Download Unit 4 ~ Chemical Reactions (Chapters 6, 7, 18)

Unit 4 ~ Chemical Reactions
(Chapters 6, 7, 18)
And you
4-1 Evidence for Chemical Reactions (Section 6.1)
Color change
gas
Energy change
Precipitate
4-2 Writing and Balancing Chemical
Equations (Sections 6.2, 6.3)
• Chemical equations express in symbols the
qualitative experiences in real life and allow
scientists to calculate quantitative
relationships between the chemical species.
• Like How???
• Word Expression:
• Sodium solid is added to liquid water,
producing aqueous sodium hydroxide and
hydrogen gas.
• Chemical Equation:
• Na(s) + H2O(l) → NaOH(aq) + H2(g)
• But notice that the reaction is as unbalanced
as Mr. Wozniak
Na(s) + H2O(l) → NaOH(aq) + H2(g)
•
•
•
•
Balanced” Chemical Equation:
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
Phase symbols:
(s) = solid
(l) = liquid
(g) = gas
(aq) = aqueous or dissolved in water
• Other symbols:
• → means “yields” or “produces”
• ∆ means “heat added”
Note:
• chemicals on the left side of the → are the
reactants
• while the right side are the products
Balancing Equations:
• Equations must be balanced to satisfy the Law
of Conservation of Matter. In other words,
there must be equal numbers of atoms of all
elements on both sides of the arrow which, in
turn, will necessitate equal mass on both sides
of the →.
• To “balance” an equation, you may only
change the coefficients in front of the
chemicals, NEVER the subscript (this would
change the substance!).
Example: KCl → K + Cl2
• Acceptable: 2KCl → 2K + Cl2
• Unacceptable: KCl2 → K + Cl2
Useful HINTS for balancing include:
• 1) Use the concept of least common multiple.
• 2) Start with elements that appear only once
on both sides of the →.
• 3) even numbers can not be changed into odd
numbers with a whole number coefficient but
odd numbers can be changed into even
numbers!!!
Practice for you!!!
•
•
•
•
____KClO
2
2
+ ____O
3 2
3 → ____KCl
Change odd to even with those oxygen atoms!
Balance those oxygen atoms
Now balance those K and Cl atoms!
Try the next one please
3
• ____C3H8 + ____O
5
4
2 → ____CO2 + ____H
2O
And another please
• ____C
2 5H10 + ____O
15 2 → ____CO
10
10
2 +
____H2O
• When balancing a hydrocarbon reaction, it can
be beneficial to start with a coefficient 2 in
front of the hydrocarbon – then reduce at the
end if needed.