chemical equations chapter 10 eugene passer chemistry department bronx community college © john...
TRANSCRIPT
Chemical Equations Chapter 10
Chemical Equations Chapter 10
Eugene PasserChemistry DepartmentBronx Community College
© John Wiley and Sons, Inc
Version 1.1
Chapter Outline8.1 The Chemical Equation
8.2 Writing and Balancing Equations8.3 What Information Does an Equation Tell Us
8.4 Types of Chemical Equations
8.5 Heat in Chemical Reactions
• Chemists use chemical equations to describe reactions they observe in the laboratory or in nature.
• Chemical equations provide us with the means to1.summarize the reaction
2.display the substances that are reacting
3.show the products
4.indicate the amounts of all component substances in a reaction.
The Chemical EquationThe Chemical Equation
• Chemical reactions always involve change.
• Atoms, molecules or ions rearrange to form new substances.
• The substances entering the reaction are called reactants.
• The substances formed in the reaction are called products.
• During reactions, chemical bonds are broken and new bonds are formed.
• A chemical equation uses the chemical symbols and formulas of the reactants and products and other symbolic terms to represent a chemical reaction.
• A chemical equation is a shorthand expression for a chemical change or reaction.
Coefficients (whole numbers) are placed in front of substances to balance the equation and to indicate the number of units (atoms, molecules, moles, or ions) of each substance that are reacting.
Al + Fe2O3 Fe + Al2O3
coefficient2 2
coefficient
Conditions required to carry out the reaction may be placed above or below the arrow.
Al + Fe2O3 Fe + Al2O3
coefficient2 2
coefficient
heat
The physical state of a substance is indicated by symbols such as (l) for liquid.
2Al(s) + Fe2O3(s) 2Fe(l) + Al2O3 (s)
All atoms present in the reactant must also be present in the products.
In a chemical reaction atoms are neither created nor destroyed.
(s) (l) (s)(s)
Symbols UsedSymbols Usedin Chemical Reactionsin Chemical Reactions
Symbols UsedSymbols Usedin Chemical Reactionsin Chemical Reactions
Al + Fe2O3 Fe + Al2O3
reactants productsAl + Fe2O3 Fe + Al2O3
Chemical Equation
iron oxygen bonds break
aluminum oxygen bonds form
(aq)symbol
aqueousmeaning
after formulalocation
placed between substances
+symbol
plusmeaning
location
symbol
heatmeaning
written above location
symbol
gas formationmeaning
after formulalocation
Writing andWriting andBalancing EquationsBalancing Equations
• To balance an equation adjust the number of atoms of each element so that they are the same on each side of the equation.
• Never change a correct formula to balance an equation.
Steps for Steps for Balancing EquationsBalancing Equations
Steps for Steps for Balancing EquationsBalancing Equations
Step 1 Identify the reaction. Write a description or word equation for the reaction.
Mercury (II) oxide decomposes to form mercury and oxygen.
mercury(II) oxide → mercury + oxygen
HgO Hg + O2
– The formulas of the reactants and products must be correct.
– The reactants are written to the left of the arrow and the products to the right of the arrow.
Step 2 Write the unbalanced (skeleton) equation.
The formulas of the reactants and products can never be changed.
HgO → Hg + O2
Step 3a Balance the equation.
– There is one mercury atom on the reactant side and one mercury atom on the product side.
– Mercury is balanced.
Element Reactant Side Product Side Hg 1 1
Step 3a Balance the equation.– Count and compare the number of atoms of each
element on both sides of the equation.
– Determine the elements that require balancing.
Element Reactant Side Product Side O 1 2
Step 3a Balance the equation.
– There are two oxygen atoms on the product side and there is one oxygen atom on the reactant side.
– Oxygen needs to be balanced.
HgO Hg + O2
Step 3b Balance the equation.– Balance each element one at a time, by
placing whole numbers (coefficients) in front of the formulas containing the unbalanced element.
– A coefficient placed before a formula multiplies every atom in the formula by that coefficient.
Element Reactant Side Product Side O 1 2
Oxygen (O) is balanced.
Step 3b Balance the equation.
• Place a 2 in front of HgO to balance O.
There are two oxygen atoms on the reactant side and there are two oxygen atoms on the product side.
HgO Hg + O22
2
Step 3c Balance the equation.
• Check all other elements after each individual element is balanced to see whether, in balancing one element, another element became unbalanced.
Element Reactant Side Product Side Hg 2 1
• Count and compare the number of mercury (Hg) atoms on both sides of the equation.
Step 3c Balance the equation.
Mercury (Hg) is not balanced.
2HgO Hg + O2
• There are two mercury atoms on the reactant side and there is one mercury atom on the product side.
2HgO Hg + O2
Step 3c Balance the equation.
• Place a 2 in front of Hg to balance mercury.
Mercury (Hg) is balanced.
There are two mercury atoms on the reactant side and there are two mercury atoms on the product side.
Element Reactant Side Product Side Hg 2 1
2
2
2HgO 2Hg + O2
Element Reactant Side Product Side Hg 2 2 O 2 2
THE EQUATION IS BALANCED
sulfuric acid + sodium hydroxide → sodium sulfate + water
Balance the Equation
There is one Na on the reactant side and there aretwo Na on the product side.
Reactant Side Product Side
SO4 1 1
Na 1 2
O 1 1
H 3 2
2
H2SO4(aq) + NaOH(aq) → Na2SO4(aq) + H2O(l)2
Place a 2 in front of NaOH to balance Na.
Balance the Equation
2
4
H2SO4(aq) + NaOH(aq) → Na2SO4(aq) + H2O(l)
There are 4 H on the reactant side and two H on the product side.
Reactant Side Product Side
SO4 1 1
Na 2 2
O 2 1
H 4 2
2
Place a 2 in front of H2O to balance H.
2
2
4
THE EQUATION IS BALANCED
butane + oxygen → carbon dioxide + water
Balance the Equation
There are now 26 O on the product side.
Reactant Side Product Side
C 8 8
H 20 20
O 2 26
13
Place a 13 in front of O2 to balance O.
26
THE EQUATION IS BALANCED C4H10 (g) + O2 (g) → CO2(g) + H2O(l)2 8 10
What Information Does What Information Does an Equation Tell Usan Equation Tell Us
The meaning of a formulais context dependent.
The formula H2O can mean:
1. 2H and 1 O atom
2. 1 molecule of water
3. 1 mol of water
4. 6.022 x 1023 molecules of water
5. 18.02 g of water
In an equation formulas can represent units of individual chemical entities or moles.
H2 + Cl2 2HCl→
1 molecule H2 1 molecule Cl2 2 molecules HCl
1 mol H2 1 mol Cl2 2 mol HCl
FormulasNumber of molecules
Number of atomsNumber of molesMolar masses
Types of Chemical Types of Chemical EquationsEquations
CombinationDecomposition
Single-DisplacementDouble-Displacement
Combustion
Combination ReactionsCombination ReactionsCombination ReactionsCombination Reactions
A + B AB
Two reactants combine to form one product.
2Ca(s) + O2(g) 2CaO(s)
Metal + Oxygen → Metal Oxide
4Al(s) + 3O2(g) 2Al2O3(s)
S(s) + O2(g) SO2(g)
Nonmetal + Oxygen → Nonmetal Oxide
N2(g) + O2(g) 2NO(g)
2K(s) + F2(g) 2KF(s)
Metal + Nonmetal → Salt
2Al(s) + 3Cl2(g) 2AlCl3(s)
Na2O(s) + H2O(l) 2NaOH(aq)
Metal Oxide + Water → Metal Hydroxide
CaO(s) + 2H2O(l) 2Ca(OH)2(aq)
SO3(g) + H2O(l) H2SO4(aq)
Nonmetal Oxide + H2O(l) → Oxy-acid
N2O5(g) + H2O(l) 2HNO3(aq)
Decomposition ReactionsDecomposition ReactionsDecomposition ReactionsDecomposition Reactions
AB A + B
A single substance breaks down togive two or more different substances.
Carbonate → CO2(g)
CaCO3(s) CaO(s) + CO2(g)
2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g)
Hydrogen Carbonate → CO2(g)
2Ag2O(s) 4Ag(s) + O2(g)
Metal Oxide → Metal + Oxygen
Metal Oxide → Metal Oxide + Oxygen
2PbO2(s) 2PbO(s) + O2(g)
Miscellaneous Reactions
2KClO3(s) 2KCl(s) + 3O2(g)
2NaNO3(s) 2NaNO2(s) + O2(g)
2H2O2(l) 2H2O(l) + O2(g)
Single Displacement Single Displacement ReactionsReactions
Single Displacement Single Displacement ReactionsReactions
A + BC AC + B
One element reacts with a compound toreplace one the elements of that compound.
Mg(s) + 2HCl(aq) H2(g) + MgCl2(aq)
2Al(s) + 3H2SO4(aq) 3H2(g) + Al2(SO4)3(aq)
salt
Metal + Acid → Hydrogen + Salt
salt
2Na(s) + 2H2O(l) H2(g) + 2NaOH(aq)
Ca(s) + 2H2O(l) H2(g) + Ca(OH)2(aq)
Metal + Water → Hydrogen + Metal Hydroxide
metal hydroxide
metal hydroxide
Metal + Water → Hydrogen + Metal Oxide
metal oxide
3Fe(s) + 4H2O(g) 4H2(g) + Fe3O4(s)
The Activity SeriesThe Activity SeriesThe Activity SeriesThe Activity Series
Metals KCaNaMgAlZnFeNiSnPbH
CuAgHg
An atom of an element in the activity series will displace an atom of an element below it from one of its compounds .
Sodium (Na) will displace an atom below it from one of its compounds.
increasing activity
Mg(s) + PbS(s) MgS(s) + Pb(s)
Metal Higher in Activity Series Displacing Metal Below It
Magnesium is above lead in the activity series.
Metals MgAlZnFeNiSnPb
Ag(s) + CuCl2(s) no reaction
Metal Lower in Activity Cannot Displace Metal Above It
Metals PbH
CuAgHg
Silver is below copper in the activity series.
Double Displacement Double Displacement ReactionsReactions
Double Displacement Double Displacement ReactionsReactions
Cl2(g) + CaBr2(s) CaCl2(aq) + Br2(aq)
Halogen Higher in Activity Series Displaces Halogen Below It
Halogens F2
Cl2
Br2
I2
Chlorine is above bromine in the activity series.
AB + CD AD + CB
Two compounds exchange partners with each other to produce two different compounds.The reaction can be thought of as an exchange of positive and negative groups.
A displaces C and combines with DB displaces D and combines with C
The Following Accompany Double Displacement Reactions
• formation of a precipitate
• release of gas bubbles
• release of heat
• formation of water
Acid Base Neutralization
HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l)
acid + base → salt + water
Formation of an Insoluble Precipitate
AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)
Pb(NO3)2(aq) + 2KI(aq) PbI2(s) + 2KNO3(aq)
Metal Oxide + Acid
CuO(s) + 2HNO3(aq) Cu(NO3)2(aq) + H2O(l)
CaO(s) + 2HCl(aq) CaCl2(s) + H2O(l)
metal oxide + acid → salt + water
Formation of a Gas
H2SO4(aq) + 2NaCN(aq) Na2SO4(aq) + 2HCN(g)
NH4Cl(aq) + NaOH(aq) NaCl(aq) + NH4OH(aq)
NH4OH(aq) NH3(g) + H2O(l)
indirect gas formation
Introduction to Stoichiometry:Introduction to Stoichiometry:The Mole-Ratio MethodThe Mole-Ratio Method
• Stoichiometry: The area of chemistry that deals with the quantitative relationships between reactants and products.
• Mole Ratio: a ratio between the moles of any two substances involved in a chemical reaction.– The coefficients used in mole ratio
expressions are derived from the coefficients used in the balanced equation.
1 mol 2 mol3 molN2 + 3H2 2NH3
2
32 mo
3 mol H
l NH
2
23 m
1 mol
ol H
N
N2 + 3H2 2NH31 mol 2 mol3 mol
• The mole ratio is used to convert the number of moles of one substance to the corresponding number of moles of another substance in a stoichiometry problem.
• The mole ratio is used in the solution of every type of stoichiometry problem.
Identify the starting substance from the data given in the problem statement. Convert the quantity of the starting substance to moles, if it is not already in moles.
1 molemoles = grams
molar mass
Step 1 Determine the number of moles of starting substance.
The number of moles of each substance in the balanced equation is indicated by the coefficient in front of each substance. Use these coefficients to set up the mole ratio.
moles of desired substance in the equationmole ratio =
moles of starting substance in the equation
Step 2 Determine the mole ratio of the desired substance to the starting substance.
moles of desired substance
in the equationmoles of desired substance = moles of starting substance
moles of starting substance
in the equation
Step 2 Determine the mole ratio of the desired substance to the starting substance.
Multiply the number of moles of starting substance (from Step 1) by the mole ratio to obtain the number of moles of desired substance.
Step 3. Calculate the desired substance in the units specified in the problem.
• If the answer is to be in moles, the calculation is complete
• If units other than moles are wanted, multiply the moles of the desired substance (from Step 2) by the appropriate factor to convert moles to the units required.
Step 3. Calculate the desired substance in the units specified in the problem.
molar mass1. To calculate : grams =gr moles x
1 moams
l
236.022 x 10 atoms2. To calculate : atoms = moles
1 moatoms
l
Step 3. Calculate the desired substance in the units specified in the problem.
236.022 x 10 molecules3. To calculate : molecules = mol moles x
1 molecules
Step 3. Calculate the desired substance in the units specified in the problem.
Limiting-Reactant and Limiting-Reactant and Yield CalculationsYield Calculations
• It is called the limiting reactant because the amount of it present is insufficient to react with the amounts of other reactants that are present.
• The limiting reactant limits the amount of product that can be formed.
• The limiting reactant is one of the reactants in a chemical reaction.
Steps Used to Determine Steps Used to Determine the Limiting Reactantthe Limiting Reactant
Steps Used to Determine Steps Used to Determine the Limiting Reactantthe Limiting Reactant
1. Calculate the amount of product (moles or grams, as needed) formed from each reactant.
2. Determine which reactant is limiting. (The reactant that gives the least amount of product is the limiting reactant; the other reactant is in excess.
3. Calculate the amount of the other reactant required to react with the limiting reactant, then subtract this amount from the starting quantity of the reactant. This gives the amount of the that substance that remains unreacted.
Reaction YieldReaction YieldReaction YieldReaction Yield
The quantities of products calculated from equations represent the maximum yield (100%) of product according to the reaction represented by the equation.
Many reactions fail to give a 100% yield of product.
This occurs because of side reactions and the fact that many reactions are reversible.
• The theoretical yield of a reaction is the calculated amount of product that can be obtained from a given amount of reactant.
• The actual yield is the amount of product finally obtained from a given amount of reactant.
• The percent yield of a reaction is the ratio of the actual yield to the theoretical yield multiplied by 100.
actual yield x 100 = percent yield
theoretical yield