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Chemical Reactions

Chemical Reactions
Chapter 7 Chemical ReactionsExperiencing Chemical Change • Chemical reactions are happening both around you and in you all the time. • Some are very simple, others are complex.  In terms of the pieces—even the simple ones are interesting. • Chemical reactions involve atom exchanges, which change the structures of molecules. • What are some examples of chemical reactions you experience Tro's "Introductory Chemistry", 2 Chapter 7Combustion Reactions • Reactions in which O is consumed by combining with 2 another substance are called combustion reactions.  Always release heat and/or other forms of energy.  Produce one or more oxygencontaining compounds. • Combustion reactions are a subclass of oxidation– reduction reactions.  aka redox reactions.  Involve the transfer of electrons between atoms. Reactants Products 3Precipitation Reactions • Some reactions involve the combining of ions resulting in formation of a material that is insoluble in water. These are called precipitation reactions.  Formation of ―bathtubring‖ • Precipitation reactions are generally done with the reactants dissolved in water to allow the ions to move more freely.  Allowing the ions to contact each other more frequently.  Resulting in the reaction occurring faster. Tro's "Introductory Chemistry", 4 Chapter 7Chemical Reactions • A chemical change produces a new substances. • Reactions involve rearrangement and exchange of atoms, producing new molecules. Elements do not transmute. Atoms combine to make new compounds. 1. Molecules can combine to make bigger molecules. 2. Molecules can decompose into smaller molecules or atoms. 3. Atoms can be exchanged between molecules or transferred to another molecule. 4. Atoms can gain or lose electrons, turning them into ions. Tro's "Introductory Chemistry", 5 Chapter 7Evidence of Chemical Reactions • Look for evidence of a new substance. • Visual clues (permanent). Color change. Precipitate formation. Solid that forms when liquid solutions are mixed. Gas bubbles. Large energy changes. Container becomes very hot or cold. Emission of light. • Other clues. New odor. Whooshing sound from a tube. Permanent new physical state. Tro's "Introductory Chemistry", 6 Chapter 7Evidence of Chemical Change Release or Absorption of Heat Emission of Light Color Change Formation of a Gas Formation of Solid Precipitate Tro's "Introductory Chemistry", 7 Chapter 7Evidence of Chemical Change, Continued • In order to be absolutely sure that a chemical reaction has taken place, you need to go down to the molecular level and analyze the structures of the molecules at the Is boiling water a chemical change beginning and end. Tro's "Introductory Chemistry", 8 Chapter 7Practice—Decide Whether Each of the Following Involve a Chemical Reaction. Yes, CO and H O combine into carbohydrates • Photosynthesis 2 2 • Heating sugar until it turns black Yes, sugar decomposing No, molecules still same • Heating ice until it turns liquid Yes, food decomposing and combining • Digestion of food with stomach acid • Dissolving sugar in water No, molecules still same • Burning of alcohol in a flambé dessert Yes, alcohol combining with O to make CO and H O 2 2 2 Tro's "Introductory Chemistry", 9 Chapter 7Chemical 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 masses of reactants used and products that can be made. Tro's "Introductory Chemistry", 10 Chapter 7Conservation of Mass • Matter cannot be created or destroyed. Therefore, the total mass cannot change. And the total mass of the reactants will be the same as the total mass of the products. • In a chemical reaction, all the atoms present at the beginning are still present at the end. If all the atoms are still there, then the mass will not change. Tro's "Introductory Chemistry", 11 Chapter 7An Example: Combustion of Methane • Methane gas burns to produce carbon dioxide gas and gaseous water. Whenever something burns it combines with O (g). 2 Tro's "Introductory Chemistry", 12 Chapter 7An Example: Combustion of Methane • Methane gas burns to produce carbon dioxide gas and gaseous water.  Whenever something burns it combines with O (g). 2 CH (g) + O (g)  CO (g) + H O(g) 4 2 2 2 What incorrect assumption was made when writing this equation This equation reads ―1 molecule of CH gas combines with 1 4 O molecule of O gas to make 1 molecule of CO gas and 1 molecule 2 2 H H O of H O gas‖. 2 + + 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 We are assuming that all reactants combine 1 molecule : 1 molecule; and that 1 molecule of each product is made – an incorrect assumptionAn Example: Combustion of Methane • To show the reaction obeys the Law of Conservation of Mass the equation must be balanced.  We adjust the numbers of molecules so there are equal numbers of atoms of each element on both sides of the arrow. CH (g) + 2 O (g)  CO (g) + 2 H O(g) 4 2 2 2 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 Tro's "Introductory Chemistry", 14 Chapter 7An Example: Combustion of Methane CH (g) + 2 O (g)  CO (g) + 2 H O(g) 4 2 2 2 • CH and O are the reactants, and CO and H O 4 2 2 2 are the products. • The (g) after the formulas tells us the state of the chemical. • The number in front of each substance tells us the numbers of those molecules in the reaction.  Called the coefficients. Tro's "Introductory Chemistry", 15 Chapter 7An Example: Combustion of Methane CH (g) + 2 O (g)  CO (g) + 2 H O(g) 4 2 2 2 • This equation is balanced, meaning that there are equal numbers of atoms of each element on the reactant and product sides.  To obtain the number of atoms of an element, multiply the subscript by the coefficient. 1  C  1 4  H  4 4  O  2 + 2 Tro's "Introductory Chemistry", 16 Chapter 7Symbols Used in Equations • Symbols used to indicate state after chemical. (g) = gas; (l) = liquid; (s) = solid. (aq) = aqueous = dissolved in water. • Energy symbols used above the arrow for decomposition reactions. D = heat.  hn = light. shock = mechanical. elec = electrical. Tro's "Introductory Chemistry", 17 Chapter 7Writing Balanced Chemical Equations 1. Write a skeletal equation by writing the formula of each reactant and product. 2. Count the number of atoms of each element on each side of the equation.  Polyatomic ions may often be counted as if they are one ―element‖. 3. Pick an element to balance.  If an element is found in only one compound on both sides, balance it first.  Metals before nonmetals.  Leave elements that are free elements somewhere in the equation until last.  Balance free elements by adjusting the coefficient where it is a free element. Tro's "Introductory Chemistry", 18 Chapter 7Writing Balanced Chemical Equations, Continued 4. Find the least common multiple (LCM) of the number of atoms on each side.  The LCM of 3 and 2 is 6. 5. Multiply each count by a factor to make it equal to the LCM. 6. Use this factor as a coefficient in the equation.  If there is already a coefficient there, multiply it by the factor.  It must go in front of entire molecules, not between atoms within a molecule. 7. Recount and repeat until balanced. Tro's "Introductory Chemistry", 19 Chapter 7Example • When magnesium metal burns in air, it produces a white, powdery compound magnesium oxide. 1. Write a skeletal equation Mg(s) + O (g)  MgO(s) 2 2. Count the number of atoms on each side. Mg(s) + O (g)  MgO(s) 2 1  Mg 1 2  O  1 Tro's "Introductory Chemistry", 20 Chapter 7Example, Continued • When magnesium metal burns in air, it produces a white, powdery compound magnesium oxide. Mg(s) + O (g)  MgO(s) 2 3. Pick an element to balance.  Avoid element in multiple compounds.  Do free elements last.  Since Mg already balanced, pick O. 4. Find the LCM of both sides 5. and multiply each side by factor so it equals LCM.  LCM of 2 and 1 is 2. Mg(s) + O (g)  MgO(s) 2 1  Mg 1 1 x x 2 2  O  1 21Example, Continued • When magnesium metal burns in air, it produces a white, powdery compound magnesium oxide. Mg(s) + O (g)  MgO(s) 2 6. Use factors as coefficients in front of the compound containing the element.  We do not write 1 as a coefficient, its understood. Mg(s) + O (g) MgO(s) 2 2 1  Mg 1 1 x 2  O  1 x 2 Tro's "Introductory Chemistry", 22 Chapter 7Example, Continued • When magnesium metal burns in air, it produces a white, powdery compound magnesium oxide. Mg(s) + O (g)  MgO(s) 2 7. Recount—Mg not balanced now—That’s OK Mg(s) + O (g)  2 MgO(s) 2 1  Mg 2 2  O  2 7. and Repeat—attacking an unbalanced element. 2 Mg(s) + O (g)  2 MgO(s) 2 2 x 1  Mg 2 2  O  2 Tro's "Introductory Chemistry", 23 Chapter 7Another Example • Under appropriate conditions at 1000°C, ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and steam 1. write the skeletal equation a) first in words  identify the state of each chemical ammonia(g) + oxygen(g) nitrogen monoxide(g) + water(g) b) then write the equation in formulas  identify diatomic elements  identify polyatomic ions  determine formulas NH (g) + O (g)  NO(g) + H O(g) 3 2 2 Tro's "Introductory Chemistry", 24 Chapter 7Another Example, Continued • Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and steam NH (g) + O (g)  NO(g) + H O(g) 3 2 2 2) count the number of atoms of on each side NH (g) + O (g)  NO(g) + H O(g) 3 2 2 1  N 1 3  H  2 2  O  1 + 1 Tro's "Introductory Chemistry", 25 Chapter 7Another Example, Continued • Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous steam NH (g) + O (g)  NO(g) + H O(g) 3 2 2 3) pick an element to balance H  avoid element in multiple compounds on same side O 4) find least common multiple of both sides (6) 5) multiply each side by factor so it equals LCM NH (g) + O (g)  NO(g) + H O(g) 3 2 2 1  N 1 2 x 3  H  2 x 3 2  O  1 + 1 Tro's "Introductory Chemistry", 26 Chapter 7Another Example, Continued • Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water NH (g) + O (g)  NO(g) + H O(g) 3 2 2 6) use factors as coefficients in front of compound containing the element 2 NH (g) + O (g)  NO(g) + H O(g) 3 3 2 2 1  N 1 2 x 3  H  2 x 3 2  O  1 + 1 Tro's "Introductory Chemistry", 27 Chapter 7Another Example, Continued 7) Recount – N O not balanced 2 NH (g) + O (g)  NO(g) + 3 H O(g) 3 2 2 2  N 1 6  H  6 2  O  1 + 3 7) and Repeat – attack the N 2 NH (g) + O (g) 2 NO(g) + 3 H O(g) 3 2 2 2  N 1 x 2 6  H  6 2  O  1 + 3 Tro's "Introductory Chemistry", 28 Chapter 7Another Example, Continued 7) Recount Again – Still not balanced and the only element left is O 2 NH (g) + O (g)  2 NO(g) + 3 H O(g) 3 2 2 2  N 2 6  H  6 2  O  2 + 3 Tro's "Introductory Chemistry", 29 Chapter 7Another Example, Continued 7) and Repeat Again  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 2 NH (g) + O (g) 2 NO(g) + 3 H O(g) 3 2 2 2  N 2 6  H  6 2  O  2 + 3  We want to make the O on the left equal 5, therefore we will multiply it by 2.5 2 NH (g) + 2.5 O (g) 2 NO(g) + 3 H O(g) 3 2 2 2  N 2 6  H  6 2.5 x 2  O  2 + 3 30Another Example, Continued 7) You can’t have a coefficient that isn’t a whole number. Multiply all the coefficients by a number to eliminate fractions  If .5, then multiply by 2; if .33, then 3; if .25, then 4 2 NH (g) + 2.5 O (g) 2 NO(g) + 3 H O(g) x 2 3 2 2 4 NH (g) + 5 O (g) 4 NO(g) + 6 H O(g) 3 2 2 4  N 4 12  H  12 10  O  4 + 6 Tro's "Introductory Chemistry", 31 Chapter 7Practice 1 When aluminum metal reacts with air, it produces a white, powdery compound called aluminum oxide. Reacting with air means reacting with O : 2 Aluminum(s) + oxygen(g) aluminum oxide(s) Al(s) + O (g)  Al O (s) 2 2 3 Tro's "Introductory Chemistry", 32 Chapter 7Practice 1, Continued When aluminum metal reacts with air, it produces a white, powdery compound called aluminum oxide. Reacting with air means reacting with O : 2 Aluminum(s) + oxygen(g) aluminum oxide(s) Al(s) + O (g)  Al O (s) 2 2 3 4 Al(s) + 3 O (g)  2 Al O (s) 2 2 3 Tro's "Introductory Chemistry", 33 Chapter 7Practice 2 Acetic acid reacts with the metal aluminum to make aqueous aluminum acetate and gaseous hydrogen. Acids are always aqueous. Metals are solid except for mercury. Tro's "Introductory Chemistry", 34 Chapter 7Practice 2, Continued Acetic acid reacts with the metal aluminum to make aqueous aluminum acetate and gaseous hydrogen. Acids are always aqueous. Metals are solid except for mercury. Al(s) + HC H O (aq)  Al(C H O ) (aq) + H (g) 2 3 2 2 3 2 3 2 2 Al(s) + 6 HC H O (aq)  2 Al(C H O ) (aq) + 3 H (g) 2 3 2 2 3 2 3 2 Tro's "Introductory Chemistry", 35 Chapter 7Practice 3 Combustion of ethyl alcohol (C H OH) in 2 5 flambé (a brandied flaming dessert). Combustion is burning, and therefore, reacts with O . 2 Combustion of compounds containing C and H always make CO (g) and H O(g) as products. 2 2 C H OH(l) + O (g)  CO (g) + H O(g) 2 5 2 2 2 Tro's "Introductory Chemistry", 36 Chapter 7Practice 3, Continued Combustion of ethyl alcohol (C H OH) in 2 5 flambé (a brandied flaming dessert). Combustion is burning, and therefore, reacts with O . 2 Combustion of compounds containing C and H always make CO (g) and H O(g) as products. 2 2 C H OH(l) + O (g)  CO (g) + H O(g) 2 5 2 2 2 C H OH(l) + 3 O (g)  2 CO (g) + 3 H O(g) 2 5 2 2 2 Tro's "Introductory Chemistry", 37 Chapter 7Practice 4 Combustion of liquid butane (C H ) in a lighter. 4 10 C H (l) + O (g)  CO (g) + H O(g) 4 10 2 2 2 Tro's "Introductory Chemistry", 38 Chapter 7Practice 4, Continued Combustion of liquid butane (C H ) in a lighter. 4 10 C H (l) + O (g)  CO (g) + H O(g) 4 10 2 2 2 2 C H (l) + 13 O (g)  8 CO (g) + 10 H O(g) 4 10 2 2 2 Tro's "Introductory Chemistry", 39 Chapter 7Aqueous Solutions • Many times, the chemicals we are reacting together are dissolved in water. Mixtures of a chemical dissolved in water are called aqueous solutions. • Dissolving the chemicals in water helps them to react together faster. The water separates the chemicals into individual molecules or ions. The separate, freefloating particles come in contact more frequently so the reaction speeds up. Tro's "Introductory Chemistry", 40 Chapter 7Predicting Whether a Reaction Will Occur in Aqueous Solution • ―Forces‖ that drive a reaction:  Formation of a solid.  Formation of water.  Formation of a gas.  Transfer of electrons. • When chemicals (dissolved in water) are mixed and one of the abovenoted forces occur, the reaction will generally happen. Tro's "Introductory Chemistry", 41 Chapter 7Dissociation • When ionic compounds dissolve in water, the anions and cations are separated from each other. This is called dissociation.  However, not all ionic compounds are soluble in water • When compounds containing polyatomic ions dissociate, the polyatomic group stays together as one ion. Tro's "Introductory Chemistry", 42 Chapter 7Dissociation, Continued • Potassium iodide dissociates in water into potassium cations and iodide anions. +1 1 KI(aq) → K (aq) + I (aq) +1 1 K I K I • Copper(II) sulfate dissociates in water into copper(II) cations and sulfate anions. +2 2 CuSO (aq) → Cu (aq) + SO (aq) 4 4 2 +2 SO Cu SO Cu 4 4 Tro's "Introductory Chemistry", 43 Chapter 7Dissociation, Continued • Potassium sulfate dissociates in water into potassium cations and sulfate anions. +1 2 K SO (aq) → 2 K (aq) + SO (aq) 2 4 4 +1 K SO K K 2 4 SO 4 +1 K Tro's "Introductory Chemistry", 44 Chapter 7Electrolytes • Electrolytes are substances whose water solution is a conductor of electricity. • All electrolytes have ions dissolved in water. Tro's "Introductory Chemistry", 45 Chapter 7Electrolytes, Continued 1. In strong electrolytes, all the electrolyte molecules or formula units are separated into ions. 2. In nonelectrolytes, none of the molecules are separated into ions. 3. In weak electrolytes, a small percentage of the molecules are separated into ions. Tro's "Introductory Chemistry", 46 Chapter 7Types of Electrolytes • Salts = Water soluble ionic compounds.  All strong electrolytes. +1 • Acids = Form H ions and anions in water solution.  In binary acids, the anion is monoatomic. In oxyacids, the anion is polyatomic.  Sour taste.  React and dissolve many metals.  Strong acid = strong electrolyte, weak acid = weak electrolyte. • Bases = Watersoluble metal hydroxides.  Bitter taste, slippery (soapy) feeling solutions. 1  Increases the OH concentration. Tro's "Introductory Chemistry", 47 Chapter 7When Will a Salt Dissolve • A compound is soluble in a liquid if it dissolves in that liquid.  NaCl is soluble in water, but AgCl is not. • A compound is insoluble if a significant amount does not dissolve in that liquid.  AgCl is insoluble in water. Though there is a very small amount dissolved, but not enough to be significant. Tro's "Introductory Chemistry", 48 Chapter 7When Will a Salt Dissolve, Continued • Predicting whether a compound will dissolve in water is not easy. • The best way to do it is to do some experiments to test whether a compound will dissolve in water, then develop some rules based on those experimental results. We call this method the empirical method. Tro's "Introductory Chemistry", 49 Chapter 7Using the Solubility Rules to Predict an Ionic Compound’s Solubility in Water + + + • First check the cation: If it is Li , Na , K , or + NH , then the compound will be soluble in water. 4  Regardless of the anion. + + + + • If the cation is not Li , Na , K , or NH , then 4 follow the rule for the anion. • If a rule says the compounds are mostly soluble, then the exceptions are insoluble. • If a rule says the compounds are mostly insoluble, then the exceptions are soluble.  Note: slightly soluble  insoluble. Tro's "Introductory Chemistry", 52 Chapter 7Determine if Each of the Following Is Soluble in Water, Continued + • KOH Soluble, because the cation is K . • AgBr Insoluble, even though most compounds with − Br are soluble, this is an exception. − • CaCl Soluble, most compounds with Cl are soluble. 2 − Soluble, because the anion is NO . • Pb(NO ) 3 2 3 Insoluble, even though most compounds with • PbSO 4 2− SO are soluble, this is an exception. 4 Tro's "Introductory Chemistry", 53 Chapter 7Precipitation Reactions • Many reactions are done by mixing aqueous solutions of electrolytes together. • When this is done, often a reaction will take place from the cations and anions in the two solutions that are exchanging. • If the ion exchange results in forming a compound that is insoluble in water, it will come out of solution as a precipitate. Tro's "Introductory Chemistry", 54 Chapter 7Precipitation Reactions, Continued 2 KI(aq) + Pb(NO ) (aq)  2 KNO (aq) + PbI (s) 3 2 3 2 Tro's "Introductory Chemistry", 55 Chapter 7Precipitation Reactions, Continued 2 KI(aq) + Pb(NO ) (aq)  2 KNO (aq) + PbI (s) 3 2 3 2 Tro's "Introductory Chemistry", 56 Chapter 7No Precipitate Formation = No Reaction KI(aq) + NaCl(aq)  KCl(aq) + NaI(aq) All ions still present,  no reaction. Tro's "Introductory Chemistry", 57 Chapter 7Determining Products • Chapter 4 illustrated how to make molecules based on charge, we use the same procedure here. 3Mg(NO ) + 2Na PO Mg (PO ) + 6NaNO 3 2 3 4 3 4 2 3 58Process for Predicting the Products of a Precipitation Reaction 1. Write the formula for the reactants and Determine what ions each aqueous reactant has. 2. Exchange ions.  (+) ion from one reactant with () ion from the other. 3. Balance charges of combined ions to get formula of each product. 4. Balance the equation.  Count atoms. 5. Determine solubility of each product in water.  Use the solubility rules.  If product is insoluble or slightly soluble, it will precipitate.  If neither product will precipitate, no reaction. Tro's "Introductory Chemistry", 59 Chapter 7Determining Products Mg(NO ) + Na PO 3 2 3 4 1. Identify charges, from group number or ion charge 2+ 1 1+ 3 Mg (NO ) + Na PO 3 2 3 4 2. Visualize charges, moving across to the product side 2+ 1 1+ 3 + + 2 3 1 1 Mg (NO ) + Na PO 3 2 3 4Determining Products 2. Visualize charges, moving across to the product side + 2 3 + 1 1 2+ 1 1+ 3 Mg (NO ) + Na PO 3 2 3 4 3. Write the ion with That charge; cation first 2+ 1 1+ 3 2+ 3 1+ 1 Mg (NO ) + Na PO Mg PO + Na NO 3 2 3 4 4 3 1+ 1+ Note, only ion and ion charges, i.e. Na not Na 3Determining Products 3. Write the ion with charge 2+ 1 1+ 3 2+ 3 1+ 1 Mg (NO ) + Na PO Mg PO + Na NO 3 2 3 4 4 3 4. Balance the charges in the new molecules Mg(NO ) + Na PO Mg (PO ) + NaNO 3 2 3 4 3 4 2 3Determining Products 4. Balance the charges in the new molecules Mg(NO ) + Na PO Mg (PO ) + NaNO 3 2 3 4 3 4 2 3 5. Balance atoms 3Mg(NO ) + 2Na PO Mg (PO ) + 6NaNO 3 2 3 4 3 4 2 3Example 7.7—When an Aqueous Solution of Sodium Carbonate Is Added to an Aqueous Solution of Copper(II) Chloride, a White Solid Forms. 1. Write the formulas of the reactants and Determine the ions present when each reactant dissociates. Na CO (aq) + CuCl (aq)  2 3 2 + 2 +2 (Na + CO ) + (Cu + Cl )  3 2. Exchange the ions. + 2 +2 + +2 2 (Na + CO ) + (Cu + Cl )  (Na + Cl ) + (Cu + CO ) 3 3 Tro's "Introductory Chemistry", 64 Chapter 7Example 7.7—When an Aqueous Solution of Sodium Carbonate Is Added to an Aqueous Solution of Copper(II) Chloride, a White Solid Forms, Continued. 3. Write the formulas of the products.  Cross charges and reduce. Na CO (aq) + CuCl (aq)  NaCl + CuCO 2 3 2 3 4. Balance the equation. Na CO (aq) + CuCl (aq)  2NaCl + CuCO 2 3 2 3 Tro's "Introductory Chemistry", 65 Chapter 7Example 7.7—When an Aqueous Solution of Sodium Carbonate Is Added to an Aqueous Solution of Copper(II) Chloride, a White Solid Forms, Continued. 5. Determine the solubility of each product. Write an (s) after the insoluble products and a (aq) after the soluble products NaCl is soluble. CuCO is insoluble. 3 Na CO (aq) + CuCl (aq)  2NaCl(aq) + CuCO (s) 2 3 2 3 Tro's "Introductory Chemistry", 66 Chapter 7Another Example Al (CO ) + MgCl 2 3 3 2 Before you start write the ions involved; charges are very critical to know when you make new molecules 3+ Al 2 CO 3 2+ Mg Cl Don’t forget to ―move‖ charges Balance charges Balance equation 67 Look up solubilityAnother Example Al (CO ) + MgCl 2 3 3 2 3+ 2 2+ Al CO Mg Cl 3 Place these ions in their respective position Al (CO ) + MgCl 2 3 3 2 3+ 2 2+ 3+ 2+ 2 Al CO Mg Cl Al Cl Mg CO 3 3 Now change partners Tro's "Introductory Chemistry", 68 Chapter 7Another Example Al (CO ) + MgCl AlCl + MgCO 2 3 3 2 3 3 3+ 2 2+ 3+ 2+ 2 Al CO Mg Cl Al Cl Mg CO 3 3 Crisscross charges to make neutral molecules Al (CO ) + MgCl AlCl + MgCO 2 3 3 2 3 3 Balance Al (CO ) + 3MgCl2AlCl + 3MgCO 2 3 3 2 3 3 Tro's "Introductory Chemistry", 69 Chapter 7Another Example Al (CO ) + 3MgCl2AlCl + 3MgCO 2 3 3 2 3 3 Determine the solubility Al (CO ) (s) + 3MgCl (aq)2AlCl (aq)+ 3MgCO (s) 2 3 3 2 3 3 Tro's "Introductory Chemistry", 70 Chapter 7Practice–Predict the Products and Balance the Equation • Na S(aq) + CaCl (aq)  2 2 • (NH ) PO (aq) + AgNO (aq)  4 3 4 3 Tro's "Introductory Chemistry", 71 Chapter 7Practice–Predict the Products and Balance the Equation, Continued • Na S(aq) + CaCl (aq)  2 2 • Na S(aq) + CaCl (aq)  2 NaCl(aq) + CaS(s) 2 2 • (NH ) PO (aq) + AgNO (aq)  4 3 4 3 • (NH ) PO (aq) + AgNO (aq)  3NH NO (aq) + 4 3 4 3 4 3 Ag PO (s) 3 4 Tro's "Introductory Chemistry", 72 Chapter 7Practice—Write an Equation for the Reaction that Takes Place when an Aqueous Solution of (NH ) SO is Mixed with an Aqueous 4 2 4 Solution of Pb(C H O ) . 2 3 2 2 Tro's "Introductory Chemistry", 73 Chapter 7Practice—Write an Equation for the Reaction that Takes Place when an Aqueous Solution of (NH ) SO is Mixed with an Aqueous 4 2 4 Solution of Pb(C H O ) , Continued. 2 3 2 2 (NH ) SO (aq) + Pb(C H O ) (aq)  2 NH C H O (aq) + PbSO (s) 4 2 4 2 3 2 2 4 2 3 2 4 Tro's "Introductory Chemistry", 74 Chapter 7Ionic Equations • Equations that describe the chemicals put into the water and the product molecules are called molecular equations. 2 KOH(aq) + Mg(NO ) (aq)  2 KNO (aq) + Mg(OH) (s) 3 2 3 2 • Equations that describe the actual dissolved species are called complete ionic equations.  Aqueous electrolytes are written as ions. Soluble salts, strong acids, strong bases.  Insoluble substances and nonelectrolytes written in molecule form. Solids, liquids, and gases are not dissolved, therefore, molecule form. +1 1 +2 1 +1 1 2K + 2OH + Mg + 2NO2K + 2NO + Mg(OH) (aq) (aq) (aq) 3 (aq) (aq) 3 (aq) 2(s) Tro's "Introductory Chemistry", 75 Chapter 7Writing Complete Ionic Equations • Rewrite the molecular equation, but dissociate strong electrolytes into individual ions.  Strong electrolytes must be aqueous.  Solids, liquids, or gases cannot be electrolytes.  All soluble ionic compounds are strong electrolytes.  Strong acids are strong electrolytes.  HCl, HNO , H SO . 3 2 4 .  Weak acids are not written in the dissociated ion form.  Molecular compounds do not have ions, leave in the molecular form. Tro's "Introductory Chemistry", 76 Chapter 7Ionic Equations • Ions that are both reactants and products are called spectator ions. +1 1 +2 1 +1 1 2K + 2OH + Mg + 2NO2K + 2NO + Mg(OH) (aq) (aq) (aq) 3 (aq) (aq) 3 (aq) 2(s) • An ionic equation in which the spectator ions are removed is called a net ionic equation. 1 +2 2OH + MgMg(OH) (aq) (aq) 2(s) Tro's "Introductory Chemistry", 77 Chapter 7Writing Net Ionic Equations • First, identify the spectator ions in the complete ionic equation. Identical ions on both sides of the equation. • Cancel out the spectator ions—the result is the net ionic equation. Tro's "Introductory Chemistry", 78 Chapter 7Summary • A molecular equation is a chemical equation showing the complete, neutral formulas for every compound in a reaction. • A complete ionic equation is a chemical equation showing all of the species as they are actually present in solution. • A net ionic equation is an equation showing only the species that actually participate in the reaction. Tro's "Introductory Chemistry", 79 Chapter 7Practice–Write the Ionic and Net Ionic Equation. K SO (aq) + Ba(NO ) (aq)  2 KNO (aq) + BaSO (s) 2 4 3 2 3 4 3Na CO (aq) + 2 FeCl (aq)  6NaCl(aq) + Fe (CO ) (s) 2 3 3 2 3 3 Tro's "Introductory Chemistry", 80 Chapter 7Practice–Write the Ionic and Net Ionic Equation. K SO (aq) + Ba(NO ) (aq)  2 KNO (aq) + BaSO (s) 2 4 3 2 3 4 + 2− 2+ − + − 2K (aq) + SO (aq) + Ba (aq) + 2NO (aq)  2K (aq) + 2NO (aq) + BaSO (s) 4 3 3 4 2+ 2 Ba (aq) + SO (aq) BaSO (s) 4 4 3Na CO (aq) + 2 FeCl (aq)  6 NaCl(aq) + Fe (CO ) (s) 2 3 3 2 3 3 + 2− 3+ − + − 6Na (aq) + 3CO (aq) + 2Fe (aq) + 6Cl (aq)  6Na (aq) + 6Cl (aq) + Fe (CO ) (s) 3 2 3 3 2− 3+ 3CO (aq) + 2Fe (aq)  Fe (CO ) (s) 3 2 3 3 Tro's "Introductory Chemistry", 81 Chapter 7Properties of Acids • Sour taste. • Change color of vegetable dyes. • React with ―active‖ metals, not noble metals.  I.e., Al, Zn, Fe, but not Cu, Ag or Au. Zn + 2 HCl ZnCl + H 2 2  Corrosive. • React with carbonates, producing CO . 2  Marble, baking soda, chalk, limestone. CaCO + 2 HCl CaCl + CO + H O 3 2 2 2 • React with bases to form ionic salts.  And often water. Tro's "Introductory Chemistry", 82 Chapter 7Common Acids Chemical name Formula Old name Strength Nitric acid HNO Aqua fortis Strong 3 Sulfuric acid H SO Vitriolic acid Strong 2 4 Hydrochloric acid HCl Muriatic acid Strong Phosphoric acid H PO Moderate 3 4 Chloric acid HClO Moderate 3 Acetic acid HC H O Vinegar Weak 2 3 2 Hydrofluoric acid HF Weak Carbonic acid H CO Soda water Weak 2 3 Boric acid H BO Weak 3 3 Tro's "Introductory Chemistry", 83 Chapter 7Properties of Bases • A.k.a. alkalis. • Taste bitter. • Feel slippery. • Change color of vegetable dyes. Different color than acid. Litmus = blue. • React with acids to form ionic salts. And often water. Neutralization. Tro's "Introductory Chemistry", 84 Chapter 7Common Bases Chemical Formula Common Strength name name Sodium NaOH Lye, Strong hydroxide caustic soda Potassium KOH Caustic potash Strong hydroxide Calcium Ca(OH) Slaked lime Strong 2 hydroxide Magnesium Mg(OH) Milk of magnesia Weak 2 hydroxide Ammonium NH OH, Ammonia water, Weak 4 hydroxide NH (aq) aqueous ammonia 3 Tro's "Introductory Chemistry", 85 Chapter 7Acid–Base Reactions • Also called neutralization reactions because the acid and base neutralize each other’s properties. +1 • In the reaction of an acid with a base, the H from the 1 acid combines with the OH from the base to make water. • The cation from the base combines with the anion from the acid to make the salt. acid + base salt + water 2 HNO (aq) + Ca(OH) (aq)  Ca(NO ) (aq) + 2 H O(l) 3 2 3 2 2 • The net ionic equation for an acidbase reaction often is: +1 1 H (aq) + OH (aq)  H O(l) 2  As long as the salt that forms is soluble in water. Tro's "Introductory Chemistry", 86 Chapter 7Process for Predicting the Products of an Acid–Base Reaction 1. Determine what ions each aqueous reactant has. 2. Exchange ions.  (+) ion from one reactant with () ion from the other. + −  H combines with OH to make water. 3. Balance charges of combined ions to get formula of the salt. 4. Balance the equation.  Count atoms. 5. Determine solubility of the salt.  Use the solubility rules.  If the salt is insoluble or slightly soluble, it will precipitate. Tro's "Introductory Chemistry", 87 Chapter 7Practice—Complete and Balance These Acid–Base Reactions. NH OH(aq) + H SO (aq)  4 2 4 Al(OH) (aq) + H SO (aq)  3 2 3 Ba(OH) (aq) + H SO (aq)  2 2 4 Tro's "Introductory Chemistry", 92 Chapter 7Practice—Complete and Balance These Acid–Base Reactions, Continued. 2 NH OH(aq) + H SO (aq) (NH ) SO (aq) + 2 H O(l) 4 2 4 4 2 4 2 2 Al(OH) (aq) + 3 H SO (aq)  Al (SO ) (s) + 6 H O(l) 3 2 3 2 3 3 2 Ba(OH) (aq) + H SO (aq)  BaSO (s) + 2 H O(l) 2 2 4 4 2 Tro's "Introductory Chemistry", 93 Chapter 7Gas Evolution Reactions • Reactions in which the driving force is the production of a material that escapes as a gas are called gas evolution reactions. • Some reactions form a gas directly from the ion exchange. K S(aq) + H SO (aq)  K SO (aq) + H S(g) 2 2 4 2 4 2 • Other reactions form a gas by the decomposition of one of the ion exchange products into a gas and water. K SO (aq) + H SO (aq)  K SO (aq) + H SO (aq) 2 3 2 4 2 4 2 3 H SO H O(l) + SO (g) 2 3 2 2 Tro's "Introductory Chemistry", 94 Chapter 7Compounds that Undergo Gas Evolving Reactions Reactant Reacting Ion Decom Gas Example exchange pose type with formed product Metal S, Acid H S No H S K S(aq) + 2HCl(aq)  n 2 2 2 metal HS 2KCl(aq) + H S(g) 2 Metal CO , Acid H CO Yes CO K CO (aq) + 2HCl(aq)  n 3 2 3 2 2 3 metal HCO 2KCl(aq) + CO (g) + H O(l) 3 2 2 Metal SO Acid H SO Yes SO K SO (aq) + 2HCl(aq)  n 3 2 3 2 2 3 metal HSO 2KCl(aq) + SO (g) + H O(l) 3 2 2 (NH ) anion Base NH OH Yes NH KOH(aq) + NH Cl(aq)  4 n 4 3 4 KCl(aq) + NH (g) + H O(l) 3 2 Tro's "Introductory Chemistry", 95 Chapter 7Practice—Complete the Following Reactions, Continued. PbS(s) + H SO (aq)  PbSO (s) + H S(g) 2 4 4 2 HNO (aq) + NaHCO (aq)  NaNO (aq) + CO (g) + H O(l) 3 3 3 2 2 Tro's "Introductory Chemistry", 101 Chapter 7Other Patterns in Reactions • The precipitation, acid–base, and gas evolving reactions all involved exchanging the ions in the solution. • Other kinds of reactions involve transferring electrons from one atom to another. These are called oxidation–reduction reactions.  Also known as redox reactions.  Reactions of materials with O are redox reactions. 2  Unlike the others, many of these reactions are not done by dissolving the reactants in water. Tro's "Introductory Chemistry", 102 Chapter 7Oxidation–Reduction Reactions • We say that the element that loses electrons in the reaction is oxidized. • And the substance that gains electrons in the reaction is reduced. • You cannot have one without the other. • In combustion, the O atoms in O are 2 reduced, and the nonO atoms in the other material are oxidized. Tro's "Introductory Chemistry", 103 Chapter 7Combustion as Redox • In the following reaction: 2 Mg(s) + O (g)  2 MgO(s) 2 • The magnesium atoms are oxidized. 0 2+ Mg Mg + 2 e • The oxygen atoms are reduced. 0 2 O + 2 e O Tro's "Introductory Chemistry", 104 Chapter 7Combustion as Redox, Continued • Even though the following reaction does not involve ion formation, electrons are still transferred. CH (g) + 2 O (g)  CO (g) + 2 H O(g) 4 2 2 2 • The carbon atoms are oxidized. 4 +4 C C + 8 e  These are not charges, they are called oxidation numbers, but they help us see the electron transfer. • The oxygen atoms are reduced. 02 O + 2 e O Tro's "Introductory Chemistry", 105 Chapter 7Summary • Redox reactions occur when: A substance reacts with O . 2 A metal combines with a nonmetal. In general, whenever electrons are transferred. Tro's "Introductory Chemistry", 106 Chapter 7Reactions of Metals with Nonmetals (Oxidation–Reduction) • Metals react with nonmetals to form ionic compounds.  Ionic compounds are solids at room temperature. • The metal loses electrons and becomes a cation.  The metal undergoes oxidation. • The nonmetal gains electrons and becomes an anion.  The nonmetal undergoes reduction. • In the reaction, electrons are transferred from the metal to the nonmetal. 2 Na(s) + Cl (g)  NaCl(s) 2 Tro's "Introductory Chemistry", 107 Chapter 7Ionic Compound Formation as Redox • In the reaction: Mg(s) + Cl (g)  MgCl (s) 2 2 • The magnesium atoms are oxidized. 0 2+ Mg Mg + 2 e • The chlorine atoms are reduced. 0 Cl + 1 e Cl Tro's "Introductory Chemistry", 108 Chapter 7Recognizing Redox Reactions • Any reaction where O is a reactant or a product is a 2 redox reaction. • Any reaction between a metal and a nonmetal is redox. • Any reaction where electrons are transferred is redox.  When a free element gets combined into a compound, it will be either oxidized or reduced. N (g) + H (g)  NH (g) 2 2 3  When a metal cation changes its charge, it will be either oxidized if its charge increases or reduced if its charge decreases. CuCl(aq) + FeCl (aq)  FeCl (aq) + CuCl (aq) 3 2 2 Tro's "Introductory Chemistry", 109 Chapter 7Practice—Decide Whether Each of the Following Reactions Is a Redox Reaction. 2 Al(s) + 3 Br (l)  2 AlBr (s) 2 3 CaSO (s) + 2 HCl(aq)  CaCl (aq) + SO (g) + H O(l) 3 2 2 2 Fe O (s) + C(s)  2 Fe(s) + 3 CO(g) 2 3 SO (g) + O (g) + H O(l)  H SO (aq) 2 2 2 2 4 Tro's "Introductory Chemistry", 110 Chapter 7Practice—Decide Whether Each of the Following Reactions Is a Redox Reaction, Continued. 2 Al(s) + 3 Br (l)  2 AlBr (s)—Yes, metal + nonmetal. 2 3 CaSO (s) + 2 HCl(aq)  CaCl (aq) + SO (g) + H O(l)—No, this 3 2 2 2 is a gas evolving reaction. Fe O (s) + C(s)  2 Fe(s) + 3 CO(g)—Yes, the Fe is reduced 2 3 and the C gets combined. SO (g) + O (g) + H O(l)  H SO (aq)—Yes, O reactant. 2 2 2 2 4 2 Tro's "Introductory Chemistry", 111 Chapter 7Combustion Reactions • Reactions in which O (g) is a 2 reactant are called combustion reactions. • Combustion reactions release lots of energy. They are exothermic. • Combustion reactions are a subclass of oxidation– reduction reactions. 2 C H (g) + 25 O (g)  16 CO (g) + 18 H O(g) 8 18 2 2 2 Tro's "Introductory Chemistry", 112 Chapter 7Combustion Products • To predict the products of a combustion reaction, combine each element in the other reactant with oxygen. Reactant Combustion product Contains C CO (g) 2 Contains H H O(g) 2 Contains S SO (g) 2 Contains N NO(g) or NO (g) 2 Contains metal M O (s) 2 n Tro's "Introductory Chemistry", 114 Chapter 7Practice—Write the Equation for Each Reaction. • Combustion of the anesthetic cyclopropane, C H . 3 6 • Combustion of the nontoxic antifreeze propylene glycol, C H O . 3 8 2 Tro's "Introductory Chemistry", 115 Chapter 7Practice—Write the Equation for Each Reaction, Continued. • Combustion of the anesthetic cyclopropane, C H (g). 3 6 2 C H (g) + 9 O (g)  6 CO (g) + 6 H O(g) 3 6 2 2 2 • Combustion of the nontoxic antifreeze propylene glycol, C H O (l). 3 8 2 2 C H O (l) + 7 O (g)  6 CO (g) + 6 H O(g) 3 6 2 2 2 2 Tro's "Introductory Chemistry", 116 Chapter 7Classifying Reactions • One way is based on the process that happens. Precipitation, neutralization, formation of a gas, or transfer of electrons. 117Classifying Reactions, Continued • Another scheme classifies reactions by what the atoms do. Type of reaction General equation Synthesis A + B  AB Decomposition AB  A + B Displacement A + BC AC + B Double displacement AB + CD AD + CB 118Synthesis Reactions • Also known as composition or combination reactions. • Two (or more) reactants combine together to make one product. Simpler substances combining together. 2 CO + O 2 CO 2 2 2 Mg + O  2 MgO 2 HgI + 2 KI  K HgI 2 2 4 Tro's "Introductory Chemistry", 119 Chapter 7Decomposition Reactions • A large molecule is broken apart into smaller molecules or its elements. Caused by addition of energy into the molecule. • Have only one reactant, make 2 or more products. elec 2 FeCl (s) 2 FeCl (l) Cl (g) 3 2 2 D 2 HgO(s)  2 Hg(l) O (g) 2 h 2 O  3 O 3 2 Tro's "Introductory Chemistry", 120 Chapter 7Decomposition of Water elec 2 H O(l)  2 H (g)  O (g) 2 2 2 Tro's "Introductory Chemistry", 121 Chapter 7Single Displacement Reactions • Reactions that involve one atom displacing another and replacing it in a compound. • In the reaction Zn(s) + 2 HCl(aq)  ZnCl (aq) + H (g), the atom Zn displaces H 2 2 from the compound. • Other examples of displacement reactions are: Fe O (s) + Al(s)  Fe(s) + Al O (s) 2 3 2 3 2 Na(s) + 2 H O(aq)  2 NaOH(aq) + H (g) 2 2 Tro's "Introductory Chemistry", 122 Chapter 7Displacement of Copper by Zinc Zn(s)  CuCl (aq) Cu(s)  ZnCl 2 2 Tro's "Introductory Chemistry", 123 Chapter 7Double Displacement Reactions • Two ionic compounds exchange ions. • May be followed by decomposition of one of the products to make a gas. qq • X Y (aq) + A B (aq)  XB + AY • Precipitation, acid–base, and gas evolving reactions are also double displacement reactions. Tro's "Introductory Chemistry", 124 Chapter 7Practice—Classify the Following Reactions as Synthesis, Decomposition, Single Displacement, or Double Displacement, Continued. 3 Mg(s) + 2 FeCl (aq)  3 MgCl (aq) + 2 Fe(s) 3 2 Single displacement. CO (g) + H O(l)  H CO (aq) 2 2 2 3 Synthesis. 3 KOH(aq) + H PO (aq)  K PO (aq) + 3 H O(l) 3 4 3 4 2 Double displacement. heat CaCO (s) CaO(s)  CO (g) 3 2 Decomposition. Tro's "Introductory Chemistry", 125 Chapter 7
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