Unit 13 Chemical Equations

Transcription

1 Unit 13 Chemical Equations INTRODUCTION The fashion in which elements or compounds interact with one another is one of the most basic parts of chemistry. Man has been concerned with chemical interactions for thousands of years. Ancient alchemists searched vainly for a reagent that would turn lead into gold. Chemical interactions that produce useful substances are the mainstay of our economy. Pharmaceuticals, building products and synthetic materials are just few of the results of chemical interactions. A large part of the study of chemistry is devoted to these chemical reactions making their qualitative and quantitative interpretations one of the most important parts in the study of chemistry. OBJECTIVES 1. The students will identify the reactants and products in a chemical reaction. 2. The student will interpret qualitatively chemical reactions. 3. The student will quantitatively interpret chemical reactions using the Law of Conservation of Mass. DISCUSSION A. Chemical Equations, Qualitative s A chemical equation is a description of the qualitative and quantitative features of a chemical reaction. In chemical reaction some substances whether elements or compounds or both interact with one another to form new compounds with different properties. In this course we will deal only with the observed outcome of a reaction. Further courses in chemistry will be necessary for you to be able to predict what the outcome of the reaction will be and why it came about. A chemical equation is the shorthand notation for a chemical reaction and is written in the following general fashion. Reactants Products reacts to form, yields, produces. Reactants are those elements or compounds that interact. There can be many reactants or just one reactant. Products are the new compounds that are formed as a result of the combination of decomposition of reactants. There may be many or just one.

2 The arrow between reactants and products indicates the direction of the reaction and is read as reacts to form, yields or produces. Some examples of chemical reactions and their qualitative interpretations follows. Example Problem (1) Equation H H 2 0 Hydrogen and Oxygen react to form water Example Problem (2) Equation H C0 H 2 +C0 2 Water plus Carbon Monoxide yields Hydrogen and Carbon Dioxide. Example Problem (3) Equation C 8 H C0 2 + H 2 0 Octane plus Oxygen produces Carbon Dioxide and water. The interpretations are all qualitative. They tell us what reacts and what the products are. B. Chemical Equations, Quantitative s Along with the qualitative part of a chemical equation there is a quantitative part. This part tells us what amounts of reactants will be needed to produce certain amounts of products. The quantitative aspect is governed by the Law of Conservation of Mass. This Law states that in a chemical reaction there is no detectable loss or gain in mass as reactants are converted to products. The atoms in the reactants simply recombine to form new compounds as products. A chemical equation that obeys the Law of Conservation of Mass is said to be balanced. An equation is balanced by inserting simple whole numbers of fractional coefficients in front of the symbols for the elements or formulas for the compounds so that there are the same number of atoms of each element on both sides of the equation, as reactants and products. The three example equations would be balanced and interpreted as follows.

3 Example Problem (4) Balanced Equation 2 H H 2 0 The equation is balanced because there are 4 Hydrogen atoms and two Oxygen atoms as reactants and 4 Hydrogen atoms and two Oxygen atoms as products. This is because two molecules of Hydrogen (H 2 ) each containing two atoms of Hydrogen would give a total of four atoms of Hydrogen. One molecule of Oxygen (0 2 ) contain two atoms of Oxygen. The products are two molecules of water contain two atoms of Hydrogen per molecule for a total of four and one atom of Oxygen for a total of two. Two molecules of Hydrogen plus one molecule of Oxygen react to form two molecules of water. Example Problem (5) Balanced Equation 1 H C0 1 H C0 2 The equation is balanced as it is written. One molecule of water plus one molecule of Carbon Monoxide yields one molecule of Hydrogen and one molecule of Carbon Monoxide. Note* As in chemical formulas the ones as coefficients may be omitted. The balanced equation can correctly be written as: Example Problem (6) H C0 H 2 + C0 2 Balanced Equation 2 C 8 H C H molecules of Octane plus 25 molecules of Oxygen react to form 16 molecules of Carbon Dioxide and 18 molecules of water. In this course you will not be asked to balance an equation but only to identify whether it is balanced or not. Balancing will be taken up in Chemistries 101 and 102.

4 C. Chemical Equations and the Mole Perhaps nowhere else is the mole concept more important than in dealings with chemical equations. This concept allows us to bridge the gap between atoms and molecules and elements and compounds. It allows us to interpret equations on a level that we can realistically deal with in the chemical laboratory. Let us review in table form how the mole concept fits in with elements and atoms. Substance Element Compound Smallest Part Atom Molecule Relative Weight Atomic Weight (AW) Molecular Weight (MW) Weight of an Arogodros number of smallest parts Number of smallest parts in one mole Gram Atomic Weight (GAW) Gram Molecular Weight (GMW) x atoms x Mass of one mole 1 mole = GAW 1 mole = GMW We can now interpret chemical equations in a mole complete fashion. Example Problem (7) Equation 2H H 2 0 Molecular 2 molecules of H molecule of molecules of H 2 0 Molecular 2 molecules of H mole of moles of H 2 0 This interpretation can be made because the mole ratio of 2:1:2 for H 2, 0 2 and H 2 0 is exactly the same as the molecule ratio of 2:1:2 for the corresponding number of molecules in the balanced equation. We have only increased the magnitude of the equation by Arogodros number. 2 molecules of H 2 0 x N = 2 moles of H 2 1 molecule of 0 2 x N = 1 mole of molecules of H 2 0 x N = 2 moles of H 2 0 Again, the important feature is that the mole ratio is exactly the same as the ratio of molecules. Two further interpretations may be made as a result of the mole concept. They include the equality between the mole and the gram atomic or gram molecular weight and the actual gram atomic or gram molecular weight of the substance.

5 Equation 2 H H 2 0 Molecular 2 molecules of H molecule of molecules of H 2 0 Mole Concept 2 moles of H mole of moles of H 2 0 Gram Molecular Weight 2 GMW s of H GMW of GMW s of H 2 0 Weight 4 grams of H grams of grams of H 2 0 The weight relationship was made by converting gram molecular weights or moles to grams and then multiplying by the coefficient. Reatants H 2 ;1 GMW = 2g = 1 mole of H ; 1 GMW = 32g = 1 mole of 0 2 Product H 2 0; 1 GMW = 1 mole of H 2 0 2H H 2 0 2g 2 mole mole H 2 32g 1 mole mole O 2 18g 2 mole 1mole H 2 O 4g of H 2 32g of g of H 2 0 The Law of Conservation at Mass is more easily seen now since 36 grams of reactants (rg of H g of 0 2 ) produced 36 grams of product H 2 0. The most convenient fashion to approach a chemical equation on a macro level is to interpret it in terms of the number of moles of reactants and products in the balanced equation and use the resulting mole ratios for calculations. The weight ratios that are found may be used, but they are a result of the mole ratios and are not as convenient to work with. Let us interpret the other two equations thoroughly.

6 Example Problem (8) Equation H C0 H 2 + C0 2 Molecular 1 molecule of H molecule of C0 1 molecule of H molecule of C0 2 Mole Concept 1 mole of H mole of C0 1 mole of H mole of C0 2 Gram Molecular Weight 1 GMW of H GMW of C0 GMW of H GMW of C0 2 Weight Ratio 18 grams of H grams of C0 2 grams of H grams of C0 2 Comparing the mole ratio to the weight ratio we see that they are not the same due to the different molecular weights of the molecules. Mole Ratio H 2 0 C0 H 2 C0 2 1 : 1 : 1 : 1 Weight Ratio 18 : 28 : 2 : 44 The reaction obeys the Law on Conservation of Mass. There are 46 grams of reactants (18 grams of H grams of C0) being converted to 46 grams of products (4 grams of H grams of C0 2 ). Example Problem (9) Equation 2 C 8 H C H 2 0 Mole Concept 2 moles of C 8 H moles of moles of C moles of H 2 0 Weight Ration 228 grams of C 8 H grams of grams of C grams of H 2 0 The next unit will deal with calculations involving balanced chemical reactions. There correct interpretation is a necessity.

7 PROBLEMS 1. Interpret the following chemical reactions: a. Identify the reactants and products b. Interpret them on a molecular level c. Interpret them on a mole level d. Interpret them on a weight level e. Show that they are balanced by conserving numbers of atoms as well as mass in grams of reactants and products 2 Na + Cl 2 2Na Cl S S0 2 N 2 + O 2 2NO N N0 2 CH 4 + Cl 2 CH 3 Cl + HCl N H 2 2 NH 3 CuSO 4 + Zn ZnSO 4 + Cu Key C0 2 + H 2 0 H 2 C0 3 NaCl = Sodium Cholride S0 2 = Sulfur Dioxide N0 = Nitric Oxide N = Dinitrogen Tetraoxide N0 2 = Nitrogen Dioxide CH 4 = Methane NH 3 = Ammonia CuS0 4 = Copper Sulfate ZnS0 4 = Zinc Sulfate H 2 C0 3 = Carbonic Acid