1. Chemical Reactions: An Introduction Chapter 6

2. 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

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

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

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 Therefore the total mass of the reactants will be the same as the total mass of the products

7. Combustion of Methane methane gas burns to produce carbon dioxide gas and liquid water –whenever something burns it combines with O 2 (g) CH 4 (g) + O 2 (g)  CO 2 (g) + H 2 O(l) H H C H H OO + O O C + O HH 1 C + 4 H + 2 O1 C + 2 O + 2 H + O 1 C + 2 H + 3 O

8. Combustion of Methane Balanced to show the reaction obeys the Law of Conservation of Mass it must be balanced CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(l) H H C H H OO + O O C + O HH OO + O HH + 1 C + 4 H + 4 O

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 equation shows the relationship between the relative numbers of molecules of reactants and products –can be used to determine mass relationships

10. Symbols Used in Equations symbols used after chemical formula to indicate state –(g) = gas; (l) = liquid; (s) = solid –(aq) = aqueous, dissolved in water

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 O 2 –Metals are solids, except for Hg which is liquid ¬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) + O 2 (g)  MgO(s)

12. Balancing by Inspection ¬Count atoms of each element apolyatomic ions may be counted as one “element” if it does not change in the reaction Al + FeSO 4  Al 2 (SO 4 ) 3 + Fe 1 SO 4 3 bif an element appears in more than one compound on the same side, count each separately and add CO + O 2  CO 2 1 + 2 O 2

13. Balancing by Inspection Pick an element to balance avoid elements from 1b Find Least Common Multiple 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

14. A bombardier beetle defending itself.

15. Examples when magnesium metal burns in air it produces a white, powdery compound magnesium oxide –burning in air means reacting with O 2 ¬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) + O 2 (g)  MgO(s)

16. Examples when magnesium metal burns in air it produces a white, powdery compound magnesium oxide –burning in air means reacting with O 2 ®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) + O 2 (g)  MgO(s) 1  Mg  1 2  O  1

17. Examples when magnesium metal burns in air it produces a white, powdery compound magnesium oxide –burning in air means reacting with O 2 ¯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) + O 2 (g)  MgO(s) 1  Mg  1 1 x 2  O  1 x 2

18. Examples when magnesium metal burns in air it produces a white, powdery compound magnesium oxide –burning in air means reacting with O 2 ±use factors as coefficients in front of compound containing the element 3if coefficient already there, multiply them together Mg(s) + O 2 (g)  2 MgO(s) 1  Mg  1 1 x 2  O  1 x 2

19. Examples when magnesium metal burns in air it produces a white, powdery compound magnesium oxide –burning in air means reacting with O 2 ²Recount Mg(s) + O 2 (g)  2 MgO(s) 1  Mg  2 2  O  2 ³Repeat 2 Mg(s) + O 2 (g)  2 MgO(s) 2 x 1  Mg  2 2  O  2

20. Examples Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water 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 NH 3 (g) + O 2 (g)  NO(g) + H 2 O(g)

21. Examples Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water ®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 NH 3 (g) + O 2 (g)  NO(g) + H 2 O(g) 1  N  1 3  H  2 2  O  1 + 1

22. Examples Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water ¯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 NH 3 (g) + O 2 (g)  NO(g) + H 2 O(g) 1  N  1 2 x 3  H  2 x 3 2  O  1 + 1

23. Examples Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water ±use factors as coefficients in front of compound containing the element 2 NH 3 (g) + O 2 (g)  NO(g) + 3 H 2 O(g) 1  N  1 2 x 3  H  2 x 3 2  O  1 + 1

24. Examples Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water ²Recount 2 NH 3 (g) + O 2 (g)  NO(g) + 3 H 2 O(g) 2  N  1 6  H  6 2  O  1 + 3 ³Repeat 2 NH 3 (g) + O 2 (g)  2 NO(g) + 3 H 2 O(g) 2  N  1 x 2 6  H  6 2  O  1 + 3

25. Examples Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water ´Recount 2 NH 3 (g) + O 2 (g)  2 NO(g) + 3 H 2 O(g) 2  N  2 6  H  6 2  O  2 + 3

26. Examples Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water µRepeat –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 NH 3 (g) + 2.5 O 2 (g)  2 NO(g) + 3 H 2 O(g) 2  N  2 6  H  6 2.5 x 2  O  2 + 3