Chapter 5 - Chemical Reactions

Chapter Five: Chemical Reactions General Features of Physical and Chemical Changes a physical change alters the physical state of a substance without changing its composition. a chemical change/reaction converts one substance into another; they include: • breaking bonds in the reactants • forming new bonds in the products Writing Chemical Equations a chemical equation uses chemical formulas and other symbols showing what reactants are the starting materials in a reaction and what products are formed. • the reactants are written on the left • the products are written on the left • coefficients show the number of molecules of a given element or compound that react or are formed the law of conservation of mass states that atoms cannot be created or destroyed in a chemical reaction. coefficients are used to balance an equation. a balanced equation has the same number of atoms of each element on both sides of the equation. How to Balance a Chemical Equation • write the equation with the correct formulas; the subscripts in a formula can never be changed to balance an equation because changing a subscript changes the identity of a compound • balance the equation with coefficients one element at a time • check to make sure that the smallest set of whole numbers is used Types of Reactions the majority of chemical reactions fall into six categories: • combination • decomposition • single replacement • double replacement • oxidation and reduction • acid-base Combination and Decomposition a combination reaction is the joining of two or more reactants to form a single product a decomposition reaction is the conversion of a single reactant to two or more products. Replacement Reactions a single replacement reaction is a reaction in which one element replaces another element in a compound to form a different compound and element as products. a double replacement reaction is a reaction in which two compounds exchange “parts” - atoms or ions to form two new compounds. General Features of Oxidation and Reduction Reactions oxidation is the loss of electrons from an atom. reduction is the gain of electrons by an atom. both processes occur together in a single reaction called an oxidation-reduction or redox reaction. a redox reaction always has two components, one that is oxidized and one that is reduced. it involves the transfer of electrons from one element to another. oxidation half reaction: reduction half reaction: a compound that is oxidized while causing another compound to be reduced is called a reducing agent. a compound that is reduced while causing another compound to be oxidized is called an oxidizing agent. Examples of Redox Reactions iron rusting a flashlight battery: alkaline batteries consist of zinc powder (Zn) and manganese dioxide (MnO2), along with a paste of sodium hydroxide (NaOH) or potassium hydroxide (KOH). when electrical contact is made, Zn atoms 4+ lose electrons, which flow toward the Mn cations in MnO2. the resulting electric current can be used to power a lightbulb, radio, or other electrical device. pacemakers: a pacemaker generates a small electrical impulse that triggers the heart to beat. today’s pacemakers sense when the heart beats normally and provide an electrical signal only when the heart rate slows. such devices are called “demand” pacemakers, and they quickly replaced earlier “fixed” rate models that continuously produced impulses to set the heart rate at a fixed value. Oxidation-Reduction Reactions for Molecules with Covalent Bonds molecules with covalent bonds have no net gain or loss of electrons. oxidation results in the: • gain of oxygen atoms • loss of hydrogen atoms reduction results in the: • loss of oxygen atoms • gain of hydrogen atoms The Mole and Avogadro’s Number molecular weight of H2O: 18.0 amu 23 18.0 g of H2O is 1 mole, or 6.02 x 10 molecules 23 6.02 x 10 is avogadro’s number. a mole is a quantity that contains 6.02 x 10 23 23 items. 1 mole of C atoms = 6.02 x 10 atoms (12.0 g), 1 mole of H2O molecules = 23 6.02 x 10 molecules (18.0 g), and 1 mole of vitamin C molecules = 6.02 x 10 23 molecules (176.12 g). it can be used as a conversion factor to relate the number of moles of a substance to the number of atoms or molecules: examples: how many molecules are in 5.0 moles of CO2? calculate the formula weight for FeSO4. 1 Fe atom, 1 S atom, and 4 O atoms 1 Fe atom x 55.85 amu = 55.85 amu 1 S atom x 32.07 amu = 32.07 amu 4 O atoms x 16.00 amu = 64.00 amu 55.85 + 32.07 + 64.00 = 151.92 amu Molar Mass the molar mass is the mass of one mole of any substance, reported in grams per mole (g/mol). the value of the molar mass of a compound in grams equals the value of its formula weight in amu. Relating Grams to Moles the molar mass relates the number of moles to the number of grams of a substance. in this way, molar mass can be used as a conversion factor. the molar mass of H2O is 18.02 g/mol, the conversion factor can be written: example: how many moles are present in 100. g of aspirin (C9H8O4, molar mass 180.2 g/mol)? original quantity: 100. g of aspirin desired quantity: ? mol of aspirin conversion factor: mol/molar mass Relating Grams to Number of Atoms or Molecules we can also use the molar mass to show the relationship between grams and number of molecules (or atoms). example: how many molecules are in a 325-mg tablet of aspirin (C9H8O4, molar mass 180.2 g/mol)? original quantity: 325 mg of aspirin desired quantity: ? molecules of aspirin conversion factors: grams/milligrams, mol/grams, molecules/mol Mole Calculations in Chemical Equations a balanced chemical equation also tells us: • the number of moles of each reactant that reacts • the number of moles of each product formed coefficients are used to form mole ratios, which can serve as conversion factors: use the mole ratios from the coefficients in the balanced equation to convert moles of one compound (A) into moles of another compound (B). examples: using the balanced chemical equation (balance if it is not already), how many moles of CO are produced from 3.5 moles of C2H6? original quantity: 3.5 mol C2H6 desired quantity: ? mol CO conversion factor: mol CO/mol C2H6 using the balanced equation, how many grams of O3 are formed from 9.0 mol of O2? original quantity: moles of O2 desired quantity: ? grams of O3 conversion factors: mol O3/mol O2 and g O3/mol O 3 ethanol (C2H6O, molar mass 46.1 g/mol) is synthesized by reacting ethylene (C2 H4, molar mass 28.1 g/mol) with water. how many grams of ethanol are formed from 14 g of ethylene? original quantity: 14 g of C2 H4 desired quantity: ? g of C2H6O conversion factors: mol C2 H4/g C2 H4, mol C2H6O/mol C2 H4, g C2H6O/mol C2H6O Percent Yield the theoretical yield is the amount of product expected from a given amount of reactant based on the coefficients in the balanced chemical equation. usually, however, the amount of product formed is less than the maximum amount of product predicted. the actual yield is the amount of product isolated from a reaction. example: if the reaction of ethylene (C2 H4) with water (H2O) to form ethanol (C2H6O) has a calculated theoretical yield of 23 g of ethanol, what is the percent yield if only 15 g of ethanol are actually formed? percent yield 15 g/23 g x 100% = 65% Limiting Reactants the limiting reactant is the reactant that is completely used up in a reaction. excess reactant + limiting reactant -> product + leftovers Determining the Limiting Reactant examples: identify the limiting reactant in this picture for the reaction and determine how much of one reactant is needed to react with the other. 2 H2 is chosen to be the original quantity, and O2 is chosen to be the unknown quantity. conversion factor: molecules O2/molecules H2 • if the amount present of the second reactant is less than what is needed, the second reactant is the limiting reagent. • if the amount present of the second reactant is greater than what is needed, the second reactant is in excess. in this reaction, only two molecules of O2 are needed, but three molecules of O2 are present, so: • O2 is present in excess • H2 is the limiting reactant using the balanced equation, determine the limiting reactant when 10.0 g of N2 (MM = 28.02 g/mol) react with 10.0 g of O2 (MM = 32.00 g/mol). convert the number of grams of each reactant into moles using the molar masses. conversion factor (when starting with N2) : mol O2/mol N2 the amount of O2 we started with (0.313 mol) is less than the amount we would need (0.357 mol), so O2 is the limiting reagent and N2 is in excess.