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Chemical Reactions: An Introduction Chapter 6

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: An Introduction Chapter 6

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  1. Chemical Reactions:An IntroductionChapter 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

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

  5. Figure 6.2: Hot and cold pack reactions

  6. Figure 6.3 (a): Chemical reactions

  7. Figure 6.3 (b): Chemical reactions

  8. Figure 6.3 (c): Chemical reactions

  9. Figure 6.3 (d): Chemical reactions

  10. 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

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

  12. 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

  13. 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)

  14. 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)

  15. 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

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

  17. 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

  18. 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

  19. 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)

  20. 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

  21. 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

  22. 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

  23. 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

  24. 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)

  25. 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

  26. 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

  27. 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

  28. 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

  29. 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

  30. 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

  31. 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

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