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Chemical Reactions: An Introduction Chapter 6. Chemical Reactions. Reactions involve chemical changes in matter resulting in new substances Reactions involve rearrangement and exchange of atoms to produce new molecules Elements are not transmuted during a reaction. Reactants  Products.

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chemical reactions
Chemical Reactions
  • Reactions involve chemical changes in matter resulting in new substances
  • Reactions involve rearrangement and exchange of atoms to produce new molecules
    • Elements are not transmuted during a reaction

Reactants  Products

evidence of chemical reactions
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
slide4

Figure 6.1: Bubbles of hydrogen and oxygen gas form when an electric current is used to decompose water

chemical equations
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
conservation of mass
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
  • Therefore the total mass of the reactants will be the same as the total mass of the products
combustion of methane

O

H

H

O

+

+

C

O

O

C

H

H

H

H

O

1 C + 4 H + 2 O

1 C + 2 O + 2 H + O

1 C + 2 H + 3 O

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)

combustion of methane balanced

O

O

O

O

H

H

H

H

+

+

+

C

C

+

H

H

O

O

O

O

H

H

1 C + 4 H + 4 O

1 C + 4 H + 4 O

Combustion of MethaneBalanced
  • to show the reaction obeys the Law of Conservation of Mass it must be balanced

CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)

writing equations
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 equation shows the relationship between the relative numbers of molecules of reactants and products
    • can be used to determine mass relationships
symbols used in equations
Symbols Used in Equations
  • symbols used after chemical formula to indicate state
    • (g) = gas; (l) = liquid; (s) = solid
    • (aq) = aqueous, dissolved in water
balancing by inspection
Balancing by Inspection

1. Count atoms of each element

  • polyatomic ions may be counted as one “element” if it does not change in the reaction

Al + FeSO4Al2(SO4)3 + Fe

1 SO4 3

  • if an element appears in more than one compound on the same side, count each separately and add

CO + O2 CO2

1 + 2 O 2

balancing by inspection18
Balancing by Inspection

2. Pick an element to balance

3. Find Least Common Multiple and factors needed to make both sides equal

4. Use factors as coefficients in equation

  • if already a coefficient then multiply by new factor

5. Recount and Repeat until balanced

examples
Examples
  • when magnesium metal burns in air it produces a white, powdery compound magnesium oxide
    • burning in air means reacting with O2

1. write the equation in words

    • identify the state of each chemical

magnesium(s) + oxygen(g) magnesium oxide(s)

2. write the equation in formulas

    • identify diatomic elements
    • identify polyatomic ions
    • determine formulas

Mg(s) + O2(g) MgO(s)

examples20
Examples
  • when magnesium metal burns in air it produces a white, powdery compound magnesium oxide
    • burning in air means reacting with O2

3. count the number of atoms 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

examples21
Examples
  • when magnesium metal burns in air it produces a white, powdery compound magnesium oxide
    • burning in air means reacting with O2

4. pick an element to balance

    • avoid element in multiple compounds

5. 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  1x 2

examples22
Examples
  • when magnesium metal burns in air it produces a white, powdery compound magnesium oxide
    • burning in air means reacting with O2

6. use factors as coefficients in front of compound containing the element

    • if coefficient already there, multiply them together

Mg(s) + O2(g) 2MgO(s)

1  Mg 1

1 x 2  O  1 x 2

examples23
Examples
  • when magnesium metal burns in air it produces a white, powdery compound magnesium oxide
    • burning in air means reacting with O2

7. Recount

Mg(s) + O2(g) 2MgO(s)

1  Mg 2

2  O  2

8. Repeat

2 Mg(s) + O2(g) 2MgO(s)

2 x 1  Mg 2

2  O  2

examples24
Examples
  • Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water

1. write the equation in words

    • identify the state of each chemical

ammonia(g) + oxygen(g) nitrogen monoxide(g) + water(g)

2. write the equation in formulas

    • identify diatomic elements
    • identify polyatomic ions
    • determine formulas

NH3(g) + O2(g) NO(g) + H2O(g)

examples25
Examples
  • Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water

3. 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

examples26
Examples
  • Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water

4. pick an element to balance

    • avoid element in multiple compounds

5. 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

examples27
Examples
  • Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water

6. 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

examples28
Examples
  • Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water

7. Recount

2 NH3(g) + O2(g) NO(g) + 3 H2O(g)

2  N 1

6  H  6

2  O  1 + 3

8. Repeat

2 NH3(g) + O2(g) 2NO(g) + 3 H2O(g)

2  N 1 x 2

6  H  6

2  O  1 + 3

examples29
Examples
  • Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water

9. Recount

2 NH3(g) + O2(g)  2 NO(g) + 3 H2O(g)

2  N 2

6  H  6

2  O  2 + 3

examples30
Examples
  • Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water

10. Repeat

    • A trick of the trade, 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) + 2.5 O2(g) 2 NO(g) + 3 H2O(g)

2  N 2

6  H  6

2.5 x 2  O  2 + 3

examples31
Examples
  • Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water

11. Multiply all the coefficients by a number to eliminate fractions

    • 0.5 x 2, 0.33 x 3, 0.25 x 4, 0.67 x 3

2 x [2 NH3(g) + 2.5 O2(g) 2 NO(g) + 3 H2O(g)]

4 NH3(g) + 5 O2(g) 4NO(g) + 6 H2O(g)

4  N 4

12  H  12

10  O  10