Chemical Reactions

Chemical Equations and Chemical Reactions

Chemical Reaction – process that causes a chemical changeReactants – starting materialsProducts –substances formedReagents –chemicals available in the lab

Symbolizing chemical reactions

Reactants products

Na + H2O NaOH + H2

Skeletal equation – not balancedEquations must be balanced – have same number of each kind of atom in reactants and products, since atoms are not created or destroyed in chemical reactions (law of conservation of mass)

2 Na + 2 H2O 2NaOH + H2, balanced

2, 2, 2, (1) are stoichiometric coefficients- coefficients are relative numbers of atoms or moles in reaction:“2 atoms/2 mol Na react with 2 molecules/2 mol H2O to produce.”Include the physical states:S – solid, precipitatel- liquidg- gasaq – in water solutionhigh temperature/heat required

Balanced equation: 2 Na(s) + 2 H2O(l) 2 NaOH(aq) + H2(g)

Balancing Chemical equations

Many equations can be balanced by simple inspection:

H2 + O2 H2O - skeletal

2H2 + O2 2H2O- with coefficients

2H2(g) + O2(g) 2H2O(l) - with physical states

Never change subscripts to balance equation:

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H2 (g) + O2(g) H2O2 - other product

Steps in balancing eqautions:

1. First balance the element(s) occurring in fewest formulas.2. Balance the element in greater number of formulas last

Butane, C4H10(g) burns in oxygen (air) to form carbon dioxide and water vapor [word equation]. Balance the equation.

C4H10 + O2 CO2 + H2O skeletal- balance C first, then H, leave O to last

O in products = 4 x 2 + 5 x 1 = 13 CO or 13/2 O2, so multiply all coefficients by 2.

2 C4H10 (g) + 13O2(g) 8CO2(g) + 10H2O(g)

Example:Ce(OH)4(aq) + H3PO4(aq) Ce3(PO4)4(s) + H2O(l)

3Ce(OH)4(aq) + 4H3(PO4)(aq) Ce3(PO4)4(s) + 12 H2O(l)

Precipitation reactions

Soluble – readily dissolves in solvent (H2O)

Insoluble – slightly/sparingly soluble – do not dissolve significantly, ≤ 0.1 M

Precipitation reaction – insoluble solid product (precipitate) is formed when two solution are mixed

AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)

Ionic/Net Ionic equations – showing ions present

Complete ionic equation – shows all ions

Ag+(aq) + NO3

(aq) + Na+

(aq) + Cl(aq) AgCl(s) + Na+(aq) + NO3

(aq)

Hydrated ions Na+(aq) ≡ Na(H2O)+

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Spectator ions – ions remaining unchanged in reaction: Na+(aq), NO3

(aq)

Delete spectator ions to give net ionic equation:

Ag+(aq) + Cl(aq) AgCl(s) shows actual change.

Writing a net ionic equation

1. Write and balance the complete ionic equation2. Cancel the spectator ions to obtain the net ionic equation

Example:Ba(NO3)2(aq) + 2 NH4IO3(aq) mixed to produce Ba(IO3)2(s) and NH4NO3(aq)

Chemical equation:Ba(NO3)2(aq) + 2NH4IO3(aq) Ba(IO3)2(s) + 2NH4NO3(aq)

Complete ionic equation:

Ba2+(aq) + 2NO3

(aq) 2NH4

+(aq) + 2IO3

(aq) Ba(IO3)2(s) + 2NH4

+(aq) + 2NO3

(aq)

Net ionic equation:

Ba2+ + 2IO3(aq) Ba(IO3)2(s)

Knowing solubility rules allows prediction of precipitation reactions

Solubility rules for common Inorganic Compounds:

A. Soluble compounds

1. Ionic salts of group 1 cations: Li+, Na+, K+, Rb+, Cs+

2. Ammonium salts, NH4+

3. Chloride (Cl), bromide (Br), iodide (I) salts except Ag+, Hg22+, Pb2+

4. Nitrate (NO3), acetate (CH3CO2), chlorate (ClO3

), perchlorate (ClO4) salts

5. Sulfates (SO42-) except with Ca2+, Sr2+, Ba2+, Pb2+, Hg2

2+, Ag+

B. Insoluble salts

1. Carbonates (CO32), chromates (CrO4

2), oxalates (C2O42), phosphates

(PO43); except with group I cations or NH4

+

2. sulfides (S2), except with group I and 2 cations or NH4+

3. Hydroxides (OH-) and oxides (O2), which becomes OH when dissolved in water) except with group 1 and 2 cations

Slightly soluble (borderline) salts include Ag2SO4, PbCl2, Ca(OH)2, Sr(OH)2, and Mg(OH)2

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Preparing precipitates:Hg2I2 is insoluble, formed in net ionic equation

Hg22+

(aq) + 2I(aq) Hg2I2(s)

- so select possible spectator ions that give initially soluble salts: NO3,

K+

Hg2(NO3)2(aq) + 2KI(aq) Hg2I2(s) + 2KNO3(aq)

Predict results of mixing (NH4)2S(aq) and CuSO4(aq); net ionic equation

CuSO4(aq) + (NH4)2S(aq) CuS(s) + (NH4)2SO4(aq)

Cu2+(aq) + SO4

2-(aq) + 2NH4

+(aq) + S2-

(aq) CuS(s) + 2NH4+

(aq) + SO42-

(aq)

Cu2+(aq) + S2-

(aq) CuS(s)

Acid-Base Reactions

-historical recognitionAcid – sour taste, turns litmus dye redBase – bitter taste, soapy feeling, turns litmus blue(poor procedures)

Acids/Bases in aqueous solutions-

An acid molecule reacts with water to produce a hydronium ion

HCl(g) + H2O(l) H3O+(aq) + Cl(aq)

-does HCl (aq) exist?

HCl, HNO3, CH3COOH are typical acids, the acidic H is the atom released as H+ to form H3O+ by adding to H2O molecule

Mineral (inorganic) acids – write acidic H first, H3AsO4

A base is a molecule or ionic compound producing OH in water: NaOH(s) already contains Na+OH and NH3(g) attracts H+ from H2O:

NaOH(s) + H2O Na+(aq) + OH-(aq)

NH3(g) + H2O(l) NH4+ + OH

(aq)

(bases)

Strong and Weak Acids/Bases

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Strong acid - effectively totally ionized/dissociated in solutionWeak acid – only partially ionized in solution

Ionization/dissociation means transfer of H+ from acid molecule to H2O forming hydronium ion, H3O+.

HCl(aq) is strong acid; effectively only H3O+, Cl ion in solution, no HCl(g) molecules existing water solution:

CH3COOH(aq) acetic acid, is weak acid, its solution containing mainly CH3COOH molecules (~ 99%) with small concentrations (~1%) of H3O+ and CH3COO ions

(equilibrium, later discussion)

Most acids are weak; only a few common strong acids: HCl(aq), HBr(aq), (Not HF(aq), HNO3(aq), H2SO4(aq) – 1st H+

only, HClO4(aq), NOT HClO3(aq)

Strong base is completely ionized in water, while weak base is only partially dissociated:

NaOH(s) completely ionizes; NH3 partially dissociates

Common strong bases – hydroxides and oxides of grop 1 and 2 metals: MOH, M = Li+, Na+, Rb+, Cs+; Ca(OH)2, Sr(OH)2, Ba(OH)2 –limited solubility

Oxide ion, O2- reacts with water to produce OH quantitatively

Common weak bases – ammonia – NH3, organic amines- RNH2, R2NH, R3N, R = -CH3 (methyl), -CH2CH3 (ethyl) etc

No NH4OH

Acidic or Basic Character of Compounds

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Most basic metal oxides from basic solutions in water, while many non-metal oxides from acidic solutions:

- SO2, SO3 and various oxides cause acidity in rain water- “Acid rain” – dissolve marble (CaCO3)

Most metallic elements form basic oxides and/or hydroxides, non metallic elements form acidic oxides/hydroxides – acidic oxides react with bases, basic oxides react with acids:

Oxides/hydroxides of some elements (particularly along metalloid diagonal) have both acidic and basic character (react with bases and acids) – called amphoteric behavior.

More easily seen as hydroxide:

Neutralization-

Neutralization is the reaction between an acid and base to produce a salt and water: acid + base salt + water

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Cation of salt comes from base, anion from acid; OH from base reacts with H+ (H3O+) from acid to produce H2O:

Net ionic equation for neutralization of strong acids with strong base

Net ionic equation for reactions of weak acids and bases:

Ionic equations show major species present:

What salt is produced from aqueous barium hydroxide and sulfuric acid?

Formation of Gases-

Some reactions of acids produce gases:

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A hydrogen ion transfer from strong acid to anion of salt producing a gas. The initial hydrogen compound (acid) may not be a gas, but rapidly decomposes in to a gas:

(test for limestone/marble)

Water itself can act as acid with a very strong base:

Redox Reactions Oxidation and Reduction – Redox reactions, oxidation-reduction reactions, are reactions in which oxidation and reduction takes place simultaneously.

Oxidation – loss of electron, increase in oxidation numberReduction – gain of electrons, decrease in oxidation number

Oxidation and reduction can be related to oxidation numbers of elements, the real or arbitrary charge assigned to an atom

For monoatomic ions, ionic charge = oxidation numberFor the free element, oxidation number is zero: He, H2, P4, S8

Mg2+, oxidation number = +2Cl, oxidation number = -1

Rules for Oxidation Numbers of Elements in Compounds

1. Oxidation number for element in the uncombined (free state) is zero2. Sum of Oxidation numbers of all the atoms in a species is equal to total

charge on the species.

H2PH4, 2 x ox. no. H + ox. No P + 4 x ox no O = -1

Na2SO4, 2 x ox. no. Na + ox no S + 4 x ox no O = 0

3. oxidation number for alkali metal, group 1 in compounds, is always +1: Li+, Na+, K+, Rb+, Cs+ (in compounds)

4. Oxidation number of alkaline earth, group 2 metals in compounds is always +2: Be2+, Mg2+, Ca2+, Sr2+, Ba2+

5. Oxidation number of F in compounds is always -1: F

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6. Oxidation number of H is usually +1, except in binary compounds with active metals when it is -1 (hydride ion); NaH, CaH2.

7. Oxidation number for O is usually -2 in compounds, except as peroxide, O2

2 (-1), superoxide, O2 (-1/2), or in binary compounds with F: OF2,

O2F2

8. Halogens, Cl, Br, I; tend to have -1 oxidation number except when combined with O or another more active halogen, F > Cl > Br > I eg. IF5, BrCl3

9. Some metal cations are reliably one charge: Al3+, Zn2+, Cd2+, Sc3+, Ni2+ (usually)

Example:S in NaHSO4:

x + 1 + - 8 = 0, x = +6

Oxidation number of P in H2PO4-:

+2 x -8H2 P O4

+1 -2X+ 2-8 = -1, x = +5

Oxidation number of Cr in K2Cr2O7

+2 2x -14K2 Cr2 O7+1 x -2

2x + 2 – 14 = 02x = 12X = 6

Oxidizing and Reducing Agents

An oxidizing agent in a redox reaction-i. received electrons from species being oxidized, thereby “causing”

oxidation;ii. Is the species being reduced;iii. Contains the element undergoing a decrease in oxidation number

A reducing agent in a redox reaction-i. donates electrons to species being reduced, thereby “causing

reduction”;ii. Is the species being oxidized:iii. Contains the element undergoing an increase in oxidation number

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Example:

Net;

Zn is oxidized, therefore it is the reducing agentCu2+ is reduced, therefore it is the oxidizing agent, or CuSO4

Example:Acidic solution of potassium dichromate reacts with iron(II) chloride solution to form iron(III) and chromium(III) ions. Identify the oxidizing and reducing agents.

(not balanced)Cr atom has been reduced: Cr in +6 oxidation state in K2Cr2O7 (+6 to +3)So K2Cr2O7 is the oxidizing agent.

Fe atom has been oxidized: Fe in +2 oxidation state in FeCl2 (+2 to +3)So FeCl2 is the reducing agent.

Balancing Simple Redox Equations

In a redox reaction, all electrons lost in an oxidation step must be gained in the reduction step; total change of reactants must equal that of products

Consider:

Atoms balance, but charges don’t:

Also,

Half-Reaction Method for Balancing Redox Reactions The oxidation and reduction processes are written as two, separated half-

reactions

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Each half-reaction is balanced separately and then combined, so that electrons lost (oxidation) = electrons gained (reduction)

Any redundancies, inconsistencies are corrected in final step

Procedure

1. Write the unbalanced half-reactions for the oxidation and reduction step2. Balance all atoms, except H and O3. Balance O by adding H2O to the opposite side of the equation4. Balance H by adding H+ (instead of cumbersome H3O+) to the appropriate

side of the equation5. For acidic solutions, can have H+

(aq), H2O(l), not OH(aq), never e

(aq) or O2(aq);

- for basic solutions can have OH(aq), H2O(l), not H+(aq) never e

(aq) or O2(aq);

so must correct for H+ from step (4). – For reactions in basic media add appropriate number of OH to each side of equation so that all H+ ions are converted to H2O molecules; ie H+ + OH = H2O. Cancel H2O molecules that are on both sides of equation.

6. Balance ionic charges by adding adequate numbers of electrons to appropriate side of equation (right side for oxidation, left side for reduction) [Check: electrons lost = increase in oxidation number, electron gained = decrease in oxidation number]

7. Equalize number of electrons lost and gained in the two half-reactions with appropriate factors (ie multiply half-reactions with factors making e lost = egained)

8. Add the two half-reactions and cancel equal amounts of species appearing on both sides of equation; check the coefficients are smallest whole-numbers

Example 1

Use half-reaction method to balance in acid medium:

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Example 2

Use half-reaction method to balance in basic solution:

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1 x (4e) = 2 x (2e)

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