Chemical Reactions

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Chemical Reactions

Natural Approach to ChemistryChapter 10

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Learning ObjectivesSections 10.1-3Characterize types of chemical

reactions (redox,acid-base, synthesis, and single and double replacement)

Apply mole concept & conservation of mass to calculate quantities

Distinguish between endothermic and exothermic processes

Sections 10.4Evaluate costs & benefits of

resourcesDiscuss technological effects

& impacts on environ- mental quality

Discuss production and use of natural resources

Create and interpret potential energy diagrams

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Be sure you know these terms:

Parts of a chemical reactionSynthesisDecompositionSingle displacementDouble displacementPolymerizationPrecipitatePolymer, polymerizationExo- and endothermic

Enthalpy of reaction/formation

Energy barrierPhotosynthesisChemical engineeringBiodegradableHazardous substancesSustainable chemistryGreen chemistry

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Chemical Reaction Assignments

• 10.1 322/1-7,31,37,38,52• 10.3 322/8-15,39-41,64,65• 10.4 3.22/16-29,43-47,66-71

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A Chemical Equation

• Represents with symbols and formulas, the identities and relative molecular or molar amounts of the reactants and products in a chemical reaction.

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Word Equation / Formula Equation

Methane + oxygen --> carbon dioxide + water

CH4(g) + 02(g) --> CO2(g) + H20(g)

Reactants Products

The above formula equation is not balanced.

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Chemical Reaction Indications

1. Production of energy as heat and/or light.2. Production of a gas.3. Formation of a precipitate – a solid produced

as a result of a chemical reaction in solution. A precipitate separates from the solution.

4. A color change.

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Chemical Equation Requirements

1. The equation must represent known facts.2. The formulas for the reactants and products must

be written correctly. DO NOT change subscripts.3. The law of conservation of mass must be satisfied.

The number of atoms of each element must be the same on each side of the yield sign. After a formula is written correctly, place coefficients in front of a formula to show conservation of mass.

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Diatomic MoleculesElement Symbol Molecular Formula Physical State at

Room Temp.Hydrogen H H2 Gas

Nitrogen N N2 Gas

Oxygen O O2 Gas

Fluorine F F2 Gas

Chlorine Cl Cl2 Gas

Bromine Br Br2 Liquid

Iodine I I2 Solid

When writing a chemical equation including any of the above elements, they are shown as diatomic molecules as in column 3 above.

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Coefficients

• When placed in front of a correctly written chemical formula, a coefficient multiplies the number of atoms of each element indicated in the formula

• 2O2 means 4 O

• 2H20 means 4 H and 2 O

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Practice

Write word and balanced chemical equations for: solid calcium reacts with solid sulfur to produce solid calcium sulfide. Include symbols for physical states.

Ca(s) + S(s) --> CaS(s)Balanced: 1 Ca on each side 1 S on each side

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Write word and balanced chemical equation for:Hydrogen gas reacts with fluorine gas to produce hydrogen

fluoride gas.H2(g) + F2(g) --> HF(g)H2(g) + F2(g) --> 2HF(g)

Solid aluminum metal reacts with aqueous zinc chloride to produce solid zinc metal and aqueous aluminum chloride.

Al(s) + ZnCl2(aq) --> Zn(s) + AlCl3(aq)Al(s) + 3 ZnCl2(aq) --> Zn(s) + 2 AlCl3(aq)2 Al(s) + 3 ZnCl2(aq) -->3 Zn(s) + 2 AlCl3(aq)

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Translate these chemical equations into sentences:CS2(l) + 3O2(g) --> CO2(g) + 2SO2(g)Liquid carbon disulfide reacts with oxygen gas to

produce carbon dioxide gas and sulfur dioxide gas.NaCl(aq) + AgNO3(aq) --> NaNO3(aq) + AgCl(s)Aqueous solutions of sodium chloride and silver

nitrate react to produce aqueous sodium nitrate and a precipitate of silver chloride.

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Write & Balance:

Hydrazine, N2H4, reacts violently with oxygen to produce gaseous nitrogen and water.

N2H4(l) + O2(g) --> N2(g) + H2O(l)Is it balanced?No. There are 4H on reactant side and 2H on

product side and O is 2/1.N2H4(l) + O2(g) --> N2(g) + 2H2O(l)N: 2/2 H:4/4 O:2/2

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Chemical Equation Indications

1. Coefficients indicate relative amounts of reactants and products (proportions, molecules, moles, grams, ratios).

H2(g) + Cl2(g) --> 2HCl(g)

1 molecule H2 : 1 molecule Cl2 : 2 molecules HCl1 mole H2 : 1 mole Cl2 : 2 moles HCl

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2. Coefficients can be used to determine relative masses of reactants and products

H2(g) + Cl2(g) --> 2HCl(g)1mol H2 x 2.02g H2 = 2.02 g H2

mol 1 mol Cl2 x 70.90 g Cl2 = 70.90 g Cl2 mol2 mol HCl x 36.46 g HCl = 72.92 g HCl mol

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• 3. The reverse reaction for a chemical equation has the same relative amounts of substances as the forward reactions.

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Balancing Chemical Equations

1. Write equation.2. Write correct chemical formulae for each

compound.3. Balance according to the Law of Conservation of

Mass by adjusting coefficients.4. Start with the element appearing in the fewest

substances. Balance free elements last.5. Count atoms to be sure the equation is balanced.

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Practice - Write word, formula, and balanced chemical equations for this reaction.

1. Magnesium and hydrochloric acid react to produce magnesium chloride and hydrogen.

Word Equation: Magnesium + hydrochloric acid --> magnesium chloride + hydrogen

Formula Equation: Mg(s) + HCl(aq) --> MgCl2(s) + H2(g)

Adjust coeffs: Mg(s) + 2HCl(aq) --> MgCl2(s) + H2(g)

Count atoms: Mg: 1/1 H:2/2 Cl:2/2

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Solid sodium combines with chlorine gas to produce solid sodium chloride.

Sodium(s) + chlorine(g) --> sodium chlorideNa(s) + Cl2(g) --> NaCl(s)Balance: Na(s) + Cl2(g) --> 2NaCl(s) 2Na(s) + Cl2(g) --> 2NaCl(s)

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Types of Chemical Reactions1. Synthesis A + X --> AX2. Decomposition AX --> A + X3. Single-displacement A + BX --> AX + B4. Double-displacement AX + BY --> AY + BX5. Combustion – a substance combines with

oxygen releasing energy as light and heat.6. Acid + Base --> Salt + Water7. Reduction/oxidation (Redox) – covered in a

later chapter

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Activity Series – elements organized according to how they react

An element can replace any element placed below it BUT It cannot replace any element above it.

Zn can replace Cu but Au cannot replace Mg

Most active metals:Li react w/cold Co Do not react w/H20.Rb H20 & acids Ni React w/acids, repla-K replacing H2. Sn cing H2. React w/O2

Ba React w/O2 Pb forming oxides.Sr forming oxides H2 React w/O2, formingCa Sb oxidesNa BiMg react w/steam CuAl (not cold H20) HgMn and acids, repla-Ag Fairly unreactive,Zn cing H2. React Pt forming oxides onlyCr with O2 forming Au indirectly.Fe oxidesCd

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Nonmetal activity series:

Most activeFClBrI

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Sample Problems – activity seriesZn(s) + H2O(l) ---> No reaction, water must be 100oC (steam) at least.Sn(s) + O2(g) -->Yes, any metal more active than Ag will react w/O2 to form an

oxide. (Sn is above Ag)2Sn(s) + O2(g) --> 2SnOCd(s) + Pb(NO3)2(aq) --> Yes, Cd is above Pb. Products: Cd(NO3)2 + PbCu(s) + HCl(aq) -->No, Cu is below H

50oC

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1. Synthesis Reaction

A + X --> AXSamples:Fe(s) + S(s) --> FeS(s)

2Mg(s) + O2(g) --> 2MgO(s)

H20 + SO3 --> H2SO4

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Synthesis with Oxides (see handout for more information!!)

CaO(s) + H2O(l) --> Ca(OH)2(s)Pollution: SO2(g) + H2O(l) --> H2SO3(aq)2H2SO3(aq) + H20(l) --> 2H2SO4(aq)Oxides:CaO(s) +SO2(g) --> CaSO3(s)

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2. Decomposition Reaction

AX --> A + X

H20(l) --------> 2H2(g) + O2(g) (electrolysis)

2HgO(s) ---> 2Hg(l) + O2(g)

electricity

D

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Decomposition of Metal OxideCaCO3 --->CaO + CO2

Decomp of Metal HydroxideCa(OH)2 ---> CaO + H2ODecomp of Metal Chlorate2KClO3 ----> 2KCl + 3O2

Decomp of AcidsH2CO3 --> CO2 + H2O (occurs at room temp)H2SO4 ---> SO3 + H2O

D

D

D

D

MnO2

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3. Single Displacement Reaction

A + BX --> AX + BOr Y + BX --> BY + X1. Fe + CuSO4 --> FeSO4 + Cu2. Cu + 2AgN03 --> Cu(NO3)2 + 2Ag3. CI2 + 2KI --> 2KCl + I2

How is 3. different from 1. or 2.?In 1.& 2. metals are being displaced.In 3. a halogen is being displaced.

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Hydrogen displaced by a metal:Mg + 2HCl --> H2 + MgCl2

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4. Double-Displacement Reaction

AX + BY --> AY + BXA,X,B, and Y in reactants are ions.AY and BX are ionic or molecular compounds.1. Formation of a Precipitate2KI(aq) + Pb(NO3)2(aq) --> PbI2(s) + 2KNO3(aq)

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Formation of a GasFeS(s) + 2HCl(aq) –> FeCl2(aq) + H2S(g)

Formation of WaterHCl(aq) + NaOH(aq) --> NaCl(aq) + H20(l)

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Combustion Reactions

2H2(g) + O2(g) --> 2H2O(g)

C3H8(g) + 5O2(g) --> 3CO2(g) + 4H20(g)

Other products are heat and light.

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Predicting Activity (use the activity series handout)

Zn(s) + H20(l) --> ?No, not hot enough. Steam needed (100oC)Ca(s) + H2O(l) --> ?Yes, Ca is above H on the chart.The products are: Ca(OH)2 + H2 (g)Pt(s) + O2(g) --> ?No.

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Cd(s) + 2HBr(aq) -->Yes, Cd is above H. Products: CdBr + H2(g) Mg(s) + steam -->Yes, Products: Mg(OH)2 + H2(g)

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Activity SeriesActivity of metals: Activity of halogen nonmetals

Li <--Most active metal F2 <-- Most active nonmetalRb react w/cold Cl2

K H20 & acids Br2

Ba replacing H2. I2

Sr React w/O2

Ca forming oxides NaMg react w/steamAl (not cold H20)Mn and acids, repla-Zn cing H2. ReactCr with O2 formingFe oxidesCdCo Do not react w/H20.Ni React w/acids, repla-Sn cing H2. React w/O2

Pb forming oxides.

H2

SbBi React w/O2, formingCu oxidesHg

Ag Fairly unreactive,Pt forming oxides onlyAu indirectly.

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Solubility Chart

There are three key components to a chemical reaction:ReactantsProductsEnergy (in or out)


Chapter 4.2 Chemical Reactions



Photosynthesis is an endothermic reaction: energy is absorbed



Photosynthesis is an endothermic reaction: energy is absorbed

Cellular respiration is an exothermic reaction: energy is released



Chapter 9.3 Properties of Solutions




Energy is absorbed from the surroundings so the

pack feels cold




Energy is released into the surroundings so the

pack feels hot




Change in enthalpy

enthalpy: the amount of energy that is released or absorbed during a chemical reaction

Reaction

Exothermic

Endothermic

Energy

is released

is absorbed

Enthalpy change (∆H, J/mole)

is a negative number∆H < 0

is a positive number∆H > 0

Chemical equation for the combustion of carbon:

Reactants Products Energy

thermochemical equation: the equation that gives the chemical reaction and the energy information of the reaction.

C(s) + O2(g) CO2(g) ∆H = –393.5 kJ


The reverse chemical reaction involves the same amount of energy, but the energy flow is reversed (“in” instead of “out”):

Enthalpy calculations

C(s) + O2(g) CO2(g) ∆H = –393.5 kJ

CO2(g) C(s) + O2(g) ∆H = +393.5 kJ


1 mole 1 mole 1 mole

2 moles 2 moles 2 moles

The combustion of twice as much carbon releases twice as much energy:


C(s) + O2(g) CO2(g) ∆H = –393.5 kJ

2C(s) + 2O2(g) 2CO2(g) ∆H = –787.0 kJ

Chemical equation for the formation of rust:

2 moles 3/2 moles 1 mole


2Fe(s) + 3/2O2(g) Fe2O3(s) ∆H = –824.2 kJ


2 moles 3/2 moles 1 mole


2Fe(s) + 3/2O2(g) Fe2O3(s) ∆H = –824.2 kJ

Rewrite the chemical equation using coefficients with the smallest whole numbers possible



4Fe(s) + 3O2(g) 2Fe2O3(s) ∆H = ?

x 22 moles 3/2 moles 1 mole

2Fe(s) + 3/2O2(g) Fe2O3(s) ∆H = –824.2 kJ


What is the enthalpy change for this reaction?



4Fe(s) + 3O2(g) 2Fe2O3(s) ∆H = –1,648.4 kJ

x 22 moles 3/2 moles 1 mole

2Fe(s) + 3/2O2(g) Fe2O3(s) ∆H = –824.2 kJ


x 2

Enthalpy of formation

This is also the chemical equation for the formation of CO2.

∆Hreaction = ∆Hformation of CO2 = –393.5 kJ

Chemical equation for the combustion of carbon:C(s) + O2(g) CO2(g) ∆H = –393.5 kJ

The formation of 1 mole of CO2 releases 393.5 kJ of energy

∆Hf (CO2) = –393.5 kJ/mole

Enthalpies of formation of some common substances

Knowing these values and the following equation, you can calculate unknown enthalpy values:

reaction f fH H products H reactantsD D D

Enthalpy of formation

Enthalpy calculationsThe complete combustion of glucose (C6H12O6) releases 2,808 kJ per mole of glucose. Calculate the enthalpy of formation of glucose.

C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(g) ∆H = –2,808 kJAsked: ∆Hf(glucose) = ? Given:

2

2

2

, 0

, 393.5

, 241.8

f

f

f

kJH O gmole

kJH CO gmolekJH H O gmole

D

D

D

From the table of enthalpies of formation

reaction f fH H products H reactantsD D DRelationships:

10.4 Chemical Reactions and Energy

Enthalpy calculationsThe complete combustion of glucose (C6H12O6) releases 2,808 kJ per mole of glucose. Calculate the enthalpy of formation of glucose.

reactants products

reaction f fprodH H H reactantsuctsD D DRelationships:

Formation of glucose

C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(g) ∆H = –2,808 kJ

6CO2(g) + 6H2O(g) C6H12O6(s) + 6O2(g) ∆H = +2,808 kJ

2 22

6 0 6 393.5 6 285.5 2,808

0 2,361 1,713 2,80

6 6

8

2,8

2,808

086reaction

f

f

f f

f f f

f

f

H glucose kJ kJ kJ kJ

H glucose k

H gl

H H

J kJ kJ kJ

H

products H reactants kJ

Hucos k

g

CO H JOH He O

l

D

D

D

D

D D

D

D

D

D

4,074 2,808

2,808 4,074

1,266f

f

ucose kJ kJ

H glucose kJ kJ

H glucose kJ

D

D

Remember to multiply by the coefficients!

reactants products




reactants products




2 2 2

2,808

6 6 6 2,808

6 0 6 393.5 6 241.8 2,808

0 2,361 1,451 2,808

reaction f f

f f f f

f

f

f

H H products H reactants kJ

H glucose H O H CO H H O kJ



H gl

D D D

D D D D D

D D

3,812 2,808

2

1,

,80

004

8 3,812

f

f

ucose kJ kJ

H glucose k

H glucose

k

J

J

k

J

D

D



Answer: ∆Hf(glucose) = –1,004 kJ/mole

Asked: ∆Hf(glucose) = ?

1 mole

The complete combustion of glucose (C6H12O6) releases 2,808 kJ per mole of glucose. Calculate the enthalpy of formation of glucose.


A B+ A B ∆H = X

A B+ A B ∆H = X

A B A B+ ∆H = –X

The reverse reaction changes the sign of ∆H

A B+ A B ∆H = X

+ ∆H = 3X

x 3

AA

A

BB

BA B

A BA B

If three times more substances are involved, ∆H is three times greater

A B+ A B ∆H = X

∆H(reaction) = ∆Hf (products) – ∆Hf (reactants)

A BA B∆H(reaction) = ∆Hf – ∆Hf + ∆Hf

Reactants Products ∆H = … kJ

Energy profile

Thermochemical equation:

Energy flow during the reaction

have stored energy

also have stored energy

Ener

gy

Progress of reaction

We can graph the change in energy as the reaction takes

place

Energy profile

Reactants Products ∆H = … kJThermochemical equation:

Energy flow during the reaction

have stored energy

also have stored energy

Activation energy

The reaction cannot start without this initial input of energy

Energy profile

C(s) + O2(g) CO2(g)

Combustion of carbon:

Wood does not spontaneously light itself up on fire

Energy profile

Reaction of sodium in water:

Na(s) + H2O(l) 2NaOH + H2(g)

Sodium reacts with water immediately (and violently) upon contact

Energy profile

Hess’s law: the overall enthalpy of a reaction (1) is the sum of the reaction enthalpies of the various steps into which a reaction can be divided (2).

Hess’s law

∆H1

∆H2

∆H3

R A

BA

B P

∆H4R P

Hess’s law: ∆H4 = ∆H1 + ∆H2 + ∆H3

Hess’s law

∆H1

∆H2

∆H3

R A

BA

B P

∆H4

1 2

R P

Hess’s law

Asked: Cgr(s) Cd(s) ∆H = ?

Given that the enthalpy of combustion for graphite (Cgr) and diamond (Cd) are –393.5 kJ/mole and –395.4 kJ/mole, respectively, calculate the enthalpy of formation of diamond from graphite.

Given: Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole Cd(s) + O2(g) CO2(g) ∆H = –395.4 kJ/mole

Relationships: Hess’s law

Strategy: Create a path that leads from Cgr to Cd.

Hess’s law


Given: Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole

Cd(s) + O2(g) CO2(g) ∆H = –395.4 kJ/mole

Cd(s) is a product in: Cgr(s) Cd(s)Cd(s) is a reactant in:

CO2(g) Cd(s) + O2(g) ∆H = +395.4 kJ/mole

Write the reverse reaction so that Cd(s) is a product, and adjust DH:



Hess’s law




Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/moleWrite the sum of the two equations:

Hess’s law





Hess’s law





Hess’s law





Cgr(s) Cd(s) ∆H = (–393.5 + 395.4) kJ/mole ∆H = +1.9 kJ/mole

Energy profile of a reaction Hess’s law

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MAKE SURE THAT YOU KNOW THE TERMS FROM SLIDE #3!!

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Chemical Reactions