1. Topic VIII: “ Chemical Reactions”. Chapter 11 (Pearson)

2. EVIDENCES OF A CHEMICAL REACTION:
1) Color change.
2) Formation of a solid (precipitate).
3) Formation of a gas (bubbles).
4) Temperature changes
(exothermic or endothermic).

3. During a chemical reaction the reactants collide breaking the chemical bonds between their particles & the atoms are rearranged forming new bonds (new substances):
H H O H H H H
H H O O O HH HH HH HH HH HH HH HH HH HH HH HH
2 H2 + O2
2 H2O
2 (2 g)
32 g
2 (18 g) +
4 g
32 g
36 g +
36 g
36 g =

4. 2 NaOH (aq) + CuSO4 (aq) → Na2SO4 (aq) +Cu(OH)2 (s)
In a chemical reaction, one or more reactants change into one or more products.
yield
Products
Reactants
2 NaOH (aq) + CuSO4 (aq) → Na2SO4 (aq) +Cu(OH)2 (s)
A chemical equation is a representation of a chemical reaction (model).
(aq)
aqueous (dissolved in water).
(g)
gas.
liquid.
(s)
solid (precipitate).
(l)

5. Law of Conservation of Mass vs Chemical equations:
H2(g) +
Br2(g) → 2
HBr (g)
2 (1 g) +
2 (79.9 g) →
(1 g g) ≠
161.8 g 2 (
80.9 g )
161.8 g =
161.8 g
The mass is conserved.
Always !

6. Therefore, this model represents a balanced chemical equation.
Insert a Coefficient 2 in front of HBr to make the number of atoms of each element equal on both sides of the equation:
H2(g) Br2(g) → HBr (g) 2
2 H
2 Br
2 H
&
2 Br
This chemical equation shows the Law of Conservation of Mass.
Therefore, this model represents a balanced chemical equation.

7. Chemical Reactions:
How could you describe the rusting of iron?
“Iron reacts with oxygen to produce iron(III) oxide (rust).” Fe
(s) + O2
(g) →
Fe2O3
(s)
“The products show completely different properties (characteristics) than the reactants.”

8. Fe(s) O2(g) → Fe2O3(s) 2 3 2 2 6 4 6

9. This is a balanced chemical equation.
Fe(s) O2(g) → Fe2O3(s) 4 3 2 4 6 4 6
This is a balanced chemical equation.

10. Balancing Chemical Equations:
Determine the number of atoms of each element in
the reactants and products. Count a polyatomic ion
as a single unit if it appears unchanged on both
sides of the equation.
2. Balance the atoms one at a time by using coefficients.
(*) Never change the subscripts in a chemical formula.
Each substance only has one correct formula.
3. Check each atom to be sure that the number is equal
on both sides of the equation.
4. Make sure all the coefficients are in the lowest
possible ratio.

11. Balancing Chemical Equations(Sequence):
1st -- Metals (such as Fe, Na, K, Mg…).
2nd -- Polyatomic ions (such as SO42-, PO43- that cross from reactants to products unchanged).
3rd -- Non-metals (such as Cl, S, Br,)
4th-- Hydrogen.
5th-- Oxygen.

12. AgNO3(aq) + Cu(s) → Cu(NO3)2(aq) + Ag(s)
1st Metals are already balanced.
2nd Nitrate ions must be balanced:
Write a coefficient 2 in front of AgNO3(aq).
3rd Balance the silver (It was affected when balancing nitrate ions) :
Write a coefficient 2 in front of Ag(s).
4th Count the number of atoms (reactants & products):
2 Ag = Ag
The equation has been balanced.
2 N = N
6 O = O
1 Cu = Cu

13. C3H8(g) + O2(g) → CO2(g) + H2O(g)
Balance the following equation:
C3H8(g) O2(g) → CO2(g) H2O(g) 5 3 4
1st Write a coefficient 3 in front of CO2
2nd Write a coefficient 4 in front of H2O
3rd Write a coefficient 5 in front of O2
Finally count the total # of each atom before & after the change (must be equal):
The equation has been balanced.
3 C = C
8 H = H
10 O = O

14. Practice: 1) Na(s) + HCl(aq) → NaCl(aq) + H2(g)

15. 1) Na(s) + 2HCl(aq) → NaCl(aq) + H2(g)

16. 1) Na(s) + 2 HCl(aq) → 2 NaCl(aq) + H2(g)

17. 1) 2 Na(s) + 2 HCl(aq) → 2 NaCl(aq) + H2(g)

18. 1) 2 Na(s) + 2 HCl(aq) → 2 NaCl(aq) + H2(g)

19. 2) Al(s) O2(g) → Al2O3(s)

20. 2) 2Al(s) O2(g) → Al2O3(s)

21. 2) 2Al(s) O2(g) → Al2O3(s)

22. 2) 2Al(s) O2(g) → 2Al2O3(s)

23. 2) 4Al(s) O2(g) → 2Al2O3(s)

24. 2) 4Al(s) O2(g) → 2Al2O3(s)
Al = 4
Al = 4
O = 6
O = 6

25. 3) C4H8 (g) + O2(g) → CO2(g) + H2O(g)

26. 3) C4H8 (g) + O2(g) → 4CO2(g) + H2O(g)

27. 3) C4H8 (g) + O2(g) → 4CO2(g) + 4 H2O(g)

28. 3) C4H8 (g) + 6O2(g) → 4CO2(g) + 4 H2O(g)

29. 3) C4H8 (g) + 6 O2(g) → 4CO2(g) + 4 H2O(g) Reactants Products