solvento b. Ideal solution obeys Raoult's Law

C. Example: Calculate VP at 25°C of soin containing 99.5g C12H220 and 300ml water. vp of water at 250 G is 23.8 torr.

99.5g C ,z Hz 2G,,=0.2964 mol C, 2 H2 2011 300 mL H2O=16.65 mol H2O

Xoon H2O

. 16.onmol

__ = 0.9828 moi

0.2907 moral6.65mer ncia H22O,, Intro

Psown = XH2O P420=60.98288)( 23.8torr)

1. = 23.4 torr A iii. Boiling Point Elevation

1. A liquid boils when its vapor pressure equals the external pressure

on its surface 2. ATg = Kom where Kb is the molal boiling point elevation

constant

. H2O

CH 118-001 Final Studyguide

10

iv. Freezing Point Depression

1. Freezing occurs when the vapor pressure of the liquid equals the

vapor pressure of the solid 2. ATF = Kem where Kf is the molal freezing point depression

constant 3. Example:

consider the effect of adding 62.0g C6H2O6 to 1.00kg H2O

KfH2O = 1,8600/m

162. Og CyH2O6x Imol =1.00 moic 6 Hz oo

62g

A TE = Kfm

= (1.860 )(1.000) =-1.8600

*Neg bc it is the change

H goes ot-1.86°C

V. Osmosis

1. Net movement of solvent towards the more concentrated solution 2. Osmotic Pressure - pressure required to stop osmosis

a. = MRT b. Isotonic – two solutions of equal osmotic pressure C. Hypotonic – lower osmotic pressure

d. Hypertonic – higher osmotic pressure g. Determination of Molar Mass Camphor mahs at 119.8°C and freezes at 40°C/m. when 0.180g of org. Substance of unknown molar mass is

SONGd

ed in 22.019 Camphor, the soin Tg = 176.4°C. what Is the molar mass ?

A TE=Kq m ATP = 179.8°C-176.700 = 3.100

diss

m-a Tf - 3.10€ 0.078 m = mor solu

Bf 40°C/m

1 kg solvent moles souvent = (m) (kg solvent) = 10.078 m] (0.02201 kg)

molar. 0.1869 Liv mass 1.4x10-3 mois

11x1029 mola

CH 118-001 Final Studyguide

11

IV. Chapter 14 - Chemical Kinetics

a. Chemical Kinetics - the study of the rate/speed of a chemical reaction

change in concentration of reactant/product b. Reaction Rate =

time C. Example: d. Average rate of a reaction

i. Gives average rate over a time interval e. Instantaneous rate of change

i. Determined by calculating the slope of the tangent to the curve at the point

of interest ii. Same regardless of which reactant or product you use

1. Negative for reactants and positive for products f. The Rate Law

i. Relationship between the rate of the reaction and the concentration of the

reactant

ii. Rate = k[A]" where k is the rate constant and n is the reaction order g. Reaction Orders

i. Zero-Order

1. The rate of reaction is independent of the concentration of reactant 2. Rate =k 3. The rate is the same at all measured initial concentrations

4. Units of ii. First-Order

1. The rate of reaction is directly proportional to the concentration of

the reactant 2. Rate = k[A] 3. When the concentration doubles, the rate doubles

4. Units of s-1 iii. Second-Order

1. The rate of the reaction is proportional to the square of the

concentration of the reactant 2. Rate = k[A] 3. When the reactant concentration doubles, the rate quadruples

4. Units of M-1 * 5-1 iv. Overall order = sum of orders h. Integrated Rate Laws – dependence of concentration on time

i. First-Order

1. n(Alt = -kt ii. Second-Order

IC RESOU

'

[A]o

[A]t iii. Zero-order

1. [A]t = -kt + [A]. i. Half-Life

i. The time required for the concentration of a reactant to fall to 12 of its

initial value

CH 118-001 Final Studyguide

12

ii. First-order

1. 0.693

ek

2. Independent on initial value iii. Zero-Order

1. [4]

j. Temperature Dependence on Reaction Rate

i. Arrhenius Equation – shows relationship between K and T

-Ea

1. k = Ae RT 2. Activation Energy (Ea) – the energy required to reach the activated

complex k. Reaction Mechanisms – the series of individual steps by which an overall

chemical reaction occurs

i. The slower step is usually the rate determining step ii. Example:

Study Soup &

2 Hzca + 2NO(g) → 2H2O(g) + N2 (9)

rate law = K[Hz] [no]2

Z NO(g) = NzOz (a) fast

H2(g) +N 20269) + H2O (9)+ N2O(g) slow N2O(g) + H2 (9) N2(g) + H2O (9) Fast

Slow step gives rate=k [Hz] [N20 2]

*Bc of Stepi we substitute so

rate= K[Hz][no]2 A

1. Catalysis

i. Catalysts increase the rate of a reaction without being consumed by

lowering the activation energy ii. Homogenous catalyst - catalyst exists in the same phase as the reactants

1. Catalytic ozone by C1 iii. Heterogenous catalyst - catalyst exists in a different phase

1. Catalysts used in catalytic converters

CH 118-001 Final Studyguide

13

V. Chapter 15 – Dynamic Equilibrium

a. Dynamic equilibrium – the condition in which the rate of the forward reaction

equals the rate of the reverse reaction

i. Concentrations do not change b. Law of Mass Action where k is the equilibrium constant

aA + bB È cctd D

KE [c] [D]"

[A]a[B]b

i. If k is greater than 1, equilibrium lies to the right and is product-favored

1. Takes a long time to reach equilibrium ii. Ifk is less than 1, equilibrium lies to the left and is reactant-favored c. Equilibrium Constant

i. If we inverse the equation, invert k ii. If we multiply the coefficients by a factor, raise k to that same factor iii. If we add 2 or more equations, multiply the corresponding equilibrium

constants d. Reaction Quotient

i. For the reaction at any time ii. IfQ=K, then it is at equilibrium iii. If Q is less than K, the equilibrium shifts to the right iv. If Q is greater than K, the equilibrium shifts to the left

V. Example: Consider the reaction and its equilibrium constant.

CAD Soup

I 2(g) + Cl2(g)

= 2 Iclag Kp=81.9

A reaction mixture contains Pr: 6.114 atm, Parz-0.102

atm, and Pici = 0.355 atm, is the reaction mixture a equilibrium ? If not, which direction will it proceed?

Qpe (ich). (0.355)

=

- (0.114) (0.102)

10.8

Piz

Priz/

QuK So it will proceed right A

CH 118-001 Final Studyguide

14

e. Le Chatelier's Principle

i. When a chemical system at equilibrium is disturbed, the system shifts in a

direction that minimizes the disturbance ii. The effect of concentration change

1. Corrects to lower the increased concentration iii. The effect of volume or pressure change

1. Increase in pressure shifts to the side with the least number of

moles of gas iv. The effect of temperature change

1. An increase in temperature drives the endothermic reaction V. Example:

Su

1 What is the effect of adding CO2 to:

CaCO3(s) Ĉ Cao (s) + CO269)

ron

shifts left

2) What is the effect of decreasing the volume :

ZUC103 (5) 2 KCl (s) + 302 (6)

omola

3 moig

ronshifts left

3) What is the effect of increasing the temp:

CaCO3(s) 2 Cao (s) + CO2(g) endo

ran shifts right

CH 118-001 Final Studyguide

15

VI. Chapter 16 - Acids and Bases

a. Definitions of Acids and Bases

Arrhenius Produces H+ ions Produces OH-ions

Tons

Acid Base

Bronsted-lowry Proton donor Proton acceptor

Lewis Electron pair acceptor Electron pair donor

59

i. Binary Acids

1. Ex: H-Y ii. Oxyacids

1. Ex: H-O-Y iii. Strong Acids

1. HCI 2. HBr 3. HI 4. HNO3 5. HC104

6. H2SO4 iv. Weak Acids

1. HF 2. HC2H302 3. HCHO2 4. H2SO3 5. H2CO3

6. H3PO4 b. Acid Ionization Constant (Ka)

i. Equilibrium constant for the ionization reaction of the weak acid in water ii. HA (aq) + H + (aq) + A - (aq)

1. Ka = [H+][A-]

[HA]

c. pH Scale

i. pH = -log[H+] ii. pOH = -log[OH-] iii. In strong acids [H3O+] = [strong acid} iv. In weak acids you have to use an ICE diagram

1. Example: Find the pH of a 0.200 M HNO3 Sorin. (HNOzcaa) + H2O(A H2 Otlaq) + NO2 (aa) 10.200

pH = -log[+ ] clox

+X

5-109(9.6410-3) El 0.200-4

= 2.02A Ka= [tBot] [No2 ] x2

0.200-* [HNO2]

X-9.6410-3, H+

+x

CH 118-001 Final Studyguide

16

d. Percent Ionization o

concentration of ionized acid

acid * 100% - initial concentration of acid e. Mixtures of Acids

i. Strong acid + weak acid

1. [H+] = [strong acid] ii. Weak acid + weak acid

1. We assume that the weaker acid will not make a significant

contribution to (H+] f. Strong Bases

i. LiOH | ii. NaOH iii. КОН iv. Sr(OH)2 V. Ca(OH)2

vi. Ba(OH)2 g. Weak Bases

i. CO3 ii. CH3NH2 iii. C2H5NH2 iv. NH3

v. HCO3 vi. C5H5N

vii. C6H5NH2 h. Relationship between Ka and Kb

i. Ka x Kb = 1.0 x 10-14

ii. pka + pKb = 14 i. Acid Base Properties of Salts

i. Anions of weak acids make a basic solution ii. Anions of strong acids do not hydrolyze iii. Cations of weak bases make an acidic solution

iv. Cations of strong bases do not hydrolyze

j. Examples: 1) weak Base Solin

Find [OH and pH of a 0.100 M NH 3 soin. o NH3 (aa) + H2O (W NH4+ (aq) toti (aa)

0.100

+x ch -X El 0.100-*

x x xo

K6=[NH4T] [OH-]

[NH3

0.100-*

pH = 11.124

X=1.33x10-3 - PADA [OH] A

[H3O+] = 7.52010-12

CH 118-001 Final Studyguide

17

BLANK

CH 118-001 Final Studyguide

18

VII. Chapter 17 – Aqueous Ionic Equilibria

a. Common Ion Effect

i. 2 substances that share a common ion

ii. Example: b. Buffer – solution that resists a small change in pH upon addition of small amounts

of acid or base

i. Combination of weak acid and its conjugate base ii. Buffer capacity – Amount of acid or base which the buffer can neutralize

before an appreciable pH change iii. Henderson-Hasselbach Equation

1. pH = pka + log

Joo [base]

[acid] 2. Examples:

what is the PH of a buffer mixture composed of 0.10M formic acid and 0.1om formate? Ka=1.8x10-4

PH=pka + log [base]

1 [acia] = -109 (1.8x10-1)+109 (0.1007

0.10M

= -109 (1.8x10-4) : 3.74 A

How much NaOzCH is needed in 0.10M son of formic acid to produce a pH of 3.80?

Ka= 1.8*10-4 PH=phatlog [base] Pka=3.74

1. Tack 3.8- 3.74 +109 [x]

[0.10M) [O2CH]= 0.1m

CH 118-001 Final Studyguide

19

c. Titration

i. Addition of standard solution of known concentration that undergoes a

specific chemical reaction with a solution of unknown concentration ii. Equivalence point – point at which stoichiometric amounts of reactants are

brought together iii. Indicators

1. pH dependent colors

a. AKA phenolphthalein (strong acid-strong base reactions)

iv. Example: Calculate the equivalence point for the titration of 50mL of 0.10M HOC 2 H 3 with 50 ml of 0.10M NaOH. Ka=1.8x10-5

HOZC2H2 + oH H2O + O2 Cz H3 1 5.010% moi 5.0x10-3 mois C1-5.0X103 moi -5.0X103 mol

5.0x10-3 mol © 5.0X10-3 moi

50 mL HO2CzH3

+ 50 m

Nach

5.0810-3 mol Oz Cz H3

0.1001 = 5.0x10-

2M -02C z H3

100 ml = 0.100L

Z

HO2CCH3 (aas + OH (aa)

-OzC2Hz (aa) + H2OLO is 5.0010-2M

+X

+x

El (5.0/10-2-x) M

1

XM

XM

Ka=1.8810-5

= 5.6x10-16

Ko = 1.0x10.04

1.8X105

X2 5.0800 2-X

5. 66x10-10-[HO2GCH 3] [OH-]

1 [ 02 C2H2] X=5.3*10-6=[OH-]

POH=-109 (5.3x10-6) = 5.28

PH=14-5.28

pH=8.72

-

A

CH 118-001 Final Studyguide

e

20

d. Solubility product constant (Ksp)

Represents dissolution of ionic compound e. Molar solubility (mol/L)

i. S= V(Ksp) ii. Example:

Calculate the molar

- PbCl2 (5) P

solubility of PbCl2 in pure water,

Pb 2+ (aq) + 2C1 caa) Ksp=[P62] (c)?

- Uw

Ono

+ 25 25

Study Soa

LOOK up Ksp = 5(25)2 = 452

1.19xio

sa 3/ Kappa - 31.12.= 1.43410-2 MA

KSP

S

Studyse

f. Common Ion Effect

i. Generally, the solubility of an ionic compound is lower in a solution with

a common ion than in pure water

ii. Example: What is the molar solubility of CaF2 in a solin containing

0.100 M NaF ?

Ca Fz (s) = Ca 2+ (aq) + 2F (aa) Ksp=[Ca 2+] [F-72

0.100 s 25

0.100+25

mo

hona

>100K up ksp: S (0.100 +28)2

:5 (0.100)2 s = usp

LE1.46x10-10 0.0100

0.0100

Study Soups

CH 118-001 Final Studyguide

21

21

VIII. Chapter 18 - Free Energy and Thermodynamics

a. First Law of Thermodynamics

i. Heat energy cannot be creater/destroyed

ii. AE = q +w b. Second law of Thermodynamics

i. For any spontaneous process, the entropy of the universe increases C. Entropy (S) - state function

i. Increases with the number of energetically equivalent ways to arrange the

components of a system to achieve a particular state ii. AS = Sfinal – Sinitial iii. Increases solid to liquid to gas iv. S = klnW where K is 1.38 x 10-23J/K

1. S is in J/K d. Third Law of Thermodynamics

i. The entropy of a perfect crystal at absolute 0 (OK) is zero e. Gibbs Free Energy

i. Spontaneity ii. - is spontaneous iii. AG = AH – TAS iv. Negative H and positive S

use 1. Spontaneous at all T v. Positive H and Negative S

1. Nonspontaneous at all T vi. Negative H and S

1. Spontaneous at low T vii. Positive H and S

1. Spontaneous at high T f. Standard State

i. AG = AG° + RTlnQ

ii. AG° = -RTINK g. Examples:

1 Calculate DG @ 250C and determine whether the

reaction is spontaneous.

CCl4 (9) C(s) +2012 (8) DH=95.7kJ

As = 142.2 J/K AG=AH-TAS

-95.7X103 J - (2986) (142.25/6)

553.3 X1033 nonspontaneous A At what I does the rxn become spontaneous?

O=AH-TAS

-95.7X103 J-(T) (142.25/6) T=673KA

Study Soupo

CH 118-001 Final Studyguide

22

2) Calculate Ahiron, us rn, and AGran

SO2(g) + O2(g) 50369) AHE -26.8 0 -395.7 (hmon

so

205.2

248.2

256.8

(Iman)

Dhians

np O Hof (Products) - Enrah of Greactants)

=-395.7 KJ - (-296.8 KJ tOKJ)

---98.914JA Asoon = 256.

85/4-(248.2 J/K + 2 (205.

25/4))

= -94.0 JIKA

A Gorxn= o Hixn-TAS Exn

= (-98.9x1035) - 27876-94.05)) :- 70.9 ing spontaneous A

3) Find DGorn

36(s) + 4 Hz (9) C3 +8 (8)

using: C 3 Hg (92 + 502 (9) 300269) + 4H2O(S) Cast + O2(gr CO2(g) 2 H2 (9) + O2(g) → 2H2O (9)

DGoxn= -2014 uJ

-394.4 kJ

-457, 1kJ

Tag

Then, AGO xn=-2365

*Inverse rani 1. ** mumply rxn 2 by 3 * double ron 3

CH 118-001 Final Studyguide

IX. Chapter 19 – Electrochemistry

a. Redox Reactions

reduction reducing

agent

Ca(s) + 2 H2O ce

Ca(OH)2 +

H2 (9)

Study Soup 6

oxidizing

agent Oxidahon i. Oxidation - loss of electrons

ii. Reduction - gain of electrons b. When balancing redox reactions, you have to balance the mass and the charge

i. Acidic Solution

1. Assign oxidation states 2. Separate into half reactions 3. Balance all except O and H 4. Balance O with H2O 5. Balance H with H+ 6. Make e- equal 7. Add half reactions 8. Example:

Balance

Al(s) + Cu 2+ (aq)

>> 37

A1 3+ (aa) + Cu(s)

37

0

oxidation : Alls) - Al 34 (aa) Reduction: Cu2+ (aq) cu(s)

Balanced by massr

2 Als) + A13"caa) +3e] 3 Ee + Cu 2+ (aa) cucs)]

= 2Al(s) + 6€ + 30421 24134 + 6 + Cu(s) = 2Al(s) + BCu2+ + 2413 + 3Cu(s) A

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ii. Basic Solution

1. Assign oxidation states 2. Separate into half reactions 3. Balance all other than 0 and H 4. Balance O with H2O 5. Balance H with H+ 6. Neutralize H+ with OH- on each side 7. Balance e 8. Make e- equal 9. Add half reactions together

10. Example: Sec example 19.3 Outpc. Voltaic Cells

i. Electrochemical cell that produces electrical current from a spontaneous

chemical reaction ii. Solid strip of zinc is placed in a Zn(NO3)2 solution to form a half cell iii. Solid strip of copper placed in a Cu(NO3)2 solution to form a second half

cell iv. Each strip reaches equilibrium with its solution

1. Zn electrode becomes negatively charged V. Salt bridge connects them so that the Zn/Zn2+ equilibrium shifts right and

oxidation occurs and the Cu/Cu2+ equilibrium shifts left and reduction occurs

1. Anode - oxidation 2. Cathode – reduction 3. Salt bridge is a pathway by which counterions can flow between

half cells without the solutions mixing

a. Contains strong electrolytes (KNO3) b. Allows for flow of ions that neutralizes charge buildup in

50 solution d. Cell Electromotive Force (EMF)

i. Eocell = Eocathode – E°anode

1. Positive for spontaneous rxn 2. Negative for nonspontaneous rxn

3. Example: e. Spontaneity

i. AG° = -nFEocell where F = 96485 C/mole ii. Example:

Find DG for Iz (s) + Zbr-caas 2I (aa) + Braces oxidation : 2Br calls to Brace3+Ze' E:1.09v Reduction: Iz 65)+ ze +2Icaa) EO=0.540

Ecells-0.55V DGE-n feocell

3-2 mote (Tampere(-0.55) -1.1X1055 not spontaneous A

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f. Equilibrium Constant

i. Eocell = 0.0592V,

lod

ii. Example: " Find K for Cu(s) + 2H+ (aq) → Cu Zt caas + H2 (8) Oxigation: Cu(s) + Cuzt caq) + 2C 6°-0.34V Reduction: 2H+ (aa) + ze + H2C9Eo-0.000

Ecell = -0.34V

Eocell =0.05924 look

K=3.3x10-12

g. Cell Potential under Nonstandard Conditions

i. Nernst Equation ii. Ecell = Eocell – 0.0592V logQ

1. When a redox rxn reaches equilibrium, Ecell = 0

iii. Example: . Find Ecen based on these two haifarans: 3 [oxidation: Cucs) Cuztcaa 0.10M) + 26- E%0.340

action : Mnoy-lag 2.0M) + 4Htcaa om +36 Mnoz (S) + 2 H2Oca

EP: 1.68V 3 Cu(s) + 2Mnoy caal tHt can 3Cu2+(99) + ZMNO2Ls) + 4H2OCO

Eocells \- 340

Ecell = 1.34V-0.0592V log

(0.010)3 (2.0)2 (1.0)8

3

=1.41VA

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h. Batteries

i. Dry Cell - don't contain large amounts of liquid water

1. Alkaline batteries – Zn oxidized in basic environment

a. Longer working and shelf life ii. Lead-Acid Storage Batteries (cars)

1. 6 electro-chemical cells wired in series for a total of 12V 2. Each cell contains a porous lead anode (ox) and a lead (IV) oxide

cathode (red) iii. Other batteries

1. Rechargeable

a. Nickel-cadmium

i. Moderate overcharge tolerance b. Nickel-metal hydride Sus

i. More environmentally friendly ii. Greater energy density

iii. Low overcharge density c. Lithium ion battery

i. Newest and most expensive

ii. Used in phones, laptops, etc iv. Fuel Cells

1. Reactants need to be constantly replenished from external source 2. Hydrogen-oxygen fuel cell

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X. Chapter 20 - Radioactivity and Nuclear Chemistry

a. Radioactivity – Emissions produced by unstable nuclei of radioactive atoms that

spontaneously decompose

Stoicuccusams patole

perior des = new ECT.

W

coretan

poden

Gosteiega CS

SE

ARA

SA Oncken we

cognaccom

StudySoup

MESSO Nome

C

atama tay

Sve race W GALERSHI

08

c. Nuclear Stability

i. Everything above atomic number 84 is unstable ii. Even protons and neutrons are generally more stable iii. Magic numbers make them stable (do not have to memorize the magic

numbers) d. Predicting the type of decay

i. Above the belt of stability = beta decay ii. Below the belt of stability – positron emission or electron capture

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iii. Atomic number over 84 = alpha decay e. Radioactive Decay Series

i. Atoms with atomic number over 83 decay in one or more steps until

stability is reached f. Nuclear Transmutation

i. Changing one element into another via decay g. Half-Life and Dating

i. All radioactive nuclei decay via first-order

1. Used in radiometric dating ii. Integrated Rate Law

1. In we = -kt

2. Example: Plutonium-236 is an alpha emitter wit=2.86 yrs. If a Sample initially contains 1.35 mg of Pu-236, what mass Of Pu-236 is present after 5 years ? Sobe

+12=0.693 K=0.693 = 0.693

☆ Ky

= 0.2423lur к

t'iz 2.86yr

NO

in

Nt

-

--kt

Ni No Nt= Noekt = 1.35 mg [e-10.2423(gr) 15yr7

Ne=0.402mg

h. Detection

i. Radiocarbon dating

1. Estimates age of artifacts/fossils

2. Carbon decays to Nitrogen ii. Uranium Lead Dating

1. Most dependable technique 2. Ratio of Uranium 238 to Lead 206

3. Used in rocks i. Nuclear Binding Energy

i. Mass Defect

1. Difference in mass ii. Nuclear Binding Energy

1. Energy corresponding to the mass defect

a. Obtained by substituting the mass defect into E=mc^2

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iii. Example: j. Nuclear fission

i. The splitting of the uranium atom k. Nuclear fusion

i. Combination of two light nuclei to form a heavier one

1. Energy of sun and stars ii. High temps

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