analytical chemistry chem 3811 chapters 1 and 3 (review)

ANALYTICAL CHEMISTRY CHEM 3811

CHAPTERS 1 AND 3 (REVIEW)

DR. AUGUSTINE OFORI AGYEMANAssistant professor of chemistryDepartment of natural sciences

Clayton state university

CHAPTERS 1 AND 3

MEASUREMENTS, SIGNIFICANT FIGURESERRORS, STOICHIOMETRY, CONCENTRATIONS

ANALYTICAL CHEMISTRY

- Deals with the separation, identification, quantification, and statistical treatment of the components of matter

Two Areas of Analytical Chemistry

Qualitative Analysis- Deals with the identification of materials in a given sample

(establishes the presence of a given substance)


- Deals with the separation, identification, quantification, and statistical treatment of the components of matter

Quantitative Analysis- Deals with the quantity (amount) of material

(establishes the amount of a substance in a sample)

- Some analytical methods offer both types of information (GC/MS)


Analytical Methods

- Gravimetry (based on weight)

- Titrimetry (based on volume)

- Electrochemical (measurement of potential, current, charge, etc)

- Spectral (the use of electromagnetic radiation)

- Chromatography (separation of materials)

- Chemometrics (statistical treatment of data)


General Steps in Chemical Analysis

- Formulating the question (to be answered through chemical measurements)

- Selecting techniques(find appropriate analytical procedures)

- Sampling(select representative material to be analyzed)

- Sample preparation(convert representative material into a suitable form for analysis)


General Steps in Chemical Analysis

- Analysis(measure the concentration of analyte in several identical portions)

(multiple samples: identically prepared from another source)(replicate samples: splits of sample from the same source)

- Reporting and interpretation(provide a complete report of results)

- Conclusion(draw conclusions that are consistent with data from results)

Measurement

- Is the determination of the dimensions, capacity, quantity, or extent of something

- Is a quantitative observation and consists of two parts: a number and a scale (called a unit)

Examplesmass, volume, temperature, pressure, length, height, time

MEASUREMENT

MEASUREMENT SYSTEMS

Two measurement systems:

English System of Units (commercial measurements): pound, quart, inch, foot, gallon

Metric System of Units (scientific measurements)SI units (Systeme International d’Unites)

liter, meter, gramMore convenient to use

FUNDAMENTAL SI UNITS

Physical Quantity

MassLengthTimeTemperatureAmount of substanceElectric currentLuminous intensity

Name of Unit

KilogramMeterSecondKelvinMoleAmpereCandela

Abbreviation

kgms (sec)KmolAcd

DERIVED SI UNITS

Physical Quantity

Force PressureEnergyPowerFrequency

Name of Unit

NewtonPascalJouleWattHertz

Abbreviation

N (m-kg/s2)Pa (N/m2; kg/(m-s2)J (N-m; m2-kg/s2)W (J/s; m2-kg/s3)Hz (1/s)

METRIC UNITS

Prefix

GigaMegaKiloDeciCentiMilli

MicroNanoPico

Femto

Abbreviation

GMkdcmµnpf

Notation

109

106

103

10-1

10-2

10-3

10-6

10-9

10-12

10-15

UNIT CONVERSIONS

24 hours = 1 day or

Length/Distance

2.54 cm = 1.00 in.12 in. = 1 ft1 yd = 3 ft1 m = 39.4 in.1 km = 0.621 mile1 km = 1000 m

Time

1 min = 60 sec1 hour = 60 min24 hours = 1 day7 days = 1 week

Volume

1 gal = 4 qt1 qt = 0.946 L1 L = .0265 gal1 mL = 0.034 fl. oz.

Mass

1 Ib = 454 g1 Ib = 16 oz1 kg = 2.20 Ib1 oz = 28.3 g

day1

hours24

hours24

day1»

Convert 34.5 mg to g

How many gallons of juice are there in 20 liters of the juice?

gallon0.50.53liter1

gallon0.0265xliters20

g10x3.54org0.0354mg1000

g1xmg34.5 2

UNIT CONVERSIONS

Convert 4.0 gallons to quarts

quarts16quarts15.9559liter0.946

quart1x

gallons0.265

liter1xgallons4.0

SIGNIFICANT FIGURES

Exact Numbers - Values with no uncertainties

- There are no uncertainties when counting objects or people(24 students, 4 chairs, 10 pencils)

- There are no uncertainties in simple fractions(1/4, 1/7, 4/7, 4/5)

Inexact Numbers - Associated with uncertainties

- Measurement has uncertainties (errors) associated with it- It is impossible to make exact measurements

SIGNIFICANT FIGURES

Measurements contain 2 types of information- Magnitude of the measurement- Uncertainty of the measurement

- Only one uncertain or estimated digit should be reported

Significant Figures digits known with certainty + one uncertain digit

RULES FOR SIGNIFICANT FIGURES

1. Nonzero integers are always significant

2. Leading zeros are not significant 0.0045 (2 sig. figs.) 0.00007895 (4 sig. figs.) The zeros simply indicate the position of the decimal point

3. Captive zeros (between nonzero digits) are always significant 1.0025 (5 sig figs.) 12000587 (8 sig figs)


4. Trailing zeros (at the right end of a number) are significant only if the number contains a decimal point 2.3400 (5 sig figs) 23400 (3 sig figs)

5. Exact numbers (not obtained from measurements) are assumed to have infinite number of significant figures


Rounding off Numbers

1. In a series of calculations, carry the extra digits through to the final result before rounding off to the required significant figures

2. If the first digit to be removed is less than 5, the preceding digit remains the same (2.53 rounds to 2.5 and 1.24 rounds to 1.2)


Rounding off Numbers

3. If the first digit to be removed is greater than 5, the preceding digit increases by 1 (2.56 rounds to 2.6 and 1.27 rounds to 1.3)

4. If the digit to be removed is exactly 5- The preceding number is increased by 1 if that results in an even number (2.55 rounds to 2.6 and 1.35000 rounds to 1.4)- The preceding number remains the same if that results in an odd number(2.45 rounds to 2.4 and 1.25000 rounds to 1.2)


Multiplication and Division

- The result contains the same number of significant figures as the measurement with the least number of significant figures

2.0456 x 4.02 = 8.223312 = 8.22

3.20014 ÷ 1.2 = 2.6667833 = 2.7

- The certainty of the calculated quantity is limited by the least certain measurement, which determines the final number of

significant figures


Addition and Subtraction

- The result contains the same number of decimal places as the measurement with the least number of decimal places

- The certainty of the calculated quantity is limited by the least certain measurement, which determines the final number of

significant figures

5.479

0.234

3.2

2.045

4.028

3.52

7.548

= 5.5= 4.03

SCIENTIFIC NOTATION

- Used to express too large or too small numbers (with many zeros)in compact form

- The product of a decimal number between 1 and 10 (the coefficient)and 10 raised to a power (exponential term)

24,000,000,000,000 = 2.4 x 1013

coefficient

Exponential term

Exponent (power)0.000000458 = 4.58 x 10-7

SCIENTIFIC NOTATION

- Provides a convenient way of writing the required number of significant figures

6300000 in 4 significant figures = 6.300 x 106

2400 in 3 significant figures = 2.40 x 103

0.0003 in 2 significant figures = 3.0 x 10-4

SCIENTIFIC NOTATION

- Add exponents when multiplying exponential terms

(5.4 x 104) x (1.23 x 102) = (5.4 x 1.23) x 10 4+2

= 6.6 x 106

- Subtract exponents when dividing exponential terms

(5.4 x 104)/(1.23 x 102) = (5.4/1.23) x 10 4-2

= 4.4 x 102

DENSITY

- The amount of mass in a unit volume of a substance

Ratio of mass to volume =Density =

UnitsSolids: grams per cubic centimeter (g/cm3)

Liquids: grams per milliliter (g/mL)Gases: grams per liter (g/L)

- Density usually changes with change in temperature

- Density of 2.3 g/mL implies 2.3 grams per 1 mL

volume

mass

For a given liquid

- Objects with density less than that of the liquid will float

- Objects with density greater than that of the liquid will sink

- Objects with density equal to that of the liquid will remainstationary (neither float nor sink)

DENSITY

The amount of mass in a unit volume of a substance

TEMPERATURE

- The degree of hotness or coldness of a body or environment

- 3 common temperature scales

Metric system Celsius and Kelvin

English system Fahrenheit

TEMPERATURE

Celsius Scale (oC) - Reference points are the boiling and freezing

points of water (0oC and 100oC) - 100 degree interval

Kelvin Scale (K) - Is the SI unit of temperature (no degree sign)

- The lowest attainable temperature on the Kelvin scale is 0 (-273 oC) referred to as the absolute zero

Fahrenheit Scale (oF)- Water freezes at 32oF and boils at 212oF

- 180 degree interval

or

10o, 40o, 60o are considered as 2 significant figures

100o is considered as 3 significant figures

32F9

5C oo 32C

5

9F oo

273KCo 273CK o or

TEMPERATURE

LOGARITHMS

n = 10a implies log n = a

- The logarithm (base 10) of n is equal to a(written as log on calculators)

log 1000 = 3 since 1000 = 103

log 0.01 = -2 since 0.01 = 10-2

log 436 = 2.6392 is the characteristic0.639 is the mantissa

- The number of digits in the mantissa should be equal tothe number of significant figures in the original number (436)

log 4368 = 3.6403 log 0.4368 = -0.3597

LOGARITHMS

ANTILOGARITHMS

n = 10a implies antilog a = n

- n is the antilogarithm of a(written as antilog or 10x or INV log on calculators)

antilog 3 = 1000 since 103 = 1000antilog -2 = 0.01 since 10-2 = 0.01

ANTILOGARITHMS

antilog 2.639 = 436

- The number of significant figures in the answer should be equalto the number of digits in the mantissa

antilog 6.65 = 4466835.922 = 4.5 x 106

antilog -3.230 = 0.0005888436 = 5.89 x 10-4

ERRORS

- Two classes of experimental errors: systematic and random

Systematic Error- Also called determinate error

- Repeatable in a series of measurements- Can be detected and corrected

Examplesuncalibrated buret, pipet, analytical balance, pH meter

power fluctuations, temperature variations

ERRORS

- Two classes of experimental errors: systematic and random

Randon Error- Also called indeterminate error

- Always present and cannot be corrected

ExamplesTaking readings from an instrument, reading between

markings (interpolation), electrical noise in instruments

Precision - Provides information on how closely individual

measurements agree with one another(measure of reproducibility of a result)

Accuracy - Refers to how closely individual measurements

agree with the true value (correct value)(systematic errors reduce the accuracy of a measurement)

- Precise measurements may NOT be accurate

- Our goal is to be accurate and precise

ERRORS

Absolute Uncertainty- The margin of uncertainty associated with a measurement

- If estimated uncertainty in a buret reading is ± 0.05 mL then absolute uncertainty = ± 0.05 mL

- If estimated uncertainty in an analytical balance is ± 0.0001 g then absolute uncertainty = ± 0.0001g

ERRORS

Relative Uncertainty- Compares absolute uncertainty with its associated measurement

- Dimensionless

tMeasuremenofMagnitude

yUncertaintAbsoluteyUncertaintRelative

Percent Relative Uncertainty = Relative Uncertainty x 100

ERRORS

001.0mL 41.45

mL 0.05yUncertaintRelative

Percent Relative Uncertainty = 0.001 x 100 = 0.1 %

For a buret reading of 41.45 ± 0.05 mL

Absolute uncertainty = ± 0.05 mL

ERRORS

- Whole numbers are placed on the left side of the formula (called coefficients) to balance the equation (subscripts remain unchanged)

- The coefficients in a chemical equation are the smallest set of wholenumbers that balance the equation

C2H5OH(l) + O2(g) 2CO2(g) + H2O(g)

2 C atoms 2 C atoms

Place the coefficient 2 in front of CO2 to balance C atoms

BALANCING CHEMICAL EQUATIONS

C2H5OH(l) + O2(g) 2CO2(g) + 3H2O(g)

(5+1)=6 H atoms 3(1x2)=6 H atoms

Place 3 in front of H2O to balance H atoms




C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(g)

1+(3x2)=7 O atoms (2x2)+3=7 O atoms

Place 3 in front of O2 to balance O atoms




C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(g)

2 C atoms(5+1)=6 H atoms

1+(3x2)=7 O atoms

2 C atoms(3x2)=6 H atoms

(2x2)+3=7 O atoms

Check to make sure equation is balancedWhen the coefficient is 1, it is not written




C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g)


- States of reactants and products- Physical states of reactants and products are represented by

(g): gas(l): liquid(s): solid

(aq): aqueous or water solution

Balance the following chemical equations

Fe(s) + O2(g) → Fe2O3(s)

C12H22O11(s) + O2(g) → CO2(g) + H2O(g)

(NH4)2Cr2O7(s) → Cr2O3(s) + N2(g) + H2O(g)


MOLAR MASS

- Add atomic masses to get the formula mass (in amu) = molar mass (in g/mol)

- That is the mass, in g, of 1 mole of the substance

1 mole = 6.02214179 x 1023 entities (atoms or molecules) Usually rounded to 6.022 x 1023 (Avogadro’s number)

This implies that 6.022 x 1023 amu = 1.00 gAtomic mass (amu) = mass of 1 atom

molar mass (g) = mass of 6.022 x 1023 atoms

MOLAR MASS

Calculate the mass of 2.4 moles of NaNO3

Molar mass NaNO3 = 22.99 + 14.01 + 3(16.00)= 85.00 g /mol NaNO3

= 204 g NaNO3

= 2.0 x 102 g NaNO3

3

333 NaNOmole1

NaNOg85.00xNaNOmole2.4NaNOg

CHEMICAL FORMULA

Consider Na2S2O3:

- Two atoms of sodium, two atoms of sulfur, and three atoms ofoxygen are present in one molecule of Na2S2O3

- Two moles of sodium, two moles of sulfur, and three moles ofoxygen are are present in one mole of Na2S2O3

CHEMICAL FORMULA

How many moles of sodium atoms, sulfur atoms, and oxygenatoms are present in 1.8 moles of a sample of Na2S2O3?

I mole of Na2S2O3 contains 2 moles of Na, 2 moles of S, and 3 moles of O

atomsNamoles3.6OSNamole1

atomsNamoles2xOSNamoles1.8atomsNamoles

322322

atomsSmoles3.6OSNamole1

atomsSmoles2xOSNamoles1.8atomsSmoles

322322

atomsOmoles5.4OSNamole1

atomsOmoles3xOSNamoles1.8atomsOmoles

322322

CHEMICAL CALCULATIONS

Calculate the number of molecules present in 0.075 g of urea,(NH2)2CO

Given mass of urea: - convert to moles of urea using molar mass

- convert to molecules of urea using Avogadro’s number

= 7.5 x 1020 molecules (NH2)2CO

CO)(NHmole1

CO)NH(molecules10x6.022x

CO)(NHg60.07

CO)(NHmole1xCO)(NHg0.075

22

2223

22

2222

CHEMICAL CALCULATIONS

How many grams of carbon are present in a 0.125 g of vitamin C,C6H8O6

Given mass of vitamin C: - convert to moles of vitamin C using molar mass

- convert to moles of C (1 mole C6H8O6 contains 6 moles C)- convert moles carbon to g carbon using atomic mass

= 0.0511 g carbon

Cmol1

Cg12.01x

OHCmol1

Cmol6x

OHCg176.14

OHCmol1xOHCg0.125

686686

686686

CHEMICAL EQUATIONS(STOICHIOMETRIC CALCULATIONS)

Given: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

- 1 molecule of C3H8 reacts with 5 molecules of O2 to produce 3 molecules of CO2 and 4 molecules of H2O

- 1 mole of C3H8 reacts with 5 moles of O2 to produce 3 moles of CO2 and 4 moles of H2O

- make sure the equation is balanced- calculate moles of propane from given mass and molar mass- determine moles of oxygen from mole ratio (stoichiometry)

- calculate mass of oxygen

= 349 g O2


2

2

83

2

83

8383 Omol1

Og32.00x

HCmol1

Omol5x

HCg44.11

HCmol1xHCg96.1


What mass of oxygen will react with 96.1 g of propane?

- make sure the equation is balanced- calculate moles of propane from given mass and molar

mass- determine moles of CO2 from mole ratio (stoichiometry)

- calculate mass of CO2

= 288 g CO2


2

2

83

2

83

8383 COmol1

COg44.01x

HCmol1

COmol3x

HCg44.11

HCmol1xHCg96.1


What mass of CO2 will be produced from 96.1 g of propane?

CONCENTRATION OF SOLUTIONS

- The amount of solute dissolved in a given quantity of solvent or solution

Molarity (M)- The number of moles of solute per liter of solution

Lsolutionofvolume

solutemolesMolarity

- A solution of 1.00 M (read as 1.00 molar) contains 1.00 mole of solute per liter of solution


Calculate the molarity of a solution made by dissolving 2.56 g ofNaCl in enough water to make 2.00 L of solution

- Calculate moles of NaCl using grams and molar mass- Convert volume of solution to liters

- Calculate molarity using moles and liters

NaClmol0.0438NaClg58.44

NaClmol1xNaClg2.56

mol/L)(orM0.0219solutionL2.00

NaClmol0.0438Molarity


After dissolving 1.56 g of NaOH in a certain volume of water, the resulting solution had a concentration of 1.60 M. Calculate the

volume of the resulting NaOH solution

- Convert grams NaOH to moles using molar mass- Calculate volume (L) using moles and molarity

NaOHmol0.0380NaOHg41.00

NaOHmol1xNaOHg1.56

solutionL0.0237NaOHmol1.60

solutionLxNaOHmol0.0380solution Volume

CONCENTRATION OF IONS

Consider

1.00 M NaCl: 1.00 M Na+ and 1.00 M Cl-

1.00 M ZnCl2: 1.00 M Zn2+ and 2.00 M Cl-

1.00 M Na2SO4: 2.00 Na+ and 1.00 M SO42-

CONCENTRATION OF IONS

Calculate the number of moles of Na+ and SO42- ions in 1.50 L

of 0.0150 M Na2SO4 solution

0.0150 M Na2SO4 solution contains:2 x 0.0150 M Na+ ions and 0.0150 M SO4

2- ions

Moles Na+ = 2 x 0.0150 M x 1.50 L = 0.0450 mol Na+

Moles SO42- = 0.0150 M x 1.50 L = 0.0225 mol SO4

2-

PERCENT COMPOSITION

100%xsolutionofmasstotal

soluteofmass(wt%)PercentMass

- May also be represented by %(m/m)

mass of solution = mass of solute + mass of solvent

PERCENT COMPOSITION

A sugar solution is made by dissolving 5.8 g of sugar in82.5 g of water. Calculate the percent by mass concentration

of sugar.

% 6.6100%xg) 82.5 g (5.8

g5.8(wt%)PercentMass

PERCENT COMPOSITION

100%xsolutionofvolumetotal

soluteofvolume(vol%)PercentVolume

- May also be represented by %(v/v) - Due to the way molecules are packed and differences in distances between molecules (bond lengths), the volume

of the resulting solution is almost always less than the sum of the volume of solute and the volume of solvent

PERCENT COMPOSITION

Calculate the volume percent of solute if 345 mL of ethyl alcohol is dissolved in enough water to produce 1257 mL

of solution

% 27.4 100%x mL 1257

mL345(vol%)PercentVolume

PARTS PER MILLION (PPM)

610xsampleofmass

substanceofmassppm

Percent can be defined as parts per hundred

1 ppm ≈ 1 µg/mL or 1 mg/L

PARTS PER MILLION (PPM)

If 0.250 L of aqueous solution with a density of 1.00 g/mL contains 13.7 μg of pesticide, express

the concentration of pesticide in ppm

ppm = µg/mL0.250 L = 250 mL

Density = 1.00 g/mL Implies mass solution = 250 g

ppm0.0548mL250

μg13.7ppm

PARTS PER BILLION (PPB)

1 ppb ≈ 1 ng/mL or 1 µg/L

910xsampleofmass

substanceofmassppb

PARTS PER MILLION (PPB)

If 0.250 L of aqueous solution with a density of 1.00 g/mL contains 13.7 μg of pesticide, express

the concentration of pesticide in ppb

ppm = µg/LVolume of solution = 0.250 L

Density = 1.00 g/mL Implies mass solution = 250 g

ppb5.48L 0.250

μg13.7ppb

DILUTION

- Consider a stock solution of concentration M1 and volume V1

- If water is added to dilute to a new concentration M2 and volume V2

- moles before dilution = moles after dilution

- Implies that M1V1 = M2V2

DILUTION

Calculate the volume of 3.50 M HCl needed to prepare 500.0 mL of 0.100 M HCl

(3.50 M)(V1) = (0.100 M)(500.0 mL)

V1 = 14.3 mL

Mole Fraction (χ)

- Fraction of moles of a component of solution


components all of moles total

component of moles

The sum of mole fractions of all components = 1

Given that the total moles of an aqueous solution of NaCl andother solutes is 1.75 mol. Calculate the mole fraction of NaCl

if the solution contains 4.56 g NaCl.


NaClmole1xNaClg4.56NaClMoles


0446.0 totalmol 1.75

NaCl mol 0.0780fraction mole

MOLALITY (m)

Moles of solute per kg of solvent

Unit: m or molal

solvent kg

solute molesm

MOLALITY (m)

What is the molality of a solution that contains 2.50 g NaCl in 100.0 g water?

- Calculate moles NaCl- Convert g water to kg water

- Divide to get molality


NaClmole1xNaClg2.50NaClmoles

NaCl428.0solventkg0.1000

NaClmol0.0428molality m

CONVERTING CONCENTRATION UNITS

Calculate the molality of a 6.75 %(m/m) solution of ethanol (C2H5OH) in water

Mass water = 100 g solution – 6.75 g ethanol = 93.25 g water

solutiong 100

ethanolg6.75ethanolpercentMass

ethanol57.1kg 0)(93.25/100

mol /46.08)(6.75

solvent kg

ethanol molmolality m


Calculate the mole fraction of a 6.75 %(m/m) solution of ethanol (C2H5OH) in water

Mass water = 100 g solution – 6.75 g ethanol = 93.25 g water

mol0.146g 46.08

mol1xethanolg6.75ethanolmoles

mol5.17g 18.02

mol1xwaterg93.25watermoles

0.0275mol5.17)(0.146

mol0.146

moles total

componentmolfractionmol


Practice Question

Given that the mole fraction of ammonia (NH3) in water is 0.088Calculate the molality of the ammonia solution


- Molarity is temperature dependent (changes with change in temperature)

- Volume increases with increase in temperature hence molarity decreases

On the other hand- Molality

- Mass percent- Mole fraction

are temperature independent

CHEMICAL EQUILIBRIUM

- Occurs when there is product build-up during a chemical reaction- The product molecules interact with one another to

re-produce reactants

Chemical Equilibrium - When the rate of product formation (forward reaction)

is equal to the rate of reactant formation (reverse reaction)

A + B C + D

forward reaction

reverse reaction

- Reactant and product concentrations are usually not equal

- Such reactions are known as reversible reactions

- Forward reaction rate decreases with time as reactants are used up

- Reverse reaction rate increases with time as

products are being formed

- Concentrations are reached when both forward and reverse rates become equal

CHEMICAL EQUILIBRIUM

EQUILIBRIUM CONSTANT

- Describes the extent of reaction in a given system

- For a chemical reaction of the form

aA + bB → cC + dD

- The equilibrium constant (Keq) is given by

ba

dc

eq [B][A]

[D][C]K

- [ ] denotes concentration in moles/liter (M)

- Product concentrations in the numerator

- Reactant concentrations in the denominator

- Concentrations are raised to the powers of the respective coefficients

- Gases (in bars) and substances in solution (in mol/L) are written in Keq expressions


- Pure solids, pure liquids (e.g. water), and solvents are not written since they are constant

- Keq changes with change in temperature

- For exothermic forward reactions (heat released) Keq decreases with increasing temperature

- For endothermic forward reactions (heat absorbed) Keq increases with increasing temperature


Large Keq

- Greater product concentrations than reactant concentrations - Equilibrium position lies to the right

Small Keq

- Smaller product concentrations than reactant concentrations - Equilibrium position lies to the left

Intermediate Keq (near unity) - Both products and reactants are in significant amounts

- Equilibrium position lies neither to the right nor to the left


- Longer arrows can be used to indicate the predominant species

- Longer forward reaction arrow for large Keq

- Longer reverse reaction arrow for small Keq

CO2 + H2O H2CO3


LE CHATELIER’S PRICIPLE

- If a stress (change of conditions) is applied to a system in equilibrium the system will readjust (change the equilibrium position) in the direction that best reduces the stress imposed

on the system

- If more products form as a result of the applied stress the equilibrium is said to have shifted to the right

- If more reactants form as a result of the applied stress the equilibrium is said to have shifted to the left

Concentration Changes

For a reaction mixture at equilibrium

- Addition of reactant(s) shifts the equilibrium position to the right

- Removal of product(s) shifts the equilibrium position to the right

- Addition of product(s) shifts the equilibrium position to the left

- Removal of reactant(s) shifts the equilibrium position to the left


Temperature Changes

Exothermic Reactions

- Heat is a product

- Increase in temperature shifts the equilibrium position to the left

- Decrease in temperature shifts the equilibrium position to the right


Temperature Changes

Endothermic Reactions

- Heat is a reactant

- Increase in temperature shifts the equilibrium position to the right

- Decrease in temperature shifts the equilibrium position to the left


Pressure Changes

- Gases must be involved in the chemical reaction

- The total number of moles of the gaseous state must change

- Equilibrium is shifted in the direction of fewer moles


Pressure Changes

Higher moles of gaseous reactants than products- Increase in pressure shifts the equilibrium position to the right- Decrease in pressure shifts the equilibrium position to the left

Higher moles of gaseous products than reactants- Increase in pressure shifts the equilibrium position to the left

- Decrease in pressure shifts the equilibrium position to the right


Pressure Changes

No change in equilibrium position occurs if

- There is no reactant nor product in the gaseous state

- Number of moles of gaseous reactants equals number of moles of gaseous products

- Pressure is increased by adding a nonreactive (inert) gas


Addition of Catalysts

- Catalysts do not change equilibrium positions

- Catalysts speed up both forward and reverse reactions so have no net effect

- Catalysts allow equilibrium to be established more quickly by lowering the activation energy


analytical chemistry chem 3811 chapters 1 and 3 (review)

Documents

analytical chemistry

chemical measurements

given sample

results measurement

identification of materials

min24 hours

mass1 ib

sourcereplicate samples