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Chapter 7 – Chemical Reactions Chemical reactions are happening both around you and in you all the time.

Chapter 7 – Chemical Reactions Chemical reactions are happening both around you and in you all the time. In a chemical reaction, one or more substances change into different substances. Reactions involve rearrangement and exchange of atoms to produce new molecules.

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Chapter 7 – Chemical Reactions Chemical reactions are happening both around you and in you all the time.

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  1. Chapter 7 – Chemical Reactions • Chemical reactions are happening both around you and in you all the time. • In a chemical reaction, one or more substances change into different substances. • Reactions involve rearrangement and exchange of atoms to produce new molecules.

  2. Evidence of a Chemical Reaction Some reactions give visible signs of a chemical change or a chemical reaction. Evidence of a chemical reaction includes • Color change • Formation of a solid • Formation of gas • Bubbles or fizzing • Change in odor • Emission of light • Emission or absorption of heat These signs are only a clue, they are not definitive evidence of a chemical reaction. Many chemical reactions occur without any obvious signs.

  3. Chemical Equations • Chemical reactions are represented by chemical equations. • In a chemical equation, the substances on the left side of the equation are called reactants and the substances on the right side are called products. • Fe + O2 Fe2O3 reactants product

  4. The abbreviations in parenthesis next to the reactants and products indicate the states of the reactants and products. In the reaction of iron with oxygen to form iron (III) oxide (rust), Fe(s) + O2(g) Fe2O3(s) the (s) indicates that iron and iron (III) oxide are solids and the (g) indicates that oxygen is a gas.

  5. In the chemical equation, Fe(s) + O2(g) Fe2O3(s) there are more iron and oxygen atoms on the product side of the equation than on the reactants side of the equation. • Because matter is neither created nor destroyed (law of conservation of mass), we need to correct this equation to ensure that we have the same number of each type of atom on both sides of the equation. • To correct an equation with different numbers of atoms on one side of the equation than the other side, we must balance the equation by adding coefficients (not subscripts) in front the formulas of the reactants and products.

  6. To balance the preceding equation, we need to put a coefficient of 4 in front of iron, a coefficient of 3 in front of oxygen, and a coefficient of 2 in front of iron (III) oxide, 4Fe(s) + 3O2(g) 2Fe2O3(s). • This equation is now a balanced equation, where the numbers of each type of atom on both sides of the equation are equal. • Adding coefficients in front of the reactants and products does not change the kinds of molecules, which would occur if the subscripts were to be changed. This is why balance is NEVER achieved by changing subscripts.

  7. The number of a particular type of atom within a chemical formula embedded in an equation is obtained by multiplying the subscript for the atom by the coefficient for the chemical formula. • If there is no coefficient or subscript, a 1 is implied. • In the reaction of iron with oxygen to form iron (III) oxide, 4Fe(s) + 3O2(g) 2Fe2O3(s) Reactants Products 4 Fe atoms (4 X 1) 4 Fe atoms (2 X 2) 6 O atoms (3 X 2) 6 O atoms (2 X 3)

  8. Rules for Balancing Equations • Make sure that the law of conservation of mass is met by making sure that the number of atoms of an element on the left side of the equation is the same as the number on the right side. • Never touch subscripts when balancing equations since that will change the composition and therefore the substance itself. • Simplify equations as much as possible.

  9. Steps for Balancing Equations • Write a skeletal equation by writing chemical formulas for each of the reactants and products. • Count the number of atoms on each side of the equation. • Note which atoms are not balanced. • Select one element to balance. • If an element is found in only one compound on both sides, balance it first. If there is more than one such element, balance metals before nonmetals. • Leave elements that are free elements somewhere in the equation until last.

  10. Steps for Balancing Equations • Balance the atom by changing the proper coefficient(s) or the number(s) preceding the proper formula(s) in the equation. • If the balanced equation contains coefficient fractions, clear these by multiplying the entire equation by the appropriate factor. 6) After balancing that atom, update the atom counts for the rest of the atoms in the equation. 7) Check that both sides have the same amount of atoms of each element. Make sure that you have included all sources of each element that you are balancing. If the equation is not balanced, go back to step 3.

  11. Example #1: Write a balanced chemical equation for the reaction of magnesium with aqueous copper (I) nitrate to form aqueous magnesium nitrate and solid copper. 1) Mg(s) + CuNO3(aq) -> Mg(NO3)2 + Cu(s) 2) Mg – 1 Mg – 1 Cu – 1 Cu – 1 N – 1 N – 2 O – 3 O - 6 3) Nitrogen and oxygen are unbalanced. 4) Nitrogen and oxygen both appear in one compound on both sides sides, so we’ll start with oxygen.

  12. Example #1: Write a balanced chemical equation for the reaction of magnesium with aqueous copper (I) nitrate to form aqueous magnesium nitrate and solid copper. 5) Mg(s) + 2 CuNO3(aq) -> Mg(NO3)2 + Cu(s) 6) Mg – 1 Mg – 1 Cu – 2 Cu – 1 N – 2 N – 2 O – 6 O - 6 • The equation is unbalanced, go • back to step 3.

  13. Example #1: Write a balanced chemical equation for the reaction of magnesium with aqueous copper (I) nitrate to form aqueous magnesium nitrate and solid copper. Second Round: 3) & 4) Copper is unbalanced 5) Mg(s) + 2 CuNO3(aq) -> Mg(NO3)2 + 2 Cu(s) 6) Mg – 1 Mg – 1 Cu – 2 Cu – 2 N – 2 N – 2 O – 6 O - 6 7) The equation is balanced.

  14. Example #2: Write a balanced chemical equation for the reaction of solid vanadium (V) oxide with hydrogen gas to form solid vanadium (III) oxide and liquid water. 1) V2O5(s) + H2(g) -> V2O3(s) + H2O(l) 2) V – 2 V – 2 O – 5 O – 4 H – 2 H – 2 3) & 4) Oxygen is unbalanced. 5) V2O5(s) + H2(g) -> V2O3(s) + 2 H2O(l) 6) V – 2 V – 2 O – 5 O – 5 H – 2 H – 4

  15. Example #2: Write a balanced chemical equation for the reaction of solid vanadium (V) oxide with hydrogen gas to form solid vanadium (III) oxide and liquid water. 7) The equation is unbalanced. Go back to Step 3. Second Round: 3) & 4) Hydrogen is unbalanced. 5) V2O5(s) + 2 H2(g) -> V2O3(s) + 2 H2O(l) 6) V – 2 V – 2 O – 5 O – 5 H – 4 H – 4 7) The equation is balanced.

  16. Example #3: Liquid ethanol, C2H5OH(l), reacts with gaseous oxygen, O2(g), to produce carbon dioxide, CO2(g) and gaseous water, H2O(g). Write a balanced equation for this reaction. 1) C2H5OH(l) + O2(g) -> CO2(g) + H2O(g) 2) C – 2 C – 1 H – 6 H – 2 O – 3 O - 3 3) Carbon and hydrogen are unbalanced. 4) Carbon and hydrogen both appear in one compound on both sides sides, so we’ll start with carbon.

  17. Example #3: Liquid ethanol, C2H5OH(l), reacts with gaseous oxygen, O2(g), to produce carbon dioxide, CO2(g) and gaseous water, H2O(g). Write a balanced equation for this reaction. 5) C2H5OH(l) + O2(g) -> 2CO2(g) + H2O(g) 6) C – 2 C – 2 H – 6 H – 2 O – 3 O - 5 • The equation is unbalanced, go • back to step 3.

  18. Example #3: Liquid ethanol, C2H5OH(l), reacts with gaseous oxygen, O2(g), to produce carbon dioxide, CO2(g) and gaseous water, H2O(g). Write a balanced equation for this reaction. Second Round: 3) Hydrogen and oxygen are unbalanced. 4) Hydrogen is in only one compound on both sides. 5) C2H5OH(l) + O2(g) -> 2CO2(g) + 3H2O(g) 6) C – 2 C – 2 H – 6 H – 6 O – 3 O - 7

  19. Example #3: Liquid ethanol, C2H5OH(l), reacts with gaseous oxygen, O2(g), to produce carbon dioxide, CO2(g) and gaseous water, H2O(g). Write a balanced equation for this reaction. 6) C – 2 C – 2 H – 6 H – 6 O – 3 O - 7 7) The equation is unbalanced. Go back to Step 3. Third Round : 3) & 4) Oxygen is unbalanced. 5) C2H5OH(l) + 3O2(g) -> 2CO2(g) + 3H2O(g) • C – 2, H - 6 C – 2, H – 6 • O – 7 O - 7 7) The equation is balanced.

  20. Aqueous Solutions • Reactions occurring in aqueous solution are among the most common and most important reactions. An aqueous solution is a homogeneous mixture of a substance with water. • When ionic compounds dissolve in water, they usually dissociate with varying degrees into ions to form aqueous solutions. NaCl(s) + H2O(l) -> NaCl(aq) or Na+(aq) + Cl-(aq) CaBr2(s) + H2O(l) -> CaBr2(aq) of Ca2+(aq) + Br-(aq) • The ions in solution give rise to the ability of these solutions to conduct electricity. • Substances that completely dissociate into ions in solution are called strong electrolytes and the resultant solutions are called strong electrolyte solutions.

  21. Solubility • A substance is soluble in a particular liquid if it dissolves in that liquid. Because strong electrolytes completely dissolve in water, they are completely soluble in water. • A substance is insoluble if it does not dissolve in the liquid. Not all ionic compounds dissolve in water. For instance, silver chloride is insoluble in water and appears as a white solid at the bottom of the water. • Solubility rules are empirical rules that have been deduced from observations of many compounds and can be used to predict whether a compound will dissolve in water or not. See Table 2.2 on page 200.

  22. Example #4: Determine whether the following compounds are soluble or insoluble. If the compounds are soluble, write the ions present in solution. • Fe(OH)3 • Na3PO4 • PbBr2 • MgSO4 • Insoluble • Soluble, Na+ and PO43- • Insoluble • Soluble, Mg2+ and SO42-

  23. Precipitation Reactions • Reactions in which a solid or precipitate is formed when two aqueous solutions are mixed are called precipitation reactions. • In a precipitation reaction, two solutions containing soluble compounds react to form an insoluble compound or a precipitate. For example, when an aqueous solution of iron (III) chloride is mixed with an aqueous solution of sodium carbonate, iron (III) carbonate, an orange precipitate, and an aqueous solution of sodium chloride are formed. 2FeCl3(aq) + 3Na2CO3(aq) -> Fe2(CO3)3(s) + 6NaCl(aq)

  24. Acid and Bases • Acidsdonate protons (H+ ions). The H+ ion associates with water to form H3O+ (a hydronium ion). For example, HCl is an acid because, in solution, it donates a proton to water: HCl(aq) + H2O(l) -> H3O+(aq) + Cl-(aq) Acids have a tendency to form H+ or H3O+. • Basesacceptprotons (H+ ions). Ammonia is a base because it accepts a proton from water: NH3(aq) + H2O(l) -> NH4+(aq) + OH-(aq) Bases have a tendency to form OH-.

  25. Acid-Base (Neutralization) Reactions • When an acid is mixed with a base, the H+(aq) from the acid combines with the OH-(aq) from the base to form H2O(l) and a salt. H2SO4(aq) + 2KOH(aq) -> 2H2O(l) + K2SO4(aq) • These reactions are called neutralization reactions because the products are neutral. In other words, the products are neither acidic nor basic.

  26. Chapter 7 Homework Due October 26th 2, 3, 6, 8, 9, 12, 30, 32, 34, 38, 40, 44, 46, 50, 54

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