chem 1151: ch. 5 chemical reactions. the chemical equation 2h 2 (g) + o 2 (g) 2h 2 o( l ) ...
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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
Reactant(s) Product(s)
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
Reactant(s) Product(s)
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
Redox e- Transfer: ExamplesRedox e- Transfer: Examples
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
Mission: Determine O.N. for each atom in the following and identify oxidizing and reducing agents.
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
Redox e- Transfer: ExamplesRedox e- Transfer: Examples
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
Mission: Determine O.N. for each atom in the following and identify oxidizing and reducing agents.
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
Seager SL, Slabaugh MR, Chemistry for Today: General, Organic and Biochemistry, 7th Edition, 2011
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
Seager SL, Slabaugh MR, Chemistry for Today: General, Organic and Biochemistry, 7th Edition, 2011
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
Example 03Example 03 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.
BaCl2(aq) + Na2CO3(aq) BaCO3(s) + 2NaCl(aq) Ba +2 +2
Cl -1 -1
Na +1 +1
CO3 -2 -2
Nonredox Double-replacement
Example 04Example 04 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.
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)
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 mol CH4(g) + 2 mol O2(g) 1 mol CO2 (g) + 2 mol H2O(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)
1 mol CH4(g) + 2 mol O2(g) 1 mol CO2 (g) + 2 mol H2O(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)
1 mol CH4(g) + 2 mol O2(g) 1 mol CO2 (g) + 2 mol H2O(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