classifying chemical reactions by what atoms...

Classifying Chemical Reactions by What Atoms Do

Classification of Reactions

4 Al (s) + 3 O2 (g) 2 Al2O3 (s)

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

C2H4 (g) + H2O2 (aq) C2H6O2 (l)

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

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

2 NaCl (s) ---------> Cl2 (g) + 2 Na (l)

Decomposition reaction Cu (s) + 2 AgNO3 (aq) ---------> 2 Ag (s) + Cu(NO3)2 (aq)

2 Al (s) + Fe2O3 (s) ---------> Al2O3 (s) + 2 Fe (l)

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

Single displacement reaction

Ba(NO3)2 (aq) + Na2SO4 (aq) ---------> BaSO4 (s) + 2 NaNO3 (aq)

PCl3 (l) + 3 AgF (s) ---------> PF3 (g) + 3 AgCl (s)

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

Double displacement reaction

Chemical Reactions Classified by

Reaction Type

Precipitation Reactions


Precipitation reactions are reactions in which a solid forms when we mix two solutions.

!1) reactions between aqueous solutions of ionic compounds 2) produce an ionic compound that is insoluble in water 3) The insoluble product is called a precipitate.


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

No Precipitate Formation = No Reaction

KI(aq) + NaCl(aq) ➜ KCl(aq) + NaI(aq)

KI(aq)

NaCl(aq)

KCl(aq) + NaI(aq)

No precipitate forms, therefore, no reaction.

Process for Predicting the Products ofa Precipitation Reaction

1. Determine which ions are present in each aqueous reactant. !2. Determine formulas of possible products. !3. Determine solubility of each potential product in water.

!4. If neither product will precipitate, write no reaction after the

arrow. !5. If any of the possible products are insoluble, write their formulas as

the products of the reaction using (s) after the formula to indicate solid. Write any soluble products with (aq) after the formula to indicate aqueous.

!6. Balance the equation.

Remember to only change coefficients, not subscripts

Practice – Predict the products and balance the equation

K2CO3(aq) + NiCl2(aq) ➜

K2CO3(aq) + NiCl2(aq) ➜ 2 KCl (?) + NiCO3(?)

K2CO3(aq) + NiCl2(aq) ➜ 2 KCl (aq) + NiCO3(s)

K2CO3(aq) + NiCl2(aq) ➜ KCl (?) + NiCO3(?)


KCl(aq) + AgNO3(aq) ➜ KNO3(?) + AgCl(?)

KCl(aq) + AgNO3(aq) ➜ KNO3(aq) + AgCl(s)

KCl(aq) + AgNO3(aq) ➜


Na2S(aq) + CaCl2(aq) ➜

Na2S(aq) + CaCl2(aq) ➜ NaCl(?) + CaS(?)

Na2S(aq) + CaCl2(aq) ➜ 2 NaCl(?) + CaS(?)

Na2S(aq) + CaCl2(aq) ➜ 2 NaCl(aq) + CaS(aq)

No Reaction !!!!!


(NH4)2SO4(aq) + Pb(C2H3O2)2(aq) ➜

(NH4)2SO4(aq) + Pb(C2H3O2)2(aq) ➜ NH4C2H3O2(?) + PbSO4(?)

(NH4)2SO4(aq) + Pb(C2H3O2)2(aq) ➜ 2 NH4C2H3O2(?) + PbSO4(?)

(NH4)2SO4(aq) + Pb(C2H3O2)2(aq) ➜ 2 NH4C2H3O2(aq) + PbSO4(s)

Ionic Equations

Equations that describe the material’s structure when dissolved are called complete ionic equations.

Aqueous strong electrolytes are written as ions.

Insoluble substances, weak electrolytes, and nonelectrolytes are

written as molecules.

Equations that describe the chemicals put into the water and the product molecules are called molecular equations.

!2 KOH(aq) + Mg(NO3)2(aq) ➜ 2 KNO3(aq) + Mg(OH)2(s)

2K+(aq) + 2OH−(aq) + Mg2+(aq) + 2NO3−(aq) ➜ 2K+(aq) + 2NO3−(aq) + Mg(OH)2(s)

Ionic Equations

Ions that are both reactants and products are called spectator ions. !

2 K+(aq) + 2 OH−(aq) + Mg2+(aq) + 2 NO3−(aq) ➜ 2 K+(aq) + 2 NO3−(aq) + Mg(OH)2(s)

An ionic equation in which the spectator ions are removed is called a net ionic equation.

2 OH−(aq) + Mg2+(aq) ➜ Mg(OH)2(s)

Write the ionic and net ionic equation

K2SO4(aq) + 2 AgNO3(aq) ➜ 2 KNO3(aq) + Ag2SO4(s)

2 Ag+(aq) + SO42−(aq) ➜ Ag2SO4(s)

2K+ (aq) + SO42-(aq) + 2Ag+ (aq) + 2NO3-(aq) ➜ 2K+ (aq) + 2NO3-(aq) + Ag2SO4(s)

Na2CO3(aq) + 2 HCl(aq) ➜ 2 NaCl(aq) + CO2(g) + H2O(l)

CO32−(aq) + 2 H+(aq) ➜ CO2(g) + H2O(l)

2Na+ (aq) + CO32-(aq) + 2H+ (aq) + 2Cl-(aq) ➜ 2Na+ (aq) + 2Cl-(aq) + CO2(g) + H2O(l)

Write the ionic and net ionic equation

Acids and Bases

Acids and Bases in Solution

Acids ionize in water to form H+ ions. (More precisely, the H+ from the acid molecule is donated

to a water molecule to form hydronium ion, H3O+)

Bases dissociate in water to form OH- ions. (Bases, such as NH3, that do not contain OH- ions,

produce OH- by pulling H+ off water molecules.) !In the reaction of an acid with a base, the H+ from the acid

combines with the OH- from the base to make water. !The cation from the base combines with the anion from the

acid to make a salt.

acid + base ➜ salt + water

Molecular Models of Selected Acids

Acid-Base ReactionsAlso called neutralization reactions because the acid and base neutralize each other’s properties

2 HNO3(aq) + Ca(OH)2(aq) ➜ Ca(NO3)2(aq) + 2 H2O(l)

Note that the cation from the base combines with the anion from the acid to make the water soluble salt.

H+(aq) + OH-(aq) ➜ H2O(l)

(as long as the salt that forms is soluble in water)

The net ionic equation for an acid-base reaction is

Common AcidsName Formula Uses Strength

Perchloric HClO explosives, catalysts Strong

Nitric HNO explosives, fertilizers, dyes, glues Strong

Sulfuric H explosives, fertilizers, dyes, glue, batteries Strong

Hydrochloric HCl metal cleaning, food prep, ore refining, stomach acid

Strong

Phosphoric H fertilizers, plastics, food preservation Moderate

Chloric HClO explosives Moderate

Acetic HC plastics, food preservation, vinegar

Weak

Hydrofluoric HF metal cleaning, glass etching Weak

Carbonic H soda water, blood buffer Weak

Hypochlorous HClO sanitizer Weak

Boric H eye wash Weak

Name Formula Common Name Uses Strength

Sodium Hydroxide NaOH Lye, Caustic Soda

soap, plastic production, petroleum refining Strong

Potassium Hydroxide KOH Caustic Potash

soap, cotton processing, electroplating Strong

Calcium Hydroxide

Ca(OH) Slaked Lime cement Strong

Sodium Bicarbonate

NaHCO Baking Soda food preparation, antacids Weak

Magnesium Hydroxide

Mg(OH) Milk of Magnesia antacids Weak

Ammonium Hydroxide

NH Ammonia Waterfertilizers, detergents,

explosives Weak

Common Bases

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

HCl(aq) NaOH(aq)

NaCl(aq) +! H2O(l)

Write the molecular, ionic, and net-ionic equation for the acid-base reaction

HNO3(aq) + Ca(OH)2(aq) ➜

2H+ (aq) + 2NO3-(aq) + Ca2+ (aq) + 2OH-(aq) ➜ Ca2+ (aq) + 2NO3-(aq) + 2H2O(l)

HNO3(aq) + Ca(OH)2(aq) ➜ Ca(NO3)2(aq) + H2O(l)

2HNO3(aq) + Ca(OH)2(aq) ➜ Ca(NO3)2(aq) + 2H2O(l)

2H+(aq) + 2OH-(aq) ➜ 2H2O(l)

HCl(aq) + Ba(OH)2(aq) ➜


HCl(aq) + Ba(OH)2(aq) ➜ BaCl2(aq) + H2O(l)

2H+(aq) + 2OH-(aq) ➜ 2H2O(l)

2H+ (aq) + 2Cl-(aq) + Ba2+ (aq) + 2OH-(aq) ➜ Ba2+ (aq) + 2Cl-(aq) + 2H2O(l)

2HCl(aq) + Ba(OH)2(aq) ➜ BaCl2(aq) + 2H2O(l)

H2SO4(aq) + Sr(OH)2(aq) ➜

2H+ (aq) + SO42-(aq) + Sr2+ (aq) + 2OH-(aq) ➜ SrSO4 (s) + 2H2O(l)

2H+(aq) + SO42-(aq) + Sr2+ (aq) + 2OH-(aq) ➜ SrSO4 (s) + 2H2O(l)

H2SO4(aq) + Sr(OH)2(aq) ➜ SrSO4(s) + 2 H2O(l)


TitrationA solution’s concentration is determined by

reacting it with another solution and using stoichiometry – this process is called titration.

!In the titration, the unknown solution is added to

a known amount of another reactant until the reaction is just completed. At this point, called the endpoint, the reactants are in their stoichiometric ratio.

!The unknown solution is added slowly from an

instrument called a burette.

Acid-Base Titrations

The difficulty is determining when there has been just enough titrant added to complete the reaction.

!In acid-base titrations, because both the

reactant and product solutions are colorless, a chemical (indicator) is added that changes color when the solution undergoes large changes in acidity/alkalinity !

At the endpoint of an acid-base titration, the number of moles of H+ equals the number of moles of OH-(equivalence point).

TitrationThe titrant is the base solution in the burette.

As the titrant is added to the flask, the H+ reacts with the OH– to form water. But there is still excess acid present so the color does not change.

At the titration’s endpoint, just enough base has been added to neutralize all the acid. At this point the indicator changes color.

Titration

The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point.

What is the concentration of the unknown HCl solution?


L

mL

L of !NaOH soln

mol of NaOH

L

mol

mol of !HCl

mol

mol

mL of NaOH soln

mL of HCl soln

L of !HCl soln Molarity =

mol of HCl

L of HCl soln

12.54 mL NaOH solution x x !! x = mol HCl in the sample

0.100 mol NaOH!1.000 L NaOH soln

0.001 L NaOH soln!1.000 mL NaOH soln

The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point.

What is the concentration of the unknown HCl solution?


1.00 mol HCl!1.00 mol NaOH 1.25 x 10-3

10.00 mL HCl solution x = L HCl soln0.001 L HCl soln!1.000 mL HCl soln

0.0100

Molarity of HCl solution = = 1.25 x 10-3 mol HCl!0.0100 L HCl soln 0.125 M

What is the concentration of NaOH solution that requires !27.5 mL to titrate 50.0 mL of 0.1015 M H2SO4 ?

L

m

L of !H2SO4 soln

mol of H2SO4

L

mol

mol of !NaOH

mol

mol

mL of H2SO4 soln

mL of NaOH soln

H2SO4 (aq) + 2 NaOH (aq) ➜ Na2SO4 (aq) + 2 H2O (l)

L of !NaOH soln

Molarity =mol of NaOH

L of NaOH soln

27.50 mL NaOH soln x = L NaOH soln

2.00 mol NaOH!1.00 mol H2SO4

50.00 mL H2SO4 soln x x !! x = mol NaOH in the sample

0.1015 mol H2SO4!1.000 L H2SO4 soln

0.001 L H2SO4 soln!1.000 mL H2SO4 soln

0.1015

0.001 L NaOH soln!1.000 mL NaOH soln

0.02750

Molarity of NaOH soln = = 0.1015 mol NaOH!0.02750 L NaOH soln 0.3691 M

What is the concentration of NaOH solution that requires !27.50 mL to titrate 50.00 mL of 0.1015 M H2SO4 ?

H2SO4 (aq) + 2 NaOH (aq) ➜ Na2SO4 (aq) + 2 H2O (l)

Gas-Evolving Reactions

Gas-Evolving Reactions

Some reactions form a gas directly from the ion exchange:

K2S(aq) + H2SO4(aq) ➜ K2SO4(aq) + H2S(g)

Other reactions form a gas by the decomposition of one of the ion exchange products into a gas and water.

K2SO3(aq) + H2SO4(aq) ➜ K2SO4(aq) + H2SO3(aq) !!

H2SO3 ➜ H2O(l) + SO2(g)

NaHCO3(aq) + HCl(aq) ➜ NaCl(aq) + CO2(g) + H2O(l)

NaHCO3(aq)

HCl(aq)

NaCl(aq) + CO2(g) + H2O(l)

Other reactions form a gas by the decomposition of one of the ion exchange products into a gas and water.

NaHCO3(aq) + HCl(aq) ➜ NaCl(aq) + H2CO3(aq) !!

H2CO3 ➜ H2O(l) + CO2(g)

Reactant Reactant Exchange Product

Gas Formed

Example

Metal sulfide Metal hydrogensulfide Acid H H K2

H

Metal carbonate Metal hydrogencarbonate Acid H CO K2

(g)

Metal sulfite Metal hydrogensulfite Acid H SO K2

SO

Ammonium salt Base NH NH KOH (aq) + NHNH

Compounds that UndergoGas-Evolving Reactions

Practice – Predict the products and balance the equations

Na2CO3(aq) + 2 HNO3(aq) ➜

2 HCl(aq) + Na2SO3(aq) ➜

H2SO4(aq) + CaS(aq) ➜

2 NaNO3(aq) + H2O (l) + CO2(g)

2 NaCl (aq) + H2O (l) + SO2 (g)

CaSO4(aq) + H2S(aq)

“2 NaNO3(aq) + H2CO3”

“2 NaCl(aq) + H2SO3”

Redox Reactions

Oxidation/Reduction Basic Definitions

Oxidation and Reduction - Symbolic Representation

Oxidation and Reduction at the Atomic Level

Oxidation/reduction reactions involve transferring electrons from one atom to another.

!Also known as redox reactions !Many involve the reaction of a substance with O2(g). !

4 Fe(s) + 3 O2(g) ➜ 2 Fe2O3(s)

Redox Reactions

Atoms in Elements-------> Ions in Compound

Combustion as Redox

2 H2(g) + O2(g) ➜ 2 H2O(g)

Redox without Combustion

2 Na(s) + Cl2(g) ➜ 2 NaCl(s)

Reactions of Metals with Nonmetals

Consider the following reactions: !

4 Na(s) + O2(g) → 2 Na2O(s) 2 Na(s) + Cl2(g) → 2 NaCl(s)

!The reactions involve a metal reacting with a nonmetal.

In addition, both reactions involve the conversion of free elements into ions. !

! Na2O = 2 Na+ + O2-

!NaCl = Na+ + Cl-

Oxidation and Reduction

To convert a free element into an ion, the atoms must gain or lose electrons (of course, if one atom loses electrons, another must accept them).

!Atoms that lose electrons are being oxidized,

atoms that gain electrons are being reduced.

2 Na(s) + Cl2(g) → 2 Na+Cl–(s) Na → Na+ + 1 e– oxidation Cl2 + 2 e– → 2 Cl– reduction

Electron BookkeepingFor reactions that are not metal + nonmetal, or do not involve

O2, we need a method for determining how the electrons are transferred.

!Chemists assign a number to each element in a reaction called

an oxidation state that allows them to determine the electron flow in the reaction.

!Even though they look like them, oxidation states

are not ion charges!

Oxidation states are imaginary charges assigned based on a set of rules.

!Ion charges are real, measurable charges.

Rules for Assigning Oxidation States (in order of priority)

1. Free elements have an oxidation state = 0. !

In Na (s), Na = 0 ; In Cl2 (g), Cl2 = 0 !

2. Monatomic ions have an oxidation state equal to their charge.

! In NaCl, Na = +1 and Cl = −1 !

3. (a) The sum of the oxidation states of all the atoms in a compound is 0.

! Na = +1 and Cl = −1 in NaCl, (+1) + (−1) = 0


3. (b) The sum of the oxidation states of all the atoms in a polyatomic ion equals the charge on the ion.

! In NO3–, N = +5 and O = −2 [3 x (2-) + 1 x (5+) = -1] !4. (a) Group I metals have an oxidation state of +1 in all their

compounds. !!

(b) Group II metals have an oxidation state of +2 in all their compounds. !


5. In their compounds, nonmetals have oxidation states according to the table below

Assign an oxidation state to each element in the following

Br2 !K+

!LiF !CO2 !SO42−

!Na2O2

Br = 0, (Rule 1)

K = +1, (Rule 2)

Li = +1, (Rule 4a) & F = −1, (Rule 5)

O = −2, (Rule 5) & C = +4, (Rule 3a)

O = −2, (Rule 5) & S = +6, (Rule 3b)

Na = +1, (Rule 4a) & O = −1 , (Rule 3a)

Determine the oxidation states of all the atoms in a propanoate polyatomic anion, C3H5O2–

There are no free elements or free ions in propanoate, so the first rule that applies is Rule 3b

(C3) + (H5) + (O2) = −1 !

Because all the atoms are nonmetals, the next rule we use is Rule 5, following the elements in order: H = +1 O = −2

(C3) + 5(+1) + 2(−2) = −1 (C3) = −2 C = −⅔ *

reduction

Oxidation and Reduction Another Definition

Oxidation occurs when an atom’s oxidation state increases during a reaction.

Reduction occurs when an atom’s oxidation state decreases during a reaction.

CH4 + 2 O2 → CO2 + 2 H2O-4 +4 0 -2

oxidation

Assign oxidation states, determine the element oxidized and reduced, and determine the oxidizing agent and

reducing agent in the following reactions:

Sn4+ + Ca → Sn2+ + Ca2+

2 F2 + S → SF4

Sn4+ is being reduced; Sn4+ is the oxidizing agent. Ca is being oxidized; Ca is the reducing agent.

F is being reduced from F0 to F-;F2 is the oxidizing agent.

S is being oxidized from S0 to S+4;S is the reducing agent.

0

0 0 S 4+

F -

reduction

Fe + MnO4− + 4 H+ → Fe3+ + MnO2 + 2 H2O

Assign oxidation states, determine the element oxidized and reduced, and determine the oxidizing agent and

reducing agent in the following reactions:

0 +3+7 +4

Fe is the reducing agent. MnO4− is the oxidizing agent.

oxidation

classifying chemical reactions by what atoms...

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