Chem 1151: Ch. 5

Chem 1151: Ch. 5

Chemical Reactions

The Chemical EquationThe Chemical Equation

2H2(g) + O2 (g) 2H2O(l)

Identifies reactant(s) and product(s) Identifies states of matter (g = gas, l = liquid, s = solid, aq = aqueous) Demonstrates law of conservation of matter in balanced equation Law of conservation of matter: Atoms are neither created nor

destroyed in a chemical reaction, but rearranged to form different molecules.

This equation, as written, describes the reaction between individual molecules (2 molecules of H2 react with 1 molecule of O2) as well as molar relationships (2 moles of H2 react with 1 mole of O2).

Reactant(s) Product(s)

Balancing The Chemical EquationBalancing The Chemical Equation

N2(g) + H2 (g) NH3(g) You cannot change the natural molecular formulas, you can only

change the coefficients indicating number of molecules or moles.

N2(g) + H2 (g) 2NH3(g) By multiplying NH3 by coefficient of 2, you now have 2N and 6H. This

balances with N2 on reactant side, but not H2.

N2(g) + 3H2 (g) 2NH3(g) Now check your work.

2 N6 H

2 N6 H


Balancing The Chemical EquationBalancing The Chemical Equation

NO(g) + O2 (g) NO2(g)

Multiply NO2 by coefficient of 2, then multiply NO by 2. Now check your work.

2 N4 O

2 N4 O


nitrogen monoxide

Reactant side has 3O, but product side has 2O.

2NO(g) + O2 (g) 2NO2(g)

Types of ReactionsTypes of Reactions

Seager SL, Slabaugh MR, Chemistry for Today: General, Organic and Biochemistry, 7th Edition, 2011

Redox ReactionsRedox Reactions

OxidationTo combine with oxygenTo lose hydrogenTo lose electronsTo increase in oxidation number

ReductionTo lose oxygenTo combine with hydrogenTo gain electronsTo decrease in oxidation number

RedoxCombination of Reduction and Oxidation

4Fe(s) + 3O2 (g) 2Fe2O3(s)

2Fe2O3(s) + 3C(s) 4Fe(s) + 3CO2(g)

Oxidation Numbers (States): Pos. or neg. numbers assigned to elements in chemical formula. Determined by the following rules:

Rule 1: The O.N. of any uncombined element is zero.Exs. Al(0), O2(0), H2(0), Br2(0), Na(0)

Rule 2: The O.N. of a simple ion is equal to the charge on the ion.Exs. Na+(+1), Mg2+(+2), Br-(-1)

Rule 3: The O.N. of group IA and IIA elements are +1 and +2, respectively.Exs. NaOH (Na = +1), CaCl2 (Ca = +2)

Rule 4: The O.N. of H is +1.Exs. HCl (H = +1), H3PO4 (H = +1)

Rule 5: The O.N. of oxygen is -2 except in peroxides where it is -1.Exs. H2O (O = -2), H2O2 (O = -1)

Rule 6: The algebraic sum of all O.N.s of all atoms in complete compound formula equals zero.

Rule 7: The algebraic sum of all O.N.s of all atoms in a polyatomic ion is equal to the charge on the ion.

Determining Oxidation NumbersDetermining Oxidation Numbers

K2CO3

CO2

2(O.N of K) + (O.N. of C) + 3(O.N. of O) = 0

2(+1) + 3(-2) = -4

2(+1) + (+4) + 3(-2) = 0

(O.N of C) + 2(O.N. of O) = 0

2(-2) = -4

(+4) + 2(-2) = 0

CH2O (O.N of C) + 2(O.N. of H) + (O.N. of O) = 0

2(+1) + (-2) = 0

(0) + 2(+1) + (-2) = 0

NO3- (O.N of N) + 3(O.N. of O) = -1

+ 3(-2) = -6

(+5) + 3(-2) = -1

Redox e- TransferRedox e- Transfer

(O.N. of S) + 2(O.N. of O) = 0

S(s) + O2(g) SO2(g)(O.N of uncombined

element = 0)(O.N of uncombined

element = 0)(+4) + 2(-2) = 0

Sulfur is oxidized (combines with O and loses 4 e-, increases O.N. by 4) Oxygen is reduced (gains 4 e-, decreases O.N. by 4)

Reduction and oxidation processes occur simultaneously

Reducing Agent: Reduces something else and becomes oxidized (loses e-) Oxidizing Agent: Oxidizes something else and becomes reduced (gains e-)

OIL RIGOxidation Is Loss (of e-)Reduction Is Gain (of e-)

Note: We can apply these concepts to covalently-bonded elements, but atoms in these compounds do not actually acquire a net charge

Redox e- Transfer: ExamplesRedox e- Transfer: Examples

4(O.N. of Al) + 6(O.N. of O) = 0

4Al(s) + 3O2(g) 2Al2O3(s)(O.N of uncombined

element = 0)(O.N of uncombined

element = 0) + 6(-2) = -12

Al is oxidized (combines with O and each Al loses 3 e- and increases O.N. by 3)

Al is reducing agent Oxygen is reduced (each of 6 O gains 2 e- and decreases O.N. by 2)

O is oxidizing agent

Mission: Determine O.N. for each atom in the following and identify oxidizing and reducing agents.

4(+3) + 6(-2) = 0


S2O82-(aq) + 2I-(aq) I2(aq) + 2SO4

2-(aq)

2(simple ion charge = -1) = -2

I: O.N. increases from -1 to 0 (loses e-, oxidized). I- Reducing agent S: O.N. decreases from +7 to +6 (gains e-, reduced). S2O8

2- Oxidizing agent


2(O.N. of S) + 8(O.N. of O) = -2 8(-2) = -16

S2O82-(aq)

2SO42-(aq)

2(+7) + 8(-2) = -2

2I-(aq)

I2(aq)

(O.N of uncombined element = 0)

2(O.N. of S) + 8(O.N. of O) = -4 8(-2) = -162(+6) + 8(-2) = -4


2KI(aq) + Cl2(aq) 2KCl(aq) + I2(aq)

2(O.N. of K) + 2(O.N. of I) = 0

I: O.N. increases from -1 to 0 (loses e-, oxidized). KI reducing agent Cl: O.N. decreases from 0 to -1 (gains e-, reduced). Cl2 Oxidizing agent


I2(aq)

2KI(aq) 2KCl(aq)2(O.N. of K) + 2(O.N. of Cl) = 02(+1) = +22(+1) + 2(-1) = 0

2(+1) = +22(+1) + 2(-1) = 0

Cl2(aq)(O.N of uncombined element = 0) (O.N of uncombined element = 0)

Types of ReactionsTypes of Reactions

A B + C Decomposition Reactions

Combination Reactions

Replacement Reactions

A + B C

A + BX B + AXAX + BY BX + AY

Single

Double

Decomposition ReactionsDecomposition Reactions

A B + C

A single substance is broken down to form 2 or more simpler substances. These may or may not be redox reactions.

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

CaCO3(s) CaO(s) + CO2(g)Note conservation of mass in balanced equations


Combination ReactionsCombination Reactions

A + B C

AKA addition or synthesis reactions. Two or more substances react to form a single substance. Reactants may be elements and/or compounds, but product is always a

compound.

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

SO3(g) + H2O(l) H2SO4(aq)

Redox combustion reaction

Nonredox reaction, formation of acid rain


Single Replacement ReactionsSingle Replacement Reactions

A + BX B + AX

Always redox reactions. Used to obtain metals from oxide ores.

3C(s) + 2Fe2O3(s) 4Fe(s) + 3CO2(g)

1(O.N. of Cu) + 1(O.N. of O) = 0 1(-2) = -2

H2 (g)

1(+2) + 1(-2) = 0

Cu(s)

CuO(s)(O.N of uncombined element = 0)

H2O(aq)(O.N of uncombined element = 0)

2(O.N. of H) + 1(O.N. of O) = 02(+1) + 1(-2) = 0

H: O.N. increases from 0 to +1 (loses e-, oxidized). I- Reducing agent Cu: O.N. decreases from +2 to 0 (gains e-, reduced). Cu Oxidizing agent

Double Replacement ReactionsDouble Replacement Reactions

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)AX + BY BX + AY

Never redox reactions. “Partner swapping”.

Example 01Example 01

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

Classify each of the reactions represented by the following equations as redox or nonredox. Further reclassify them as decomposition, combination, single-replacement, or double-replacement reactions.

SO2(g) + H2O(l) H2SO3(aq) Combination (addition)

S +4 +4

O -2 -2 -2

H +1 +1

Nonredox Combination (addition)

Example 02Example 02 Classify each of the reactions represented by the following equations as redox

or nonredox. Further reclassify them as decomposition, combination, single-replacement, or double-replacement reactions.

2K(s) + 2H2O(l) 2KOH(aq) + H2(g)

K 0 +1

O -2 -2

H +1 +1 0

K: O.N. increases from 0 to +1 (loses e-, oxidized). I- Reducing agent H: O.N. decreases from +1 to 0 (gains e-, reduced). Oxidizing agent Single replacement



BaCl2(aq) + Na2CO3(aq) BaCO3(s) + 2NaCl(aq) Ba +2 +2

Cl -1 -1

Na +1 +1

CO3 -2 -2

Nonredox Double-replacement



Ca +2 +2

C +4 +4

O -2 -2 -2

Nonredox Decomposition

CaCO3(s) CaO(s) + CO2(g)

Ionic EquationsIonic Equations Ionic compounds and some polar covalent compounds may dissociate in water

H+, Na+, K+, Mg2+, Ca2+, Fe2+, Fe3+, Ag1+, Pb2+, F-, Cl-, Br-, Br-, OH-, NH4+, SO4

2-, SO3

2-, PO43-, NO2

-, NO3-, CO3

2-

Reactions can be represented by total ionic equation or net ionic equation Net Ionic Equation: The actual chemistry that happens

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)Molecular equation

H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) Na+(aq) + Cl-(aq) + H2O(l)Total ionic equation

Net ionic equation

H+(aq) + OH-(aq) H2O(l)

In the total ionic equation, Na+ and Cl- appear as both reactants and products These are spectator ions

Spectator ions: Do not participate in reaction excluded from net ionic equation

Ionic Equations: Example 01Ionic Equations: Example 01

Na2SO4(aq) + BaCl2(aq) BaSO4(s) + 2NaCl(aq)

Molecular equation

2Na+(aq) + SO42-(aq) + Ba2+(aq) + 2Cl-(aq) BaSO4(s) + 2Na+(aq) + 2Cl-(aq)

Total ionic equation

Net ionic equation

SO42-(aq) + Ba2+(aq) BaSO4(s)

Ionic Equations: Example 02Ionic Equations: Example 02

CaCl2(aq) + Na2CO3(aq) 2NaCl(aq) + CaCO3(s)

Molecular equation

Ca2+(aq) + 2Cl-(aq) + 2Na+(aq) + CO32-(aq) CaCO3(s) + 2Cl-(aq) + 2Na+(aq)

Total ionic equation

Net ionic equation

Ca2+(aq) + CO32-(aq) CaCO3(s)

Energy and ReactionsEnergy and Reactions All chemical reactions have associated energy changes Energy: Ability to do work or produce change Energy may be in one of the following forms:

Sound Light Electricity Motion **Heat

Exothermic: Heat is released to surroundings Combustion Acid/base neutralization

Endothermic: Heat is absorbed from surroundings Water evaporation Chemical cold packs

2H2(g) + O2(g) H2O(g) + energy (heat released)

http://en.wikipedia.org/wiki/File:Close-up_of_mole.jpg

The Mole and Chemical EquationsThe Mole and Chemical Equations The Mole can be used to calculate mass relationships in chemical reactions. Stoichiometry: Study of mass relationships Coefficients apply to moles or molecules, but not mass.

CH4(g) + 2O2(g) CO2 (g) + 2H2O(l) combustion reaction

1 mol CH4(g) + 2 mol O2(g) 1 mol CO2 (g) + 2 mol H2O(l)

This relationship tells us that we need twice as many moles of O2 compared with CH4 to produce 1 mol CO2 and 2 moles of water.

It also enables us to figure out how many grams of each substance we would need based on atomic weight/molecular weight (e.g., g/mol)

MW of CH4 = 16.0 g/molMW of O2 = 32.0 g/molMW of CO2 = 44.0 g/molMW of H2O = 18.0 g/mol

16.0 g CH4(g) + 64.0 g O2(g) 44.0 g CO2 (g) + 36.0 g H2O(l)Mass of CO2 is greater than H2O but only ½

the moles

Applying Stoichiometry to Specific QuestionsApplying Stoichiometry to Specific Questions

How many moles of O2 are required to react with 1.72 mol CH4?

CH4(g) + 2O2(g) CO2 (g) + 2H2O(l)


1.72 mol CH4 = ? mol O2

How many grams of H2O will be produced when 1.09 mol of CH4 reacts with an excess of O2?

1.09 mol CH4 = ? Grams H2O

Applying Stoichiometry to Specific QuestionsApplying Stoichiometry to Specific QuestionsCH4(g) + 2O2(g) CO2 (g) + 2H2O(l)


How many grams of O2 must react with excess CH4 to produce 8.42 g CO2?

8.42 g CO2 = ? Grams O2

Problems 5.38, 5.42, 5.48, 5.50

The Limiting ReactantThe Limiting Reactant

CH4(g) + 2O2(g) CO2 (g) + 2H2O(l)


Reaction will only continue as long as all necessary reactants are present Limiting reactant is the one that gives the least amount of product

In a combustion reaction of 20.0 g methane and 100.0 g oxygen, how much carbon dioxide (mass) would be produced?

Molar ratio CH4(g):CO2 (g) is 1:1 Molar ratio O2(g):CO2 (g) is 2:1

In this reaction, how much O2 would remain unreacted?

g O2 used in reaction

100.0 g – 80.0 g = 20.0 g O2 unused in reaction

Reaction YieldsReaction Yields

Theoretical Yield: How much product you expect based on calculations from molecular equation

Actual Yield: How much product is actually recovered.Usually less than theoreticalCauses:

Experimental/Human error Side reactions

Examples of Reaction YieldsExamples of Reaction YieldsIn an experiment, 17.0 g of product is obtained from a reaction with a calculated theoretical yield of 34.0 g. What is percentage yield?

In the following reaction, 510.0 g of CaCO3 is heated and produces 235.0 g CaO. What is the percentage yield?

CaCO3 (s) CaO (s) + CO2 (g)MW of CaCO3 = 100.1 g/molMW of CaO = 56.1 g/molMW of CO2 = 44.0 g/mol

Chem 1151: Ch. 5

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