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Chapter 6 Chemical Reactions: An Introduction

Chapter 6 Chemical Reactions: An Introduction. 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. Evidence of Chemical Reactions.

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Chapter 6 Chemical Reactions: An Introduction

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  1. Chapter 6Chemical 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

  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

  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

  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.

  7. 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)

  8. 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 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)

  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

  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

  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.

  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 SO4 3 • If an element appears in more than one compound on the same side, count each element separately and add. CO + O2 CO2 1 + 2 O 2

  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.

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

  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

  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

  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

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

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

  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. 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

  24. 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

  25. 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

  26. Example #2 (cont.) • Recount 2 NH3(g) + O2(g) NO(g) + 3 H2O(g) 2  N 1 6  H  6 2  O  1 + 3 • Repeat 2 NH3(g) + O2(g) 2 NO(g) + 3 H2O(g) 2  N 1 x 2 6  H  6 2  O  1 + 3

  27. 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

  28. 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) + 2.5 O2(g) 2 NO(g) + 3 H2O(g) 2  N 2 6  H  6 2.5 x 2  O  2 + 3

  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) + 2.5 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

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