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© 2015 Pearson Education, Ltd.
5. Chemical Reactions
© 2015 Pearson Education, Ltd.
• A process in which one or more substances is
changed into one or more new substances.
• In a chemical reaction, the substances
present before the chemical change are called
reactants.
• The substances present after the change are
called products.
Chemical Reaction
© 2015 Pearson Education, Ltd.
© 2015 Pearson Education, Ltd.
3- Replacement reaction
• is a reaction in which one element is substituted for
another element in a compound.
• A + BC → B + AC
• 2K (s) + 2HCl (aq) → 2KCl (aq) + H2 (g)
4 - Combustion reaction
A combustion reaction is a reaction in which a
substance reacts with oxygen gas, releasing energy
in the form of light and heat. Combustion reactions
must involve O2 as one reactant
© 2015 Pearson Education, Ltd.
Chemical Reactions in Automobiles: Combustion
A combustion reaction: In an automobile engine, hydrocarbons
such as octane (C8H18) from gasoline combine with oxygen from
the air and react to form CO2 and H2O.
Combustion reactions are a subcategory of oxidation–reduction
reactions, in which electrons are transferred from one
substance to another.
© 2015 Pearson Education, Ltd.
• In chemical reactions, however, the changes
in mass are so minute that they can be
ignored. So we can say mass is neither
created nor destroyed in a chemical
reaction.
• Suppose that we burn 58 g of butane in a lighter. It
will react with 208 g of oxygen to form 176 g of
carbon dioxide and 90 g of water.
© 2015 Pearson Education, Ltd. 7
Chemical Equations
It is a way to represent the chemical reaction.
It shows us:
• The chemical symbols of reactants and
products
• The physical states of reactants and products–
(s), (l), (g), (aq.)
• Balanced equation (same number of atoms on
each side)
2H2 (g) + O2 (g) 2H2O (l)
Reactants
(starting materials)
Products
( materials formed) 7
© 2015 Pearson Education, Ltd. 8
Balancing Chemical Equations
The number of atoms of each element must be the same on both
sides of the equation.
C2H6 + O2 CO2 + H2O
Reactants Products
2 C 1 C
6 H 2 H
2 O 3 O
C2H6 + O2 CO2 + H2O
Reactants Products
4 C 4 C
12 H 12 H
14 O 14 O
C2H6 + O2 CO2 + H2O
2
7/2 3
2 7 4 6
© 2015 Pearson Education, Ltd.
Balance the following equations
9
2
2
2
2 2 2
2
2 3
2 2
3
2
2
2
2
2
2
8
8
© 2015 Pearson Education, Ltd. 10
The quantitative study of reactants and
products in a chemical reaction
CH4 + O2  CO2 + H2O
Stoichiometry
2 2
10
Mass and mole Relationships
in Chemical Reactions
© 2015 Pearson Education, Ltd. 11
The Mole Method:
Stoichiometric coefficients in a chemical equation
can be interpreted as the number of moles of each
substance.
N2 (g) + 3 H2 (g)  2NH3 (g)
N2
1 mole
6.022 ×1023 molecules
H2
3 mole
3×6.022 ×1023 molecules
NH3
2 mole
2×6.022 ×1023 molecules
11
© 2015 Pearson Education, Ltd. 12
N2 (g) + 3 H2 (g)  2NH3 (g)
Recall that the coefficient on N2 is 1 but is not
explicitly written in the reaction Coefficients:
N2 = 1
H2 = 3
NH3 = 2
Using the coefficients we can write mole ratios
Mole Ratios
Definition: mole ratio gives the relative amounts of
reactants and products
© 2015 Pearson Education, Ltd. 13
13
N2 (g) + 3H2 (g)  2NH3 (g)
© 2015 Pearson Education, Ltd. 14
MOLE to MOLE Stoichiometry
14
N2 (g) + 3H2 (g)  2NH3 (g)
© 2015 Pearson Education, Ltd. 15
Silicon tetrachloride (SiCl4) can be prepared by heating Si
in chlorine gas:
Si (s) + 2Cl2 (g) SiCl4 (l)
In one reaction, 0.5 mole of SiCl4 is produced. How many
moles of molecular chlorine were used in the reaction?
Cl2 : SiCl4
2 mol : 1 mol
?? : 0.5 mol
n of Cl2 used = (0.507 molSiCl4 ×( 2 molCl2 /1 molSiCl4)
= 1.0 mol of Cl2
15
Example
© 2015 Pearson Education, Ltd.
Convert 85.0 g to moles:
If 85.0 g of CH4 is consumed by a person over a certain period,
what is the mass of CO2 produced?
CH4 + 2O2  CO2 + 2H2O
16
16
Example
𝑤𝑡 (𝑔) = n x Mw (g/mol)
n (CH4) = wt/Mw = (85.0/16.04) = 5.30 mol(CH4)
CH4 + 2O2  CO2 + 2H2O
1 mole of CH4 1 mole of CO2
5.30 moles ??
n (CO2) = 5.30 mol(CH4) × (1mol (CO2) /1mol (CH4) ) = 5.30
mol(CO2)
wt (CO2) = n × Mw = 5.30 mol(CO2)× (44 g (CO2) / 1mol (CO2) )
= 233.2 g (CO2) =2.33×102 g (CO2)
 
 
mol
g
Mw
g
wt
n
/


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5- Chemical Reactions.pptx

  • 1. © 2015 Pearson Education, Ltd. 5. Chemical Reactions
  • 2. © 2015 Pearson Education, Ltd. • A process in which one or more substances is changed into one or more new substances. • In a chemical reaction, the substances present before the chemical change are called reactants. • The substances present after the change are called products. Chemical Reaction
  • 3. © 2015 Pearson Education, Ltd.
  • 4. © 2015 Pearson Education, Ltd. 3- Replacement reaction • is a reaction in which one element is substituted for another element in a compound. • A + BC → B + AC • 2K (s) + 2HCl (aq) → 2KCl (aq) + H2 (g) 4 - Combustion reaction A combustion reaction is a reaction in which a substance reacts with oxygen gas, releasing energy in the form of light and heat. Combustion reactions must involve O2 as one reactant
  • 5. © 2015 Pearson Education, Ltd. Chemical Reactions in Automobiles: Combustion A combustion reaction: In an automobile engine, hydrocarbons such as octane (C8H18) from gasoline combine with oxygen from the air and react to form CO2 and H2O. Combustion reactions are a subcategory of oxidation–reduction reactions, in which electrons are transferred from one substance to another.
  • 6. © 2015 Pearson Education, Ltd. • In chemical reactions, however, the changes in mass are so minute that they can be ignored. So we can say mass is neither created nor destroyed in a chemical reaction. • Suppose that we burn 58 g of butane in a lighter. It will react with 208 g of oxygen to form 176 g of carbon dioxide and 90 g of water.
  • 7. © 2015 Pearson Education, Ltd. 7 Chemical Equations It is a way to represent the chemical reaction. It shows us: • The chemical symbols of reactants and products • The physical states of reactants and products– (s), (l), (g), (aq.) • Balanced equation (same number of atoms on each side) 2H2 (g) + O2 (g) 2H2O (l) Reactants (starting materials) Products ( materials formed) 7
  • 8. © 2015 Pearson Education, Ltd. 8 Balancing Chemical Equations The number of atoms of each element must be the same on both sides of the equation. C2H6 + O2 CO2 + H2O Reactants Products 2 C 1 C 6 H 2 H 2 O 3 O C2H6 + O2 CO2 + H2O Reactants Products 4 C 4 C 12 H 12 H 14 O 14 O C2H6 + O2 CO2 + H2O 2 7/2 3 2 7 4 6
  • 9. © 2015 Pearson Education, Ltd. Balance the following equations 9 2 2 2 2 2 2 2 2 3 2 2 3 2 2 2 2 2 2 8 8
  • 10. © 2015 Pearson Education, Ltd. 10 The quantitative study of reactants and products in a chemical reaction CH4 + O2  CO2 + H2O Stoichiometry 2 2 10 Mass and mole Relationships in Chemical Reactions
  • 11. © 2015 Pearson Education, Ltd. 11 The Mole Method: Stoichiometric coefficients in a chemical equation can be interpreted as the number of moles of each substance. N2 (g) + 3 H2 (g)  2NH3 (g) N2 1 mole 6.022 ×1023 molecules H2 3 mole 3×6.022 ×1023 molecules NH3 2 mole 2×6.022 ×1023 molecules 11
  • 12. © 2015 Pearson Education, Ltd. 12 N2 (g) + 3 H2 (g)  2NH3 (g) Recall that the coefficient on N2 is 1 but is not explicitly written in the reaction Coefficients: N2 = 1 H2 = 3 NH3 = 2 Using the coefficients we can write mole ratios Mole Ratios Definition: mole ratio gives the relative amounts of reactants and products
  • 13. © 2015 Pearson Education, Ltd. 13 13 N2 (g) + 3H2 (g)  2NH3 (g)
  • 14. © 2015 Pearson Education, Ltd. 14 MOLE to MOLE Stoichiometry 14 N2 (g) + 3H2 (g)  2NH3 (g)
  • 15. © 2015 Pearson Education, Ltd. 15 Silicon tetrachloride (SiCl4) can be prepared by heating Si in chlorine gas: Si (s) + 2Cl2 (g) SiCl4 (l) In one reaction, 0.5 mole of SiCl4 is produced. How many moles of molecular chlorine were used in the reaction? Cl2 : SiCl4 2 mol : 1 mol ?? : 0.5 mol n of Cl2 used = (0.507 molSiCl4 ×( 2 molCl2 /1 molSiCl4) = 1.0 mol of Cl2 15 Example
  • 16. © 2015 Pearson Education, Ltd. Convert 85.0 g to moles: If 85.0 g of CH4 is consumed by a person over a certain period, what is the mass of CO2 produced? CH4 + 2O2  CO2 + 2H2O 16 16 Example 𝑤𝑡 (𝑔) = n x Mw (g/mol) n (CH4) = wt/Mw = (85.0/16.04) = 5.30 mol(CH4) CH4 + 2O2  CO2 + 2H2O 1 mole of CH4 1 mole of CO2 5.30 moles ?? n (CO2) = 5.30 mol(CH4) × (1mol (CO2) /1mol (CH4) ) = 5.30 mol(CO2) wt (CO2) = n × Mw = 5.30 mol(CO2)× (44 g (CO2) / 1mol (CO2) ) = 233.2 g (CO2) =2.33×102 g (CO2)     mol g Mw g wt n / 