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Download Introductory Chemistry: A Foundation FOURTH EDITION by Steven
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Chemical Reactions: An Introduction Chapter 6 1 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 2 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 3 Figure 6.1: Bubbles of hydrogen and oxygen gas form when an electric current is used to decompose water 4 Figure 6.2: Hot and cold pack reactions 5 Figure 6.3 (a): Chemical reactions 6 Figure 6.3 (b): Chemical reactions 7 Figure 6.3 (c): Chemical reactions 8 Figure 6.3 (d): Chemical reactions 9 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 10 Figure 6.4: The reaction between methane and oxygen to give water and carbon dioxide 11 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 12 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) O H H C H H + O O C + O H H O 1C+4H + 2O 1C+2O +2H+O 1C+2H+3O 13 Combustion of Methane Balanced • 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) H H C H H O + O C + O 1C + 4H + 4O O O O O + H H + O H H 1C + 4H + 4O 14 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 15 Symbols Used in Equations • symbols used after chemical formula to indicate state – (g) = gas; (l) = liquid; (s) = solid – (aq) = aqueous, dissolved in water 16 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 17 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 18 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) 19 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 2O1 20 Examples • 4. 5. when magnesium metal burns in air it produces a white, powdery compound magnesium oxide – burning in air means reacting with O2 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 1x2O1x2 21 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) 2 MgO(s) 1 Mg 1 1x2O1x2 22 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) 2 MgO(s) 1 Mg 2 2O2 8. Repeat 2 Mg(s) + O2(g) 2 MgO(s) 2 x 1 Mg 2 2O2 23 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) 24 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 3H2 2O1+1 25 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 2x3H2x3 2O1+1 26 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 2x3H2x3 2O1+1 27 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 6H6 2O1+3 8. Repeat 2 NH3(g) + O2(g) 2 NO(g) + 3 H2O(g) 2 N 1 x 2 6H6 2O1+3 28 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 6H6 2O2+3 29 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 6H6 2.5 x 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 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) 4 NO(g) + 6 H2O(g) 4 N 4 12 H 12 10 O 10 31