1. Chapter 6 Chemical Reactions: An Introduction

2. Chemical Reactions
Reactions involve chemical changes in matter that result in new substances.
Reactions involve rearrangement and exchange of atoms to produce new molecules.
Reactants  Products 2

3. Evidence of Chemical Reactions
A chemical change occurs when new substances are made.
Visual clues (permanent):
Color change, precipitate formation, gas bubbles, flames, heat release, cooling, light
Other clues:
New odor, permanent new state 3

4. Evidence of Chemical Reactions (cont.)

5. Chemical Equations Shorthand way of describing a reaction
Provides information about the reaction:
Formulas of reactants and products
States of reactants and products
Relative numbers of reactant and product molecules that are required
Can be used to determine weights of reactants used and of products that can be made 4

6. Conservation of Mass Matter cannot be created or destroyed.
In a chemical reaction, all the atoms present at the beginning are still present at the end.
Therefore, the total mass cannot change. 5

7. CH4(g) + O2(g)  CO2(g) + H2O(l)
Combustion of Methane
Methane gas burns to produce carbon dioxide gas and liquid water
Whenever something burns, it combines with O2(g).
CH4(g) + O2(g)  CO2(g) + H2O(l) H C O +
1 C + 4 H O
1 C + 2 O H + O
1 C + 2 H + 3 O 6

8. Combustion of Methane Balanced
To show a reaction obeys the Law of Conservation of Mass, it must be balanced.
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l) H C O +
1 C H O
1 C H O 7

9. Writing Equations Use proper formulas for each reactant and product.
Proper equation should be balanced.
Obey Law of Conservation of Mass.
All elements on reactants side also on product side.
Equal numbers of atoms of each element on reactant side as on product side
Balanced equations show the relationship between the relative numbers of molecules of reactants and products.
Can be used to determine mass relationships 8

10. Symbols Used in Equations
Symbols used after chemical formula to indicate state:
(g) = gas; (l) = liquid; (s) = solid
(aq) = aqueous, dissolved in water
e. g. NH3(aq) indicates ammonia dissolved in water 9

11. Sample – Recognizing Reactants and Products
When magnesium metal burns in air it produces a white, powdery compound, magnesium oxide.
Burning in air means reacting with O2
Metals are solids, except for Hg, which is liquid. 10

12. Recognizing Reactants and Products (cont.)
Write the equation in words
Identify the state of each chemical
magnesium(s) + oxygen(g) magnesium oxide(s)
Write the equation in formulas
Identify diatomic elements
Identify polyatomic ions
Determine formulas
Mg(s) + O2(g)  MgO(s) (unbalanced)

13. Balancing by Inspection
Count atoms of each element
Polyatomic ions may be counted as one “element” if they do not change in the reaction.
Al + FeSO4 Al2(SO4)3 + Fe
1 SO
If an element appears in more than one compound on the same side, count each element separately and add.
CO + O2  CO2
O 2 11

14. Balancing by Inspection (cont.)
Pick an element to balance.
Avoid elements from 1b
Find least common multiple (LCM) and factors needed to make both sides equal.
Use factors as coefficients in equation.
If already a coefficient, then multiply by new factor
Recount and repeat until balanced. 12

15. Example #1
When magnesium metal burns in air it produces a white, powdery compound, magnesium oxide.
Burning in air means reacting with O2 13

16. Example #1 (cont.) Write the equation in words.
Identify the state of each chemical
magnesium(s) + oxygen(g) magnesium oxide(s)
Write the equation in formulas.
Identify diatomic elements
Identify polyatomic ions
Determine formulas
Mg(s) + O2(g)  MgO(s) (unbalanced)

17. Example #1 (cont.) Count the number of atoms of on each side
Count polyatomic groups as one “element” if on both sides
Split count of element if in more than one compound on one side
Mg(s) + O2(g)  MgO(s)
1  Mg 1
2  O  1 14

18. Example #1 (cont.) Pick an element to balance
Avoid element in multiple compounds
Find least common multiple of both sides & multiply each side by factor so it equals LCM
Mg(s) + O2(g)  MgO(s)
1  Mg 1
1 x 2  O  1 x 2 15

19. Example #1 (cont.)
Use factors as coefficients in front of compound containing the element
If coefficient is already there, multiply them together
Mg(s) + O2(g)  2 MgO(s)
1  Mg 1 x 2
1 x 2  O  1 x 2 16

20. Example #1 (cont.) Recount Mg(s) + O2(g)  2 MgO(s) 1  Mg 2
Repeat
2 Mg(s) + O2(g)  2 MgO(s)
2 x 1  Mg 2 17

21. Example #2
Under appropriate conditions, at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water. 18

22. Example #2 (cont.) Write the equation in words.
Identify the state of each chemical
ammonia(g) + oxygen(g) nitrogen monoxide(g) + water(g)
Write the equation in formulas.
Identify diatomic elements
Identify polyatomic ions
Determine formulas
NH3(g) + O2(g)  NO(g) + H2O(g)

23. NH3(g) + O2(g)  NO(g) + H2O(g)
Example #2 (cont.)
Count the number of atoms of on each side.
Count polyatomic groups as one “element” if on both sides
Split count of element if in more than one compound on one side
NH3(g) + O2(g)  NO(g) + H2O(g)
1  N 1
3  H  2
2  O  1 + 1 19

24. NH3(g) + O2(g)  NO(g) + H2O(g)
Example #2 (cont.)
Pick an element to balance
Avoid elements in multiple compounds
Find least common multiple of both sides & multiply each side by factor so it equals LCM
NH3(g) + O2(g)  NO(g) + H2O(g)
1  N 1
2 x 3  H  2 x 3
2  O  1 + 1 20

25. 2 NH3(g) + O2(g)  NO(g) + 3 H2O(g)
Example #2 (cont.)
Use factors as coefficients in front of compound containing the element.
2 NH3(g) + O2(g)  NO(g) + 3 H2O(g)
1  N 1
2 x 3  H  2 x 3
2  O  1 + 1 21

26. Example #2 (cont.) Recount Repeat 2 NH3(g) + O2(g)  NO(g) + 3 H2O(g)
2  N 1 x 2 22

27. 2 NH3(g) + O2(g)  2 NO(g) + 3 H2O(g)
Example #2 (cont.)
Recount
2 NH3(g) + O2(g)  2 NO(g) + 3 H2O(g)
2  N 2
6  H  6
2  O  2 + 3 23

28. 2 NH3(g) + 2.5 O2(g) 2 NO(g) + 3 H2O(g)
Example #2 (cont.)
Repeat
When you are forced to attack an element that is in 3 or more compounds, find where it is uncombined. You can find a factor to make it any amount you want, even if that factor is a fraction.
We want to make the O on the left equal 5, therefore we will multiply it by 2.5
2 NH3(g) O2(g) 2 NO(g) + 3 H2O(g)
2  N 2
6  H  6
2.5 x 2  O  2 + 3 24

29. Example #2 (cont.)
Multiply all the coefficients by a number to eliminate fractions
x.5  2, x.33  3, x.25  4, x.67  3
2 x [2 NH3(g) O2(g) 2 NO(g) + 3 H2O(g)]
4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)
4  N 4
12  H  12
10  O  10 25