ch 4: chemical reactions renee y. becker valencia community college chm 1045 1

CH 4: Chemical Reactions

Renee Y. Becker

Valencia Community College

CHM 1045

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Solutions

• Solute – solid in liquid or lowest mass quantity of substance

• Solvent- liquid solute is dissolved in or highest mass quantity of substance

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Solution Concentrations

• Concentration: allows us to measure out a specific number of moles of a compound by measuring the mass or volume of a solution.

• Molarity(M) = Moles of Solute

Liters of Solution

moles = M•L L = moles/M

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Example 1: Solution Concentrations

• How many moles of solute are present in 125 mL of 0.20 M NaHCO3?

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• How many grams of solute would you use to prepare 500.00 mL of 1.25 M NaOH?

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Solution Concentrations

• Dilution: the process of reducing a solution’s concentration by adding more solvent.

Moles of solute(constant) = Molarity x Volume

Mi • Vi = Mf • Vf Vf = (Mi • Vi) / Mf

Mf = (Mi • Vi) / Vf

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• What volume of 18.0 M H2SO4 is required to prepare 250.0 mL of 0.500 M H2SO4?

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• What is the final concentration if 75.0 mL of

3.50 M glucose is diluted to a volume of 400.0 mL?

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Solution Stoichiometry

• Titration: a technique for determining the concentration of a solution

– Standard solution: known concentration

– If you have a known volume of standard

solution and use it to titrate a known volume of an unknown concentrated solution you can calculate to find the number of moles in the unknown and therefore find it’s concentration

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Titration

• When doing a titration you add titrant (standard solution) to the analyte (unknown concentration solution) until the endpoint or the equivalence point is reached. This point is when you have equal moles of titrant and analyte, from the volume of the titrant and analyte used and the molarity of the titrant, you can find the molarity of the analyte

– Endpoint- based on an indicator – Indicator- a substance that changes color in a specific pH

range– Equivalence point- not based on an indicator, usually a pH

meter– Use Manalyte• Vanalyte = Mtitrant • Vtitrant

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Example 5: Solution Stoichiometry

• A 25.0 mL sample of vinegar (dilute CH3CO2H) is titrated and found to react with 94.7 mL of a 0.200 M NaOH. What is the molarity of the acetic acid solution?

NaOH(aq) + CH3CO2H(aq) CH3CO2Na(aq) + H2O(l)

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Oxidation–Reduction Reactions

• Assigning Oxidation Numbers: All atoms have an “oxidation number” regardless of whether it carries an ionic charge.

1. An atom in its elemental state has an oxidation number of zero.

Elemental state as indicated by single elements with no charge. Exception: diatomics H2 N2 O2 F2 Cl2 Br2 and I2

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2. An atom in a monatomic ion has an oxidation number identical to its charge.

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3. An atom in a polyatomic ion or in a molecular compound usually has the same oxidation number it would have if it were a monatomic ion.

A. Hydrogen can be either +1 or –1.

B. Oxygen usually has an oxidation number of –2.

In peroxides, oxygen is –1.

C. Halogens usually have an oxidation number of –1.

• When bonded to oxygen, chlorine, bromine, and iodine have positive oxidation numbers.

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4. The sum of the oxidation numbers must be zero

for a neutral compound and must be equal to the

net charge for a polyatomic ion.

A. H2SO4 neutral atom, no net charge

SO42- sulfate polyatomic ion

[SO4]2- [Sx O42-] = -2

X + -8 = -2

X = 6 so sulfur has an oxidation # of +615


B. ClO4– , net charge of -1

[ClO4]-1 [Clx O42-] = -1

X + -8 = -1

X = 7 so the oxidation number of chloride is +7

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Example 6: Oxidation–Reduction Reactions

Assign oxidation numbers to each atom in the following:

A. CdS F. VOCl3

B. AlH3 G. HNO3

C. Na2Cr2O7 H. FeSO4

D. SnCl4 I. Fe2O3

E. MnO4– J. V2O3

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Electrolytes in Solution• Electrolytes: Dissolve in

water to produce ionic

solutions.

• Nonelectrolytes: Do not

form ions when they

dissolve in water.

a) NaCl sol’n conducts electricity, completes circuit (charged particles)

b) C6H12O6 does not 18

Electrolytes in Solution• Dissociation:

The process by which a compound splits up to form ions in the solution.

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Electrolytes in Solution

• Strong Electrolyte: Total dissociation when dissolved in water.

• Weak Electrolyte: Partial dissociation when dissolved in water.

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Types of Reactions

1. Precipitation

2. Acid-base neutralization

3. Oxidation-reduction (redox)

4. Double replacement

5. Single replacement

6. Combination

7. Decomposition

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Types of Chemical Reactions

• Precipitation Reactions: A process in which an insoluble solid precipitate drops out of the solution.

• Most precipitation reactions occur when the anions and cations of two ionic compounds change partners. (double replacement)

Pb(NO3)2(aq) + 2 KI(aq) 2 KNO3(aq) + PbI2(s)

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Solubility Rules & Precipitation

• Allow you to predict whether a reactant or a product is a precipitate.

• Soluble compounds are those which dissolve to more than 0.01 M.

• There are three basic classes of salts:

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1. Salts which are always soluble:

• All alkali metal salts: Cs+, Rb+, K+, Na+, Li+

• All ammonium ion (NH4+) salts

• All salts of the NO3–, ClO3

–, ClO4–, C2H3O2

–,

and HCO3– ions

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2. Salts which are soluble with exceptions:

• Cl–, Br–, I– ion salts except with Ag+, Pb2+, &

Hg22+

• SO42– ion salts except with Ag+, Pb2+, Hg2

2+,

Ca2+, Sr2+, & Ba2+

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3. Salts which are insoluble with exceptions:

• O2– & OH– ion salts except with the alkali metal ions, and Ca2+, Sr2+, & Ba2+ ions

• CO32–, PO4

3–, S2–, CrO42–, & SO3

2– ion salts

except with the alkali metal ions and the ammonium ion

• If not listed the compound is probably insoluble

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Example 7: Solubility Rules & Precipitation

• Predict the solubility of the following in water:

(a) CdCO3

(b) MgO

(c) Na2S

(d) PbSO4

(e) (NH4)3PO4

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Example 8: Solubility Rules & Precipitation

• Write the balanced reaction and predict whether a precipitate will form for:

(a) NiCl2 (aq) + (NH4)2S (aq)

(b) Na2CrO4 (aq) + Pb(NO3)2 (aq)

(c) AgClO4 (aq) + CaBr2 (aq)

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Equations

• Molecular equation – Balanced reaction

2 FeBr3(aq) + 3 Pb(NO3)2(aq) 2 Fe(NO3)3(aq) + 3 PbBr2(s)

• Complete ionic equation – All broken up into ions (only aqueous solutions)

2 Fe3+(aq) + 6 Br-

(aq) + 3 Pb2+(aq) + 6 NO3

-(aq) 2 Fe3+

(aq) + 6 NO3-(aq) + 3 PbBr2(s)

• Net ionic equation – Cancel out spectator ions

3 Pb2+(aq) + 6 Br-

(aq) 3 PbBr2(s) 29

Net Ionic Equations for Precipitation Reactions

• Write net ionic equation for the following reaction:

2 AgNO3(aq) + Na2CrO4(aq) Ag2CrO4(s) + 2 NaNO3(aq)

1. Is it balanced? If not do it! (molecular equation)

2. Separate all aqueous sol’n into ions (complete ionic equation)

3. Cancel out spectator ions on both sides

4. Rewrite (net ionic equation)

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Example 9:

• Write the ME, CIE, and NIE for the following reaction

Na2CrO4 (aq) + Pb(NO3)2 (aq) NaNO3(aq) + PbCrO4(s)

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• Acid–Base Neutralization: A process in which an acid reacts with a base to yield water plus an ionic compound called a salt.

• The driving force of this reaction is the formation of the stable water molecule.

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

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Acid–Base Concepts

• Arrhenius Acid:

A substance which dissociates in water to form hydrogen ions (H+).

• Arrhenius Base:

A substance that dissociates in, or reacts with, water to form hydroxide ions (OH–).

Limitations: Has to be an aqueous solution and doesn’t account for the basicity of substances like NH3.

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• Brønsted Acid: Can donate protons (H+) to another substance.

• Brønsted Base: Can accept protons (H+) from another substance. (NH3)

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Example 10: Conjugate acid-base pairs

For the following reactions label the acid, base, conjugate acid, and conjugate base.

CH3CO2H(aq) + H2O(l) H3O+(aq) + CH3CO2

-(aq)

NH3(aq) + H2O(l) NH4+

(aq) + OH-(aq)

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• Lewis Acid: Electron pair acceptor. Al3+, H+, BF3.

• Lewis Base: Electron pair donor. H2O, NH3, O2–.

• Bond formed is called a coordinate bond or dative bond.

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Example 11

Which of the following is a Bronsted-Lowry base but not an Arrhenius base?

1. NaOH

2. NH3

3. Mg(OH)2

4. KOH

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Acids and Bases

• Strong acid - st. electrolyte, almost completely dissociates in water– HCl, H2SO4, HNO3, HClO4, HI, HBr

• Weak acid - wk. electrolyte, does not dissociate well in water– HF, HCN, CH3CO2H

• Strong base - st. electrolyte, almost completely dissociates in water– Metal hydroxides

• Weak base - does not dissociate well in water

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Other Weak bases – trimethyl ammonia N(CH3)3, C5H5N pyridine, ammonium hydroxide NH4OH, H2O water

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ME, CIE, NIE for Acids/Bases

Strong Acid Strong Base

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

Complete Ionic Equation:

H+ + Cl- + Na+ + OH- H2O(l) + Na+ + Cl-

Net Ionic Equation:

H+ + OH- H2O(l)

or

H3O+ + OH- 2 H2O(l)

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ME, CIE, NIE for Acids/Bases

Weak Acid Strong Base

ME: HF(aq) + NaOH(aq) H2O(l) + NaF(aq)

Complete Ionic Equation:

HF + Na+ + OH- H2O(l) + Na+ + F-

Net Ionic Equation:

HF + OH- H2O(l) + F-

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Example 12: ME, CIE, NIE for Acids/Bases

Write ME, CIE and NIE for the following:

(a) NaOH(aq) + CH3CO2H(aq)

(b) HCl(aq) + NH3(aq)

• NaOH strong base will dissociate well

• CH3CO2H weak acid doesn’t dissociate well

• HCl is a strong acid and therefore a strong electrolyte

• NH3 is a weak base and is a weak electrolyte42


• Double Replacement: These are reactions where

two reactants just exchange parts. (double

displacement)

AX + BY AY + BX

BaCl2(aq) + K2SO4(aq) BaSO4(s) + 2 KCl(aq)

This is also a ppt reaction, if I ask you what type of reaction is

it, what is the best answer??43


• Oxidation–Reduction (Redox) Reaction: A process in which one or more electrons are transferred between reaction partners.

• The driving force of this reaction is the decrease in electrical potential.

Mg(s) + I2(g) MgI2(s)

Oxidation : Mg0 Mg2+ + 2 electrons

Reduction: I20 + 2 electrons I2

1-

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Example 12:

Which of the following is not an acid-base neutralization reaction?

1. HCl(aq) + NaOH(s) NaCl(aq) + H2O(l)

2. 2 HF(aq) + Mg(OH)2(aq) MgF2(aq) + 2 H2O(l)

3. Pb(NO3)2(aq) + 2 KI(aq) PbI2 (s) + 2 KNO3(aq)

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• Redox reactions are those involving the oxidation and reduction of species.

• Oxidation and reduction must occur together. They cannot exist alone.

Fe2+ + Cu0 Fe0 + Cu2+

Reduced: Iron gained 2 electrons Fe2+ + 2 e Fe0

Oxidized: Copper lost 2 electrons Cu0 Cu2+ + 2e

• Remember that electrons are negative so if you gain electrons your oxidation # decreases and if you lose electrons your oxidation # increases

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Fe2+ + Cu0 Fe0 + Cu2+

• Fe2+ gains electrons, is reduced, and we call it an oxidizing agent

– Oxidizing agent is a species that can gain electrons and this facilitates in the oxidation of another species. (electron deficient)

• Cu0 loses electrons, is oxidized, and we call it a reducing agent

– Reducing agent is a species that can lose electrons and this facilitates in the reduction of another species. (electron rich)

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Example 13:

Which is a reduction half reaction?

1. Fe Fe2+ + 2e

2. Fe2+ Fe3+ + 1e

3. Fe Fe3+ + 3e

4. Fe3+ + 1e Fe2+

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Example 14: Oxidation–Reduction Reactions

For each of the following, identify which species is the reducing agent and which is the oxidizing agent.

A) Ca(s) + 2 H+(aq) Ca2+

(aq) + H2(g)

B) 2 Fe2+(aq) + Cl2(aq) 2 Fe3+

(aq) + 2 Cl–(aq)

C) SnO2(s) + 2 C(s) Sn(s) + 2 CO(g)

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Balancing Redox Reactions

• Half-Reaction Method: Allows you to focus on

the transfer of electrons. This is important when

considering batteries and other aspects of

electrochemistry.

• The key to this method is to realize that the overall

reaction can be broken into two parts, or half-

reactions. (oxidation half and reduction half)

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Balance for an acidic solution:

MnO4–

(aq) + Br–(aq) Mn2+

(aq) + Br2(aq)

1. Determine oxidation and reduction half-reactions:

Oxidation half-reaction: Br–(aq) Br20(aq)

Reduction half-reaction: MnO4–(aq) Mn2+(aq)

2. Balance for atoms other than H and O:

Oxidation: 2 Br–(aq) Br2(aq)

Reduction: MnO4–(aq) Mn2+(aq)

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3. Balance for oxygen by adding H2O to the side with less oxygen


Reduction: MnO4–(aq) Mn2+(aq) + 4 H2O(l)

4. Balance for hydrogen by adding H+ to the side with less hydrogens


Reduction: MnO4–(aq) + 8 H+(aq) Mn2+(aq) + 4 H2O(l)

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5. Balance for charge by adding electrons (e–):

Oxidation: 2 Br–(aq) Br2(aq) + 2 e–

Reduction: MnO4–(aq) + 8 H+(aq) + 5 e– Mn2+(aq) + 4 H2O(l)

6. Balance for numbers of electrons by multiplying:

Oxidation: 5[2 Br–(aq) Br2(aq) + 2 e–]

Reduction: 2[MnO4–(aq) + 8 H+(aq) + 5 e– Mn2+(aq) + 4 H2O(l)]

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7. Combine and cancel to form one equation:

Oxidation: 10 Br–(aq) 5 Br2(aq) + 10 e–

Reduction: 2 MnO4–(aq) + 16 H+(aq) + 10 e– 2 Mn2+(aq) + 8 H2O(l)

2 MnO4–(aq) + 10 Br–(aq) + 16 H+(aq) 2 Mn2+(aq) + 5 Br2(aq) + 8 H2O(l)

We will not be balancing in basic solutions!! (until CHM 1046)

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Example 15: Balancing Redox Reactions

Balance the following in an acidic sol’n

NO3–

(aq) + Cu(s) NO(g) + Cu2+ (aq)

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Types of Reactions

• A single-replacement reaction is a a reaction where a more active metal displaces another, less active metal in a compound.

• If a metal precedes another in the activity series, it will undergo a single-replacement reaction:

Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s)

FeSO4(aq) + Cu(s) → NR

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Activity Series

• Metals that are most reactive appear first in the activity series.

• Metals that are least reactive appear last in the activity series.

• The relative activity series is:

Li > K > Ba > Sr > Ca > Na > Mg > Al > Mn > Zn > Fe > Cd > Co > Ni > Sn > Pb > (H) > Cu > Ag > Hg > Au

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Types of Reactions

• A combination reaction is a reaction where two simpler substances are combined into a more complex compound.

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Types of Reactions

• In a decomposition reaction, a single compound is broken down into simpler substances.

• Heat or light is usually required to start a decomposition reaction. Ionic compounds containing oxygen often decompose into a metal and oxygen gas.

2 HgO(s) → 2 Hg(l) + O2(g)

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ch 4: chemical reactions renee y. becker valencia community college chm 1045 1

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