2. Evidence for a Chemical Reaction
2) Temperature change (increase or decrea
se) to the surroundings.
3. Evidence for a Chemical Reaction
3) Formation of a gas (bubbling or an odor)
other than boiling.
4. Evidence for a Chemical Reaction
4) Color change (due to the formation of a n
ew substance).
5. Evidence for a Chemical Reaction
5) Formation of a precipitate (a new solid for
ms) from the reaction of two aqueous solut
ions.
6. Word Equations
• Statements that indicate the reactants and
products in a chemical reaction.
• Ex. Iron (s) + chlorine (g) iron (III) chloride (s)
• This is read as:
“Solid iron and chlorine gas react (combine) to produce soli
d iron (III) chloride”
7. Translating Word Equations to Skel
eton Equations
• A skeleton equation uses chemical formulas rath
er than words to identify the reactants and produ
cts of a chemical reaction.
• The word equation
Iron (s) + chlorine (g) iron (III) chloride (s)
• The skeleton equation
Fe(s) + Cl2(g) FeCl3 (s)
A skeleton equation is not yet “balanced” by coefficients!
8. One more example…
• 6 Na (s) + Fe2O3 (s) 3 Na2O (s) + 2 Fe (s)
– The numbers preceding the chemical formulae are c
oefficients. They are used to balance the reaction.
– The numbers within the chemical formulae are subs
cripts.
– You can read the above balanced reaction as:
• “6 atoms of solid sodium plus 1 formula unit of solid iro
n (III) oxide yields 3 formula units of solid sodium oxide
and 2 atoms of solid iron” or…
• “6 moles of solid sodium plus 1 mole of solid iron (III) o
xide yields 3 moles of solid sodium oxide plus 2 moles
of solid iron”
• Chemical reactions can never be read in terms of gram
s, only in terms of particles or groups of particles (mole
s).
9. Conservation of Mass
During a chemical reaction, atoms are neither
created nor destroyed (Conservation of Mas
s).
Hydrogen and oxygen gas react to form water
:
H2 (g) + O2 (g) H2O (l)
10. Conservation of Mass
H2 (g) + O2 (g) H2O (l)
What is wrong with this equation above? Doesn’t i
t appear that one oxygen atom “went missing”?
According to conservation of mass, the proper way
to write this reaction is:
2H2 (g) + 1O2 (g) 2H2O (l)
The red coefficients represent the # of molecules (
or the # of moles) of each reactant or product.
11. Not All Properties are Conserved
During Chemical Reactions!
CONSERVED NOT CONSERVED
Mass
Types of atoms
Number of each atom
Color
Physical state (solid, liquid,
gas)
Volume
Number of moles of reacta
nts/products
13. There are 5 basic types….
• Single Replacement (Displacement)
(Redox)
• Double Replacement (Displacement)
(Metathesis)
• Synthesis (Combination)
• Decomposition
• Combustion
14. A single uncombined
element replaces an
other element in an i
onic compound.
There are two reacta
nts and two products
.
1) SINGLE REPLACEMENT REA
CTION
Ex: Zn + CuSO4 ZnSO4 + Cu
15. Single Replacement Reactions
Single replacement reactions have the gene
ral form, A + BC AC + B.
Question: Do all single replacement reaction
s actually occur?
Answer: Not necessarily…
16. Single Replacement Reactions
Examine the reaction:
Zn + CuSO4 ZnSO4 + Cu
This reaction does occur!’
Now let’s try:
Cu + ZnSO4 No Reaction
Conclusion: Zn will replace Cu in solutio
n, but not vice versa!
17. Single Replacement Reactions
How do we know which reactions will occur
and which ones will not?
We look at the “activity series”.
Elements with higher activities replace elem
ents with lower activities during a single-r
eplacement reaction, but not vice-versa.
20. Predicting the Products of Single R
eplacement Reactions
1) Write the reactants.
2) Identify the cation and anion of the reactant
that is a compound.
3) Use the activity series to see if the single ele
ment will replace one of the elements in the com
pound. If no reaction will occur, just write “N
R” for the products and you are done.
4) Identify the reactant that is the element. Det
ermine its charge when it becomes an ion.
5) Perform criss-cross to predict the new comp
ound on the products side of the reaction.
6) Write both new products.
7) Balance the reaction.
21. Single Replacement Between Meta
ls and Water
• Some metals have a higher activity than hydroge
n and can replace it in a single replacement reac
tion. In these reactions, you may think of water (
H2O) as H(OH).
• Ex: Na + H2O ?
Na + HOH ?
Na + H+OH- Na+OH- + H
2Na + 2H2O 2NaOH + H2
22. Parts of two aqueou
s ionic compounds s
witch places to form
two new compound
s.
There are two react
ants and two produc
ts.
2) DOUBLE REPLACEMENT REA
CTION
Example:
AgNO3 + NaCl
AgCl + NaNO3
23. Double Replacement Reactions
The general form of a double replacement reaction is:
AB + CD AD + CB
Just like single replacement reactions, not all double replac
ement reactions actually occur.
We can experimentally attempt a D.R. reaction. The reacti
on occurs if:
1) A solid precipitate is produced, or
2) A gas is produced, or
3) Water is produced.
4) If none of the above are produced and both products a
re (aq), then there is no reaction (NR)!
25. How do you determine if one of the products
of a double replacement reaction will be a p
recipitate?
• Use the solubility rules….
Soluble compounds
These compounds break down when put in water.
Example: In water, NaCl Na1+ and Cl1-.
We say that NaCl…
has dissolved.
is soluble.
forms an aqueous solution (aq).
26. The Solubility Rules
Insoluble compounds
These compounds do NOT
break down when put in
water.
Example: In water, CaCO3
does NOT break down int
o Ca2+ and CO3
2- ions.
The CaCO3 stays as a solid
, (s) or (ppt).
This is fortunate for many s
ea-creatures!
Seashells are made of CaCO3!
27. The Solubility Rules
You do not have to memorize these rules, b
ut you do have to know how to use them t
o determine if a product is a precipitate.
See the chart on the next slide…..
Let’s check NaCl and CaCO3… Are these c
ompounds soluble or insoluble in aqueous
solution?
29. Predicting the Products of Double
Replacement Reactions…
Step Example
1) Write the two reactants (both are ionic compoun
ds)
2) Identify the cations and anions in both of the co
mpound reactants
3) Pair up each cation with the anion from the othe
r compound
(i.e. – switch the cations)
4) Write the formula for each product using the cri
ss-cross method
5) Write the complete equation for the double repl
acement reaction
6) Balance the equation.
7) Use the solubility rules chart to figure out which
product is a precipitate (s) and which product is an
aqueous solution (aq). If both products are (aq) it i
s really not a reaction.
30. Two or more simple substances
(the reactants) combine to form
a more complex substance (the
product).
3) SYNTHESIS REACTION
Ex: 2Mg + O2 2MgO
31. SYNTHESIS REACTION
Types of synthesis:
a)Element A + Element B Compound
Na(s) + Cl2 (g) 2NaCl(s)
a)Element + Compound A Compound B
O2(g) + 2SO2(g) 2SO3(g)
a)Compound A + Compound B Compound C
CaO(s) + H2O(l) Ca(OH)2 (s)
32. Synthesis Reactions (cont’d)
• Metallic and nonmetallic elements react to form ionic co
mpounds. The resultant compound should be charge b
alanced by the criss-cross method.
Ex. 4Li + O2 2Li2O
• Nonmetals react with each other to form covalent (molec
ular) compounds. You should be able to draw a valid L
ewis Structure for the product.
2H2 + O2 2H2O
or
H2 + O2 H2O2
But NOT
H2 + O2 2OH
33. A more complex substance (the
reactant) breaks down into two o
r more simple parts (products).
Synthesis and decomposition re
actions are opposites.
4) DECOMPOSITION REACTION
Ex: 2H2O 2H2 + O2
Electrolysis of W
ater
34. DECOMPOSITION REACTIONS (
Cont’d)
Decomposition of a compound produces two or
more elements and/or compounds
The products are always simpler than the reactan
t.
Gases are often produced (H2, N2, O2, CO2, etc.)
in the decomposition of covalent compounds.
Ionic compounds may be decomposed into pure
elements by using electricity (electrolysis). This is
how pure metals are obtained from salts.
35. The Decomposition of Water by Ele
ctrolysis
2H2O 2H2 + O2
An electrical curre
nt can be used to c
hemically separate
water into oxygen
gas and hydrogen
gas. Notice that tw
ice as much hydro
gen is produced co
mpared to oxygen!
36. Electrolysis of Molten Sodium Chlori
de Many pure metals are
obtained by using elec
trolysis to separate me
tallic salts (ex. NaCl is
used to obtain pure Na
).
37. 5) COMBUSTION REACTIONS
a) All involve oxygen (O2) as a reactant,
combining with another substance
b) All combustion reactions are are exot
hermic
c) Complete combustion of a hydrocarb
on always produces CO2 and H2O
d) Incomplete combustion of a hydrocar
bon will produce CO and possibly C (
black carbon soot) as well
Ex: CH4 + 2O2 => CO2 + 2H2O (complete combustion – blue flame)
Ex: CH4 + 1.5O2 => CO + 2H2O (incomplete combustion – yellow flame)
Ex: CH4 + O2 => C + 2H2O (incomplete combustion – yellow flame, soot)
38. Combustion (cont’d)
• Any synthesis reaction which involves O2 as a re
actant is also considered to be a combustion rea
ction!
Ex. 2Mg + O2 2MgO
(metal oxide)
This is called the combustion of magnesium or t
he synthesis of magnesium oxide. The combusti
on of a metal always produces a metal oxide (in
this case, magnesium oxide). Make sure the
metal product is criss-crossed correctly!
43. Rules for Counting Atoms
1)Coefficients propagate to the right through the e
ntire compound, whether or not parentheses are
present.
2) Subscripts affect only the element to the left of t
he subscript, unless…
3) If a subscript occurs to the right of a parenthese
s, the subscript propagates to the left through th
e parentheses.
4) When a coefficient and subscript “meet”, you m
ust multiply the two.
44. Examples of Counting Atoms
SnO2 + 2H2 → Sn + 2H2O
2 C4H10 + 13 O2 → 8 CO2 + 10 H2O
Cu + 2AgNO3 → Cu(NO3)2 + 2Ag
3Pb(NO3)2 + 2AlCl3 → 3PbCl2 + 2Al(NO3)3
45. Example: An Exothermic Reaction
The “Smashing” Thermite Reaction:
2Al(s) + Fe2O3 (s) 2Fe (s) + Al2O3 (s)
Reaction Progress
Chemical
Potential
Energ
y
(H)
46. Example: An Endothermic Reaction
Ba(OH)28H2O (s) + 2NH4(NO3) (s) Ba
(NO3)2 (aq) + 2NH3 (g) + 10 H2O (l)
Reaction Progress
Chemical
Potential
Energ
y
(H)
47. Bond Enthalpy Calculations
Example : Calculate the enthalpy change (H0
rxn)
for the reaction N2 + 3 H2 2 NH3
Bonds broken (energy in)
1 N≡N: = 945
3 H-H: 3(435) = 1305
Total = 2250 kJ/mol
Bonds formed (energy out)
2x3 = 6 N-H: 6 (390) = - 2340 kJ/mol
H0
rxn = [energy used for breaking bonds] + [energy released in forming bonds]
Net enthalpy change (H0
rxn)
= + 2250 + (-2340) = - 90 kJ/mol (exothermic reaction)
H - H
H - H
H - H
You may have
to draw a Lew
is Structure to
know what ty
pe of bonds a
re present!
