Chemical Reactions: An Introduction Chapter 6 - PowerPoint PPT Presentation

Chemical Reactions:An IntroductionChapter 6

• 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

• 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

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

Figure 6.2: Hot and cold pack reactions

Figure 6.3 (a): Chemical reactions

Figure 6.3 (b): Chemical reactions

Figure 6.3 (c): Chemical reactions

Figure 6.3 (d): Chemical reactions

• 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

Figure 6.4: The reaction between methane and oxygen to give water and carbon dioxide

• 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

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

• 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)

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

• 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)

• 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 after chemical formula to indicate state
• (g) = gas; (l) = liquid; (s) = solid
• (aq) = aqueous, dissolved in water

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

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

• 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)

• 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

• 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

• 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

• 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

• 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)

• 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

• 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

• 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

• 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

• 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

• 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

• 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