C3H8 + O2 CO2 + H2O C3H8 + O2 → CO2 + H2O, Study notes of Organic Chemistry

Download C3H8 + O2 CO2 + H2O C3H8 + O2 → CO2 + H2O and more Study notes Organic Chemistry in PDF only on Docsity! Balancing Chemical Equations Page 1 of 6 Preview The lectures in this unit cover balancing & classifying chemical reactions: including redox, acid- base, and precipitation reactions. This lecture covers how to express and balance chemical equations and introduces the three general classes of chemical reactions. Chemical Reactions I. Expressing Chemical Reactions We express chemical changes using chemical equations. In doing so we use our chemical formulas to express how atoms rearrange from reactants to products. Equations are read from left to right: with the reactant(s) on the left, an arrow which tells what direction the reaction occurs in the middle, and product(s) on the right. For example, propane reacts with oxygen to form carbon dioxide and water. We write the reaction as follows: C3H8 + O2 → CO2 + H2O Reactants Products Before we continue, please note that elemental oxygen exists as a dimer: O2. There are seven elements that exist naturally as diatomic species, meaning they exist as dimers of themselves. They are elemental hydrogen, nitrogen, oxygen, fluorine, chlorine, bromine, and iodine. You must remember these in order to use the correct form of these elements in chemical reactions! II. Balancing Chemical Equations You might have been concerned with the chemical equation shown above. Let’s look at it again: C3H8 + O2 → CO2 + H2O We know that the correct formula for propane is C3H8; and we know that when things burn they consume oxygen and produce carbon dioxide and water. Each of the formulas given are correct. But if you look closely to the equation as written it appears there are more carbon atoms on one side than on the other. This is true of the oxygen atoms and hydrogen atoms as well! In other words, the equation is not balanced. Balancing is a tool that helps us determine how many reactant molecules react to form how many product molecules. We balance an equation by systematically adding molecules or atoms on each side by inserting a coefficient in front of the atomic symbol or molecular formula. We continue to add or change these coefficients as needed until our reaction has the lowest ratio of reactants and products possible while still being completely balanced. Balancing Chemical Equations Page 2 of 6 In the case of the previous reaction, we note that one molecule of C3H8 provides three carbon atoms on the left hand side; So we must have three carbon atoms on the right hand side to balance the number of carbon atoms. We put a 3 in front of the formula on the right that contains carbon, as follows: 1 C3H8 + O2 → 3 CO2 + H2O I have included the 1 in front of the C3H8 molecule so you can see the relationship between the two molecules clearly. However, I need not include the 1 in front of the C3H8 molecule. If there is no coefficient shown it is assumed the value is 1. Please note that we do NOT change the formula of the CO2 compound. We cannot arbitrarily change that formula to C3O2 to satisfy the number of carbon atoms required. Nor do we change the formula to C3O6, even though the coefficient of 3 does indicate there are now 3 carbon atoms and 6 oxygen atoms represented. That would change the identity of the product molecule. Instead we are merely stating that one molecule of C3H8 must be capable of forming three molecules of CO2 according to the reaction given, because there are three carbon atoms in each of these expressions. Next we can balance the hydrogen atoms. The C3H8 molecule on the left hand side includes 8 hydrogen atoms. So we must have 8 hydrogen atoms in the products to balance the reaction. Because there are two hydrogen atoms in every molecule of H2O, it needs a coefficient of 4 to represent a total balanced number of 8 hydrogen atoms: C3H8 + O2 → 3 CO2 + 4 H2O Finally we can balance our oxygen atoms. There are a total of 10 oxygen atoms on the right hand side of the reaction, given by 3 CO2 molecules and 4 H2O molecules. In order to get 10 oxygen atoms on the left, I need a coefficient of 5 in front of the oxygen molecule: C3H8 + 5 O2 → 3 CO2 + 4 H2O Now our equation is balanced. It shows 3 carbon atoms, 8 hydrogen atoms, and 10 oxygen atoms on each side. Let’s try balancing another reaction: FeCl3 → Fe + Cl2 We have 3 chlorine atoms on the left and two on the right. In this case, the easiest way to balance them is to take the lowest common multiple for the number of chlorine atoms on each side. The lowest common multiple is 6 since one side has 2 and the other has 3. So we put a 2 in front of the FeCl3 and a 3 in front of the Cl2. That balances our chlorine atoms like so: 2 FeCl3 → Fe + 3 Cl2 Balancing Chemical Equations Page 5 of 6 III. Classifying Chemical Reactions There are many thousands of chemical reactions known, so it becomes useful to try and broadly categorize them into general classes of reactions. The following are three general classification types that help us understand chemical reactions. 1) Oxidation-Reduction reactions occur when elements either gain or lose electrons. If the charge on an atom changes, it has undergone oxidation or reduction and that process or reaction is an oxidation-reduction reaction. All reactions involving covalent compounds are oxidation-reduction reactions. Examples: Mg + O2 → MgO Al2O3 → Al + O2 C4H10O + O2 → CO2 + H2O Reactions where neutral elements make ionic compounds (also known as synthesis or combination reactions). Reactions where ionic compounds revert to the neutral elements (also known as decomposition reactions). Reactions where atoms in covalent compounds are rearranged. (The example here is also known as a combustion reaction). 2) Precipitation reactions occur when soluble elements in an aqueous solution combine to form an insoluble compound. Here we use phases to show what is happening. The soluble elements are given a designation of (aq) to indicate they are dissolved in aqueous, or water-based, solution. The insoluble compound is given a designation of (s) to indicate it is a solid and thus not part of the solution. Examples: Pb(NO3)2(aq) + Na2CO3(aq) → PbCO3(s) + NaNO3(aq) In this reaction solid PbCO3 is the insoluble solid formed. It is the precipitate. K2SO4(aq) + BaCl2(aq) → KCl(aq) + BaSO4(s) In this reaction solid BaSO4 is the insoluble solid formed. It is the precipitate. Balancing Chemical Equations Page 6 of 6 3) Acid-Base reactions occur when H+ ions in solution and OH- ions in solution combine to make liquid water as a product. These ions may be part of ionic compounds. The key is to look for the H2O(l) as a product, and then look for the H+ and OH- ions in the reactants. Examples: Fe(OH)2(aq) + H2SO4(aq) → FeSO4(aq) + H2O(l) In this reaction, OH- ions from Fe(OH)2 combine with H+ ions from H2SO4 to form H2O. HCl(aq) + Mg(OH)2(aq) → MgCl2(aq) + H2O(l) In this reaction, H+ ions from HCl combine with OH- ions from Mg(OH)2 to form H2O. --------------------------------------------------------------------------------------------------------------------- Take a minute to complete section II of your Chemical Equations worksheet. This has you classify a few chemical reactions according to the descriptions given above. Balance them as well for extra practice! ---------------------------------------------------------------------------------------------------------------------