Unit 5 Notes Chemistry I CP Page 1

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1 Unit 5 Notes Chemistry I CP Page 1 Chemistry Annotation Guide If you are NOT using the following annotation, put in your key to the left of each item. Your Key Mr. T s Key Items to be annotated Circle Box Write DEF Underline headings and subheadings. key content vocabulary next to definitions (sometimes #2 and #3 will be the same and in that case I expect to see a box AND DEF) important ideas (include captions for visuals). Mr. T s or Write Eq Underline Write Write in the margin next to an important equation. key steps to solving an example problem, or write those steps in the margin. any questions in the margin. a learning objective next to each subheading. Please note that I have annotated the 1 st section for you to help you get a better idea of what my expectations are. Please take some time and see what I have done and use this as a guide for you annotations. Using this handout effectively The studying and annotations and other homework assignments are designed to require a maximum of 45 minutes to an hour of your time. This time limit, however, does not include things like test make-up points or extra credit activities. Students are expected to plan their time. If students wait until the last minute to complete assignments that are assigned multiple days in advance, then the time required to complete the assignment may require more than the 1 hour limit. This document is a work in progress. I have made an attempt to put in as much about the knowledge you need for this unit of study as I can. This document, however, DOES NOT contain everything that you need to know to make 100% on the Unit 4 test. You must also rely on the knowledge that you should have acquired in previous units of study, classroom explanations, your notes, the textbook sections that were assigned for study, and in some cases creative thinking and problem-solving skills. Equations and Reactions This document is a work in progress. I have made an attempt to put in as much about the knowledge you need for this unit of study as I can. This document, however, DOES NOT contain everything that you need to know to make 100% on the Unit 5 test. You must also rely on the knowledge that you should have acquired in prerequisite classes and previous units of study, classroom explanations, your notes, handouts and worksheets assigned to you to study or complete, and in some cases creative thinking and problem-solving skills. Unit 5: Chemistry I Honors The Standards in This Unit According to the South Carolina Science Standard C-4, students will demonstrate an understanding of the types, the causes, and the effects of reactions. Most of the material in this unit will come from standard.

2 Unit 5 Notes Chemistry I CP Page 2 Def Def Def Under standard South Carolina Science Standards C-4 is a list of indicators and under each of those indicators are supporting documents. All the material in this section stems from these indicators and supporting documents. Key questions for this next section: 1) What is a reaction? 2) What is an unbalanced equation? 3) What is a balanced equation? 4) What is a complete ionic equation? 5) Why don t we show solid compounds in ionic form? 6) What do the symbols (aq) and (s) mean? 7) What is the difference between a complete ionic equation and a net ionic equation? Introduction to types of equations Since this unit is all about equations let s define a few things first. A equation is a symbolic representation of a reaction. A equation can be balanced or unbalanced. An unbalanced equation. NH 3 + O 2 NO + H 2 O In this unbalanced equation, the formulas of the s show how many atoms of each element there are in each compound, but the number of atoms on the left side of the equation (reactants) do not equal the number of atoms on the other side (products). There are 3 hydrogens on the left but only 2 on the right. A balanced equation uses coefficients in front of the formulas in the equation to balance the number of atoms on both sides of the equation. A balanced equation. 4NH 3 + 5O 2 4NO + 6H 2 O Your learning objective for this section: After studying this section I should know how to recognize and I should know the difference between 4 kinds of equations: 1) an unbalanced equation, 2) a balanced equation, 3) a complete ionic equation, and 4) a net ionic equation. What is a learning objective? A learning objective is an outcome statement that captures specifically what knowledge, skills, and attitudes that you should be able to exhibit after having studied and annotated the section. In what form should it be written? After studying this section, I will be able to. For example: After studying this section I should be able to figure out whether a substance is ionic or covalent by its properties and to write and recognize molecular formulas, empirical formulas and structural formulas. In this balanced equation the atoms in the compound formulas on the left side of the equation (reactants) equals the number of atoms in the compound formulas on the other side (products). There are 2 nitrogen atoms on each side, 6 hydrogen atoms on each side, and 8 oxygen atoms on each side. In addition to these complete unbalanced and balanced equations, there are other kinds of equations that we will learn about in this unit. An ionic equation shows the monoatomic and polyatomic ions in a solution. There are 2 types of ionic equations:

3 Unit 5 Notes Chemistry I CP Page 3 Def A complete ionic equation A complete ionic equation shows all the ions as separate formulas in the equation as well as any compounds for which there is no separation of ions. The separate formulas represent the ions from ionic compounds that separate in a solution. For an ionic equation to occur some of the reactants and products must be ionic compounds and they must take place in a way that breaks the ionic compound into separate ions. This occurs in water. For example, in this reaction 3(NH 4 ) 2 CO 3 (aq) + 2CrCl 3 (aq) Cr 2 (CO 3 ) 3 (s) + 6NH 4 Cl(aq) the ionic reactants and products listed as aqueous (aq) are those that have dissolved in water and have separated into distinct ions. When some ionic compounds dissolve in water, they for the most part have split into separate ions. Other ionic compounds do not separate. Here is how the complete ionic equation is written: Def Def 6NH 4 (aq) + 3CO 2 3 (aq) + 2Cr 3+ (aq) + 6Cl (aq) Cr 2 (CO 3 ) 3 (s) + 6NH 4 (aq) + 6Cl (aq) You should take note that the chromium(iii) carbonate product [Cr 2 (CO 3 ) 3 (s)] is listed as a solid (s). This means that for the most part it does not dissolve and it does not dissolve because for the most part it does not separate into distinct ions. Later we will use solubility rules to predict what ionic compounds will for the most part separate into distinct ions and dissolve, and what ionic compounds will not. A complete ionic equation such as the one above that shows a) all the unseperated compounds and ions (separated compounds) formed from the starting reagents and b) all the unseperated compounds and ions when the reaction is complete. A net ionic equation A net ionic equation eliminates spectator ions (spectator ions are those in the complete ionic equation that start out as ions and remain as ions when the reaction is done). The complete ionic equation above becomes a net ionic equation, therefore, when we get rid of the ions that are exactly the same on both sides of the equation. Here is what the complete ionic equation above looks like when the spectator ions are removed leaving only the net ionic equation: Product 3CO 2 3 (aq) + 2Cr 3+ (aq) Cr 2 (CO 3 ) 3 (s) Question for the teacher: What are solubility rules? What does solubility mean? Will I have to memorize these rules or are they in the test references? Question for the teacher: Why do we need to know about complete and net ionic equations? How are they used?

4 Unit 5 Notes Chemistry I CP Page 4 Indicator C-4.1 Indicator C-4.1 in the South Carolina Science Standards students should be able to analyze and balance equations for simple synthesis, decomposition, single replacement, double replacement, and combustion reactions. Key questions for this next section: Your learning objective for this section: 8) What do the symbols (g) and (l) mean? 9) The symbols (aq) and (s) can be used on the same formula. What change has occurred when these different symbols are used? 10) The symbols (g), (l), and (s) can be used on the same formula. For example water can be represented as H 2 O(s), H 2 O(l), or H 2 O(g). What change has occurred when these different symbols are used 11) What do the symbols A, B, C, and D represent in a symbolic reaction? 12) What is a single replacement reaction? What happens to the reactants in a single replacement reaction? 13) What symbolic reaction illustrates the 2 forms of a single replacement reaction? 14) Is the order of the A, B, C, and D important? Why or why not? When are they important and when are they not important? 15) In a real equation, how do you determine that a single replacement reaction is possible? 16) What does the term and mean? What are these parts when ionic compounds are involved in a reaction? What are these parts when molecular compounds are involved in a reaction? 17) What is a double replacement reaction? What happens to the reactants in a double replacement reaction? 18) In a real equation, how do you determine that a double replacement reaction is possible? 19) What is a decomposition reaction? What happens to the reactants in a decomposition reaction? 20) In a real equation, how do you recognize a decomposition reaction? 21) What common term is used for the more scientific term combustion? 22) What is the broad definition of a combustion reaction? 23) What kind of reactants will we be using in combustion reactions? What reactant will ALWAYS be present in a combustion reaction? What is the proper formula for the reactant that is ALWAYS present? 24) What happens to the reactants in a combustion reaction? 25) In a real equation, how do you recognize a combustion reaction? 26) What products are ALWAYS present in a combustion reaction using a hydrocarbon fuel? What is the proper formula for the products that are ALWAYS present? One of the products is most often thought of as a liquid? What is that product and why is it a gas in a combustion reaction? How does it work? 27) In this class what system must you use to balance equations?

5 Unit 5 Notes Chemistry I CP Page 5 Classifying equations Prerequisite knowledge In Physical Science, students were supposed to have learned to: Apply a procedure to balance equations for a simple synthesis or decomposition reaction. (PS- 4.9) Recognize simple equations (including single replacement and double replacement) as being balanced or not balanced. (PS-4.10) In this unit of chemistry students should: Classify typical equations based on the composition of the reactants Single replacement Double replacement Synthesis (composition) Decomposition Combustion Balance any reaction when given the reactants and the products, including the notations used to indicate the phase of the substance. Here are some examples of the notations used to indicate the phase of the substance : Cl 2 (g) H 2 O(l) NaCl(s) The (g) indicates that this substance is a gas. Cl 2 is the symbol for molecular chlorine. So Cl 2 (g) is the symbol for chlorine gas. The (l) indicates that this substance is a liquid. H 2 O is the symbol for water. So H 2 O(l) is the symbol for water as a liquid. The (s) indicates that this substance is a solid. NaCl is the symbol for sodium chloride (table salt). NaCl(s) is the symbol for sodium chloride as a solid. NaCl(aq) The (aq) indicates that this substance is a dissolved in water. NaCl is the symbol for sodium chloride (table salt). NaCl(aq) is the symbol for sodium chloride dissolved in water. We are about to learn about different forms of reactions. Before we do that, let s make sure that we understand some symbols. When classifying reactions, we use the variables A, B, C, and D. These letters do NOT represent any particular element. They represent compounds or parts of a compound. They could represent atoms, molecules, or ions depending on the context.

6 Unit 5 Notes Chemistry I CP Page 6 Let s break each of the classifications down separately: Recognizing single replacement equations Classify typical equations based on the composition of the reactants Single replacement In a single replacement reaction you can think of the reactants (the s listed on the left side of the equation) as a and a. In a single replacement reaction the replaces one of the 2 parts of the. In a single replacement reaction metals replace metal ions, metals replace hydrogen cations (H + ), or halogens replace monoatomic halogen anions. You may recall that monoatomic halogen anions always have a 1 charge. Here are 2 illustrations of single replacement reactions: Illustration reaction #1: A + BC AC + B Illustration reaction #2: A + BC BA + C The order of the s on the left side of the arrow (the reactants) is not important. Reactant A can be written 1 st or BC can be written 1 st. Likewise the order of the products is not important. AC can be written 1 st or B can be written 1 st in the products of illustration reaction #1. Also, products BA can be written 1 st or C can be written 1 st in the products of illustration reaction #2. Students need to be able to look at the reactants in a potential reaction and determine my visual inspection/analysis that the reaction has the potential to be a single replacement reaction. Example: Illustration reaction #1: A + BC AC + B Example reaction #1: 3Mg(s) + CrBr 6 (aq) 3MgBr 2 (aq) + 6Cr(s) Note that the 1 st in the reaction above is a. There is only one element. The 2 nd is a. It is made of 2 elements. So, the 2 nd has the potential to separate. You need to understand and remember that the potential to separate does not mean that it will always separate. Part of learning to be a high school chemist is learning to recognize when the potential to separate is higher or lower. In this case the potential is high and we can tell this because the symbol (aq) tells you that CrBr 6 has already separated into these separate ions in water: Cr 6+ and Br. While the order of the s in the equation is not important (as long as you don t switch the products and reactants), the order in which you write the 2 parts of the compounds is. The more positive part is almost always written 1 st. When metals replace metals in single replacement reactions, the metals must always be written before the nonmetal or the polyatomic anion. If a single replacement reaction is written in which a halogen replaces a halogen anion, then the halogen will go on the back because it will be the more

7 Unit 5 Notes Chemistry I CP Page 7 negative part of the formula. Example: Illustration reaction #1: A + BC BA + C Example reaction #2: Cl 2 (g) + NaBr(aq) NaCl(aq) + Br 2 (l) Notice that Cl 2 is a even though there are 2 atoms in this elemental molecule. Chemicals are not s just because there are 2 atoms. Even s that have 3 elements may only be s. To understand this, you might want to review how we put atoms and ions together to form compounds in the last unit. Ionic compounds tend to separate and rearrange themselves in reactions as ions. Molecular compounds tend to separate and rearrange themselves in reactions as atoms. Don t forget that the phrase tend to means that there are exceptions to this guideline. Note: I have balanced these equations for you in these examples, but you will have to learn to balance equations on your own. How you will be expected to do that will be shown to you later. Recognizing double replacement equations Double replacement In a double replacement reaction you can think of the reactants (the s listed on the left side of the equation) as a and a. In a double replacement reaction the front and back parts of the first changes places with the 2 parts of the second. In a double replacement the reactants and products are almost always ionic compounds. Here are 2 illustration forms of double replacement reactions: Illustration reaction #2: AB + CD AD + CB Illustration reaction #2: AB + CD CB + AD As with the single replacement reactions, the order of the s in the equation is not important (as long as you don t switch the reactants and products) but the order of the parts of the compound are. The more positive part of the formula must be written first in the formula. This means that the front part in either reactant formula will remain the front part in the product formula. Example: Illustration reaction #2: AB + CD CB + AD Example reaction #3: Ba(OH) 2 (aq) + MgCl 2 (aq) Mg(OH) 2 (s) + BaCl 2 (aq) Note: Again, this equation has been balanced for you in this example, but you will have to

8 Unit 5 Notes Chemistry I CP Page 8 learn to balance equations on your own. How you will be expected to do that will be shown to you later. Recognizing synthesis equations Synthesis (composition) You can think of a synthesis reaction as 2 or more reactants combining into a single or at the very least a smaller number of products. Here is an illustration form of a synthesis reaction: Illustration reaction #3: A + B AB As with the previous reactions, the order of the s in the reactants side of the equation (the left side) is not important but the order of the parts of the compound are. The more positive part or more electronegative part) of the formula must be written first in the formula. This means that the front part in either reactant formula will remain the front part in the product formula. Example: Illustration reaction #3: A + B AB Example reaction #4: N 2 (g) + 3H 2 (g) 2NH 3 (g) Note: Again I have balanced this equation for you in this example, but you will have to learn to balance equations on your own. How you will be expected to do that will be shown to you later. Recognizing decomposition equations Decomposition You can think of a decomposition reaction as the opposite of synthesis; 2 or more reactants that combine into one product. Here is an illustration form of a decomposition reaction: Illustration reaction #4: AB A + B As with the previous reactions, the order of the s in the products side (right side) of the equation is not important (as long as you don t switch reactants and products) but the order of the parts of the compounds is. The more positive part or more electronegative part) of the formula must be written first in the formula. This means that the front part in either reactant formula will remain the front part in the product formula.

9 Unit 5 Notes Chemistry I CP Page 9 Example: Illustration reaction #4: AB A + B Example reaction #5: 2H 2 O 2 (l) O 2 (g) + 2H 2 O(l) Note: Again I have balanced this equation for you in this example, but you will have to learn to balance equations on your own. How you will be expected to do that will be shown to you later. Recognizing combustion equations Combustion The common or non-scientific term for combustion is burning. When we say that a piece of paper or wood is burned, we are saying that the paper or wood is going through combustion and the implication is that the hydrocarbons in paper or wood are reacting with oxygen to produce carbon dioxide and oxygen. When any substance is burned in air, then that substance is reacting with oxygen. Combustion is a complex set of reactions. Its broadest definition is any reaction between a fuel and an oxidant accompanied by the production of heat or both heat and light in the form of either a glow or flames. In this class, we will only deal with the simplest of combustion reactions; those between a hydrocarbon or organic fuel and oxygen from the air and which produce carbon dioxide gas, water (usually as a gas), light, and heat. Here is an illustration form of single replacement reaction: Hydrocarbon or organic fuel + O 2 (g) CO 2 (g) + H 2 O(g) water Contains From the Carbon Usually carbon, air or from dioxide this is a hydrogen, and the gas sometimes atmosphere because oxygen of the heat + heat + light Sometimes symbolized Δ As with the earlier reactions, the order of the s in the equation is not important (as long as you don t switch the reactants and products) but the order of the parts of the compound are. Oxygen, carbon dioxide, and water have standard formulas. The hydrocarbon can be written in a condensed structural formula or a molecular formula (C x H y O z ). Example: C 2 H 4 (g) + 3O 2 (g) 2CO 2 (g) + 2H 2 O(g) + heat + light The reactions in combustion reactions using organic fuels will have many things that are exactly alike: a) atmospheric oxygen is ALWAYS one of the reactions, atmospheric oxygen ALWAYS has the formula O 2 (g) b) carbon dioxide gas will ALWAYS be a product, carbon dioxide gas will ALWAYS have the formula CO 2 (g) Sometimes symbolized hν

10 Unit 5 Notes Chemistry I CP Page 10 c) water in the gas phase (also known as steam) will ALWAYS be a product, water gas (steam) will ALWAYS have the formula H 2 O(g). The reason that water in combustion reactions is is ALWAYS steam is that combustion reactions always produce 2 other products; light (sometimes symbolized hυ) and heat (sometimes symbolized Δ), and the heat results in water being in the gas phase. Note: Again I have balanced this equation for you in this example, but you will have to learn to balance equations on your own. How you will be expected to do that will be shown to you later. Physical state symbols for the elements and compounds in a equation Students should be able to balance any reaction when given the reactants and the products, including the notations used to indicate the phase of the substance. The 4 physical state symbols that you need to know are: g for gas, l for liquid, s for solid, and aq for aqueous (which means dissolved in water). These symbols are always written in a parenthesis following the symbol of an element, ion, or compound. Here are some examples: Cl 2 (g) chlorine gas H 2 O(l) water as a liquid NaCl(s) sodium chloride as a solid NaCl(aq) sodium chloride dissolved in water So, the symbol (g) means that the immediately before it is a gas, the symbol (l) means that the immediately before it is a liquid, the symbol (s) means that the immediately before it is a solid, and the symbol (aq) means that the immediately before it is an aqueous solution which means it is dissolved in water. Balancing equations In this class we will always use a process called an atom inventory (AI) to balance an equation. Here s how it works. Step 1: List the elements in the equation under the unbalanced equation in an atom inventory chart. C 2 H 4 O 2 (l) + O 2 (g) CO 2 (g) + H 2 O(g) C H O

11 Unit 5 Notes Chemistry I CP Page 11 Step 2: Count the atoms and insert them in the AI chart. You have to count the atoms on either side of the equation separately. C 2 H 4 O 2 (l) + O 2 (g) CO 2 (g) + H 2 O(g) C 2 1 H 4 2 O 4 3 Step 3: Save oxygen for last and hydrogen for next to last, but put coefficients in front of the formulas in the equation to try to get the numbers of atoms to be the same on both sides of the equation and in the chart. In this case, since we re saving oxygen for last, and hydrogen for next to last, the only place to start is with carbon. Since there are 2 carbons on the left and 1 on the right, putting a coefficient of 2 in front CO 2 and ought to balance the carbons. Once the coefficients are added, you must recalculate the number of atoms and enter those numbers into your AI chart. C 2 H 4 O 2 (l) + O 2 (g) 2CO 2 (g) + H 2 O(g) C H 4 2 O Step 4: Now we keep entering coefficients until we get all the atoms balanced. In this case, entering a coefficient of 2 in front of H 2 O (water) and a 2 in front of O 2 (oxygen) should do it. Once the coefficients are added, you must recalculate the number of atoms and enter those numbers into your AI chart. C 2 H 4 O 2 (l) + 2O 2 (g) 2CO 2 (g) + 2H 2 O(g) C H O So now the formulas and the overall equation are balanced. One last tip for balancing equations that contain atmospheric oxygen: O 2 (g). Often you will reach a point where the number of oxygen atoms on one side of the equation is an odd number and the number of oxygen atoms on the other side is an even number. The quick solution is to go through the entire equation and multiply all the coefficients by 2 except the one in front of atmospheric oxygen. You can then recalculate the number of atoms and determining the number of oxygen atoms that are needed and enter the correct coefficient.

12 Unit 5 Notes Chemistry I CP Page 12 Practice Problems: Classify and balance the following unbalanced equations using the atom inventory process. 1. Fe 2 O 3 (s) + H 2 (g) Fe(s) + H 2 O(g) 2. H 2 O 2 (aq) H 2 O(l) + O 2 (g) 3. C 3 H 8 (l) + O 2 (g) CO 2 (g) + H 2 O(g) 4. Na 2 CO 3 (aq) + HCl (aq) NaCl(aq) + H 2 CO 3 (aq) 5. N 2 O 5 (aq) + H 2 O(l) HNO 3 (aq)

13 Unit 5 Notes Chemistry I CP Page 13 Indicator C-4.2 According to Indicator C-4.2 in the South Carolina Science Standards students should be able to predict the products of acid-base neutralization and combustion reactions. Key questions for this next section: 28) How do you recognize that a single replacement reaction is possible? 29) What is the activity series of the elements and how is it used? 30) What is the driving force for a single replacement reaction? 31) When do metals replace metal ions in a compound? 32) When does hydrogen replace metal ions in a compound? 33) What steps should you use for balancing a equation? 34) When do you write NR on the right side of a single replacement equation? 35) How do you recognize that a double replacement reaction is possible? 36) What are the solubility rules and how are they used? 37) What are the driving forces for a double replacement reaction? 38) When double replacement reaction occur and when do they NOT occur? 39) When do you write NR on the right side of a double replacement equation? 40) What is a neutralization reaction? 41) What is a combustion reaction? 42) How do you recognize that a combustion reaction is likely to occur? 43) What are the products of a combustion reaction? Prerequisite knowledge In Physical Science, students were supposed to have learned to: Your learning objective for this section: Classify various solutions as acids or bases according to their physical properties, properties (including neutralization and reaction with metals), generalized formulas, and ph (using ph meters, ph paper, and litmus paper). (PS-3.8)

14 Unit 5 Notes Chemistry I CP Page 14 Predicting the products of reactions In this unit of chemistry students should: Predict the products for the following classifications of reactions when given the reactants. Single replacement Students should know how to use a table showing The Activity Series of the Elements Replacement of a metal with a more active metal Replacement of hydrogen in an acid by a metal Double replacement Synthesis Students should know how to use a table of solubility rules Formation of a precipitate including Formula equation Ionic equation Net ionic equation Neutralization Combustion of hydrocarbons Let s break each of these classifications of reactions/equations down separately: Predicting the products of a single replacement reaction Single replacement As was stated earlier, in a single replacement reaction you can think of the reactants (the s listed on the left side of the equation) as a and a. In a single replacement reaction the replaces one of the 2 parts of the. In a single replacement reaction metals replace metal ions, metals replace hydrogen cations (H + ), or halogens replace monoatomic halogen anions. You may recall that monoatomic halogen anions always have a 1 charge. Here are 2 illustration forms of single replacement reactions: Illustration reaction #1: A + BC AC + B Illustration reaction #2: A + BC BA + C The order of the s is not important. Reactant A can be written 1 st or BC can be written 1 st. Likewise the order of the products is not important. AC can be written 1 st or B can be written 1 st in the products of illustration reaction #1. Also, products BA can be written 1 st or C can be written 1 st in the products of illustration reaction #2. Students need to be able to look at the reactants in a potential reaction and determine my

15 Unit 5 Notes Chemistry I CP Page 15 visual inspection/analysis that the reaction has the potential to be a single replacement reaction. Example: If you see s in the form Illustration reaction #1: A + BC Example reaction #6: Zn(s) + Fe(NO 3 ) 3 (aq) you should be able to determine that a single replacement reaction is possible. Before going on to the next subject I would like to point out what the symbols (s) and (aq) should be communicating to you. (s) means that the listed immediately in front of it is a solid. (aq) means that the listed immediately in front of it is a solution in water. The combination of these 2 symbols in this equation should communicate to you that the reaction is taking place in water with zinc starting out as a solid and iron(iii) nitrate starting out as a solution. I would also like to remind students that they must be able to recognize polyatomic ions and to determine the charge on a mono-atomic cation when the metal can have more than one oxidation number. Iron(III) nitrate is a compound made of an Fe 3+ iron(iii) cations and 3 NO 3 nitrate ions. Being able to predict the likely products accurately will depend on your being able to determine both the charge on the cation and the charge on the anion.

16 These elements will react with oxygen to form oxides These elements will react with acids to form hydrogen gas (H2) These elements will react with steam to form hydrogen gas (H2) These elements will react with water and form hydrogen gas (H2) Unit 5 Notes Chemistry I CP Page 16 Lastly, students will need to remember how to write a balanced formula. Students should know how to use a table showing The Activity Series of the Elements The Activity Series of the Elements gives chemists a tool for making predictions about the likely or possible products in a single replacement reaction. With slight modifications, here is the Activity Series that students in this class will be using in class, in homework, and on tests (it is already a part of the set of references that students were given earlier in the semester): Elemental form Activity Series of Metals Compound form Li Li + Rb Rb + K K + Ba Ba 2+ Sr Sr 2+ Ca Ca 2+ Na Na + Mg Mg 2+ Al Al 3+ Mn Mn 2+ Zn Zn 2+ Cr Cr 2+ Fe Fe 2+ Cd Cd 2+ Co Co 2+ Ni Ni 2+ Sn Sn 2+ Pb Pb 2+ [H 2 ] H + Included for comparison Activity Sb Sb 3+ Bi Bi 2+ Cu Cu 2+ Elemental form Series of Halogens Compound form Hg Hg 2+ F 2 F Ag Ag + Cl 2 Cl Pt Pt + Br 2 Br Au Au 3+ I 2 I 1. Any metal higher in the activity series will displace another metal in a single displacement reaction. 2. Metals above water may react with water rather than a metal compound. In the illustration and example reaction/equation shown above, zinc (Zn) is the. If you look at where zinc is on The Activity Series and where iron (Fe) is you will see that zinc is above iron. Take a second and put your fingers on both iron and zinc to help yourself physically and visually reference where they are relative to each other. Elements that are higher on The Activity Series will replace elements that are part of ionic compounds that are also lower on The Activity Series. I ve used the term element here, but in reality the element in an ionic compound is really an ion in the compound. In this

17 Unit 5 Notes Chemistry I CP Page 17 case the iron(iii) nitrate compound each the iron atom in has 3 electrons and, therefore is not really iron the atom anymore it s iron(iii) the cation with a 3+ charge (Fe 3+ ). Driving force for a single replacement reaction: A single replacement reaction will NOT occur without a driving force. In this class, the driving force will be an element that is by itself in a equation that is higher on the activity series than an element in a compound. An element that is higher on the activity series is said to be more active than an element that is lower. Without a driving force, students at this point in the semester will assume that the reaction does not occur and the student must write NR and only NR on the right side of the equation. Replacement of a metal with a more active metal In the example reaction above ( Example reaction #1) zinc is a more reactive metal than iron, so it will replace iron in the iron(iii) nitrate compound. Students should be able to examine the potential reactants of zinc and iron(iii) nitrate and using The Activity Series predict that zinc will replace iron in this fashion: Illustration reaction #1: A + BC AC + B Example reaction #1: Zn(s) + Fe(NO 3 ) 3 (aq) Zn(NO 3 ) 2 (aq) + Fe(s) To be successful in performing the prediction above, students must have mastered the knowledge and skill necessary to balance formulas that were taught in an earlier unit. By looking up zinc on a table of oxidation numbers (this was one of the references that students were given to use in class, on homework, and on tests), students should be able to determine that zinc has only one oxidation number; a 2+. By looking up nitrate on a table of common polyatomic ions (this, too, was one of the references that students were given to use in class, on homework, and on tests), students should be able to determine that nitrate has a charge and oxidation number of 1. To produce a balanced compound (a compound with a neutral charge or no charge overall), you need a compound with 1 zinc ion and 2 nitrate ions. So, the balanced formula for a compound of zinc nitrate is Zn(NO 3 ) 2. Recall that the subscript 2 that follows the nitrate ion (which is in parentheses) is how chemists show that there are 2 nitrate ions. Recall also that this is like algebra, but we re just using a different notation system. Now, we re not really done yet. Recall from indicator 4.1 that you should be able to balance the equation, and we haven t done that yet. We ve balanced the formulas as part of our prediction on the possible products, but now we have to balance the overall reaction. Recall that we use a system called Atom Inventory (AI) to balance an equation.

18 Unit 5 Notes Chemistry I CP Page 18 Step 1: List the elements in the equation under the unbalanced equation in an atom inventory chart. Zn(s) + Fe(NO 3 ) 3 (aq) Zn(NO 3 ) 2 (aq) + Fe(s) Zn Fe N O Step 2: Count the atoms and insert them in the AI chart. Remember that you have to count the atoms on either side of the equation separately. Zn(s) + Fe(NO 3 ) 3 (aq) Zn(NO 3 ) 2 (aq) + Fe(s) Zn 1 1 Fe 1 1 N 3 2 O 9 6 Step 3: Save oxygen for last an hydrogen for next to last, but put coefficients in front of the formulas in the equation to try to get the numbers of atoms to be the same on both sides of the equation and in the chart. In this case, zinc and iron are balanced, and since we re saving oxygen for last, the only place to start is with nitrogen. Since there are 3 nitrogens on the left and 2 on the right, putting a coefficient of 2 in front of Fe(NO 3 ) 3 and a coefficient of 2 in from of Zn(NO 3 )2 ought to balance the nitrogens. Once the coefficients are added, you must recalculate the number of atoms and enter those numbers into your AI chart. Zn(s) + 2Fe(NO 3 ) 3 (aq) 3Zn(NO 3 ) 2 (aq) + Fe(s) Zn Fe N O

19 Unit 5 Notes Chemistry I CP Page 19 Step 4: Now we keep entering coefficients until we get all the atoms balanced. In this case, entering a coefficient of 3 in front of Zn (zinc) and a 2 in front of Fe (iron) should do it. Once the coefficients are added, you must recalculate the number of atoms and enter those numbers into your AI chart. 3Zn(s) + 2Fe(NO 3 ) 3 (aq) 3Zn(NO 3 ) 2 (aq) + 2Fe(s) Zn Fe N O So now the formulas and the overall equation are balanced. So what happens when the metal that is the is NOT higher on The Activity Series than the metal cation in the. The answer is, Nothing. That does not mean that you just don t write an equation. It means that you write a reaction with no s on the right side. Instead of products you write NR for no reaction. Here s an example: Illustration reaction #1: A + BC NR Example reaction #7: Ni(s) + AlCl 3 (aq) NR No Reaction Replacement of hydrogen in an acid by a metal Go back and look at The Activity Series of Elements and take note of where H 2 (hydrogen gas) is located. All of the metals above H 2 will replace a hydrogen cation (H + ) in an acid producing hydrogen gas. Even water becomes an acid when it reacts with the most reactive of the metals. We need to step back at this point and explain what an acid is. In the last unit when we were naming acids, we said that a non-organic acid has one or more hydrogen on the front of the formula and that organic acids have this group O C on the end of a carbon or hydrocarbon chain. That s just what an acid formula looks like. It s not really as acid until it gives away hydrogen cations (H + ), and acids usually do this in water. Recall that a very small number of water molecules will autoionize to form hydrogen cations (H + ). H O H So this means that water IS an acid. Further, in the presence of a very reactive metal OH H + + [ O H ]

20 Unit 5 Notes Chemistry I CP Page 20 a lot of water molecules will ionize. The sodium gives up electrons to the hydrogen ions allowing them to covalently bond and for hydrogen gas [H 2 (g)] leaving behind sodium cations and hydroxide ions. The result is a more basic solution. So, if we take a metal that is above H 2 on The Activity Series and put that metal in an acid, a single replacement reaction takes place. Here s an example using chloric acid and iron (we ll assume that the product will include an iron(iii) cation): Illustration reaction #1: A + BC AC + B Example reaction #8: Fe(s) + HClO 3 (aq) Fe(ClO 3 ) 3 (aq) + H 2 (g) Example of a single replacement (or single displacement) reaction that will NOT occur: The following is an example an equation with s that will not react because there is no driving force: Fe(s) + Zn(NO 3 ) 2 (aq) NR The reason that these s will not react is that zinc (Zn) is higher on the activity series than is iron (Fe). When there is no reaction, you MUST write NR on the right side of the arrow and you MUST NOT write an formulas there. Notice in the Activity Series of Halogens that fluorine (F), chlorine (Cl), bromine (Br), and iodine (I) in their elemental forms are all diatomic. Diatomic means that they have the formula F 2, Cl 2, Br 2, and I 2 and these are molecules in which these atoms covalently bond with each other. So, you MUST know that these elements almost never appear in nature in single, unbounded form. Even in their pure, elemental form, the atoms of these elements are covalently bonded in pairs. In addition, you MUST know that under normal conditions fluorine and chlorine will be gases. So, not only must you know to write the pure elemental form of fluorine and chlorine as F 2 and Cl 2, you must also write their physical states as (g). So, in equations, fluorine and chlorine will be written in this form: F 2 (g) and Cl 2 (g). Bromine, on the other hand, is usually a liquid: Br 2 (l). Under normal conditions, iodine can be both a gas or a solid, so for now you will have to be given the physical state for iodine. While we are on the subject of gases, you MUST know that the following pure elements will also be diatomic gases under normal conditions: hydrogen H 2 (g), nitrogen - N 2 (g), and oxygen - O 2 (g). You MUST also know that under normal conditions carbon dioxide is also a gas: CO 2 (g). You MUST also know that in their pure state and under normal conditions, the noble gases will be monoatomic gases: helium He(g), neon Ne(g), argon Ar(g), krypton Kr(g), and radon Rn(g). Predicting the products of a double replacement reaction Double replacement Students should know how to use a table of solubility rules A set of solubility rules is shown below. This is the set of solubility rules that students in this class will be using in class, in homework, and on tests (it is already a part of the

21 Unit 5 Notes Chemistry I CP Page 21 set of references that students were given earlier in the semester): SOLUBILITY RULES: SOLUBLE: All Nitrates, Acetates, Ammonium, Chlorates, Perchlorates, and Group I salts All Chlorides, Bromides, and Iodides, except Silver, Lead, and Mercury(I) All Fluorides except Group II, Lead(II), and Iron(III) All Sulfates except Calcium, Strontium, Barium, Mercury, Lead(II), and Silver INSOLUBLE: All Carbonates and Phosphates except Group I and Ammonium All Hydroxides except Group I, Strontium, and Barium All Sulfides except Group I, II, and Ammonium All Oxides except Group I INSOLUBLE means a precipitate forms when equal volumes of 0.10 M solutions or greater are mixed The solubility rules are used to determine when 2 dissolved salts will react together to form a solid product. A that forms a solid in water is called a precipitate. In the solubility rules, insoluble compounds are considered solids. Driving force for a double replacement reaction: A double replacement reaction will NOT occur without a driving force. In this class, the formation of a solid (as a product) is one of 3 driving forces for a double replacement reaction. The other 2 are the formation of a gas or water (formation means it s a product; on the right side of the equation). You will learn more about the formation of water later. Without a driving force, students at this point in the semester will assume that the reaction does not occur and the student must write NR and only NR on the right side of the equation. In the example reaction below, IF a reaction is to occur, then the front part of sodium hydroxide (NaOH) must trade places with the back part of iron(iii) nitrate [Fe(NO 3 ) 3 ]. Illustration reaction #1: AB + CD AD + CB Example reaction #9: NaOH(aq) + Fe(NO 3 ) 3 (aq) The POSSIBLE products of this reaction, then, are Illustration reaction #1: AB + CD AD + CB Example reaction #9: NaOH(aq) + Fe(NO 3 ) 3 (aq) NaNO 3 + Fe(OH) 3 Now, look at the solubility rules. Note that ALL nitrate compounds are soluble. If they are soluble, they are aqueous. Hydroxides, on the other hand, are mostly insoluble and are therefore solids. There are a handful of exceptions. By exceptions I mean that some hydroxides are soluble (including hydroxides of Group I elements, Strontium, and Barium). Since iron(iii) hydroxide is not an exception then it must be insoluble. If it s insoluble (meaning it forms a solid) then that provides a driving force for the reaction.

22 Unit 5 Notes Chemistry I CP Page 22 Now you can write the symbols to indicate soluble (aq) and insoluble (s) compounds and balance the equation. Remember that you must use an atom inventory (AI) to balance the equation. Illustration reaction #1: AB + CD AD + CB Example reaction #9: 3NaOH(aq) + Fe(NO 3 ) 3 (aq) 3NaNO 3 (aq) + Fe(OH) 3 (s) Recall that to be successful in performing the prediction above, students must have mastered the knowledge and skill necessary to balance formulas that were taught in an earlier unit. Example of a double replacement reaction that will NOT occur: The following is an example an equation with s that will not react because there is no driving force: Fe(CH 3 CO 2 ) 3 (aq) + Zn(NO 3 ) 2 (aq) NR The reason that these s will not react is that the expected products, Fe(NO 3 ) 3 and Zn(CH 3 CO 2 ) 2 are both soluble and therefore aqueous. Since no solid, or water, or a gas is formed, there is no driving force. Another way to think about this is this: if all the products of a double replacement reaction are aqueous, there is no driving force and therefore there is no reaction (NR). Formation of a precipitate including Formula equation The equations that we ve been writing so far are formula equations. They show all the formulas of the compounds. Complete ionic equation (or simply ionic equation ) Complete ionic equations simply take all the soluble salts in the equation and break them up into ions. Only the aqueous salts (which can also be called aqueous or soluble ionic compounds) get broken up into separate ions. Example: 3NaOH(aq) + Fe(NO 3 ) 3 (aq) 3NaNO 3 (aq) + Fe(OH) 3 (s) 3Na + (aq) + 3OH (aq) + Fe 3+ (aq) + 3NO 3 (aq) 3Na + (aq) + 3NO 3 (aq) + Fe(OH) 3 (s) Note that iron(iii) hydroxide is a solid, so it doesn t get broken up into separate ions. Also take note that the resulting ions have the same physical state symbol as the ionic compound formula unit. Here s an example of a how to create a complete ionic equation from a single replacement reaction. The solid metals and gases do not get broken up, just as the precipitates do not get broken up.

23 Unit 5 Notes Chemistry I CP Page 23 Standard Equation 2Fe(s) + 6HClO 3 (aq) 2Fe(ClO 3 ) 3 (aq) + 3H 2 (g) Complete ionic equation 2Fe(s) + 6H + (aq) + 6ClO 3 (aq) 2Fe 3+ (aq) + 6ClO 3 (aq) + 3H 2 (g) Note that atmospheric hydrogen (H 2 ) is a gas, so it doesn t get broken up; and that iron metal (Fe) is a solid, so it doesn t get broken up. Also, take note that the resulting ions have the same physical state symbol as the ionic compound formula unit. You also need to know that pure elements like iron (Fe) and hydrogen (H 2 ) have a zero (0) oxidation number. So, it is acceptable to write the complete ionic equation in this way. In an ionic equation, you may write metals in their elemental form with NO oxidation number or with a 0 oxidation number, but you MUST NOT write the pure elemental form of a metal with a positive oxidation number of any kind. 2Fe 0 (s) + 6H + (aq) + 6ClO 3 (aq) 2Fe 3+ (aq) + 6ClO 3 (aq) + 3H 2 0 (g) Net ionic equation Net ionic equations simply take all the ions that are the same on both sides of the equation and cancels them out. Here s how it might look: Standard Equation 3NaOH(aq) + Fe(NO 3 ) 3 (aq) 3NaNO 3 (aq) + Fe(OH) 3 (s) Complete ionic equation 3Na+(aq) + 3OH (aq) + Fe 3+ (aq) + 3NO 3 (aq) 3Na + (aq) + 3NO 3 (aq) + Fe(OH) 3 (s) Canceling ions 3Na+(aq) + 3OH (aq) + Fe 3+ (aq) + 3NO 3 (aq) 3Na + (aq) + 3NO 3 (aq) + Fe(OH) 3 (s) Resulting net ionic equation: Net ionic equation 3OH (aq) + Fe 3+ (aq) Fe(OH) 3 (s) Take note that the ions OH (aq) and Fe 3+ (aq) are not the same thing as Fe(OH) 3 (s), so they cannot cancel each other out. OH (aq) and Fe 3+ (aq) are separate ions and they are aqueous, whereas Fe(OH) 3 (s) doesn t break up into separate ions and it s a solid. Also note that Here s an example of a how to create a complete ionic equation from a single replacement reaction. Standard Equation 2Fe(s) + 6HClO 3 (aq) 2Fe(ClO 3 ) 3 (aq) + 3H 2 (g) Complete ionic equation Net ionic equation 2Fe(s) + 6H + (aq) + 3ClO 3 (aq) 2Fe 3+ (aq) + 3ClO 3 (aq) + 3H 2 (g) 2Fe(s) + 6H + (aq) + 3ClO 3 (aq) 2Fe 3+ (aq) + 3ClO 3 (aq) + 3H 2 (g) 2Fe(s) + 6H + (aq) 2Fe 3+ (aq) + 3H 2 (g) -or- 2Fe 0 (s) + 6H + (aq) 2Fe 3+ (aq) + 3H 0 2 (g) The solid metals and gases do not get broken up, just as the precipitates do not get broken up. Take note, also, that pure elements like iron (Fe) and hydrogen (H 2 ) have a zero (0) oxidation number. So Fe(s) is NOT the same thing as Fe 3+ (aq) and they cannot cancel each other out. The ions that get canceled are called spectator ions.

24 Unit 5 Notes Chemistry I CP Page 24 Neutralization or acid-base reactions Neutralization reactions are another form of a double replacement reaction. It is also an acid-base reaction. There are 3 major definitions of acids and bases, but for the acid-base reactions in this unit we will be using the Arrhenius definition. In an Arrhenius acid-base reaction acids will have hydrogen on the front and bases will have hydroxide anions on the back of their formulas. By treating these reactions just as we would any other double replacement reaction we will always form water as a product. In these acid- base reactions water will be assumed to be in the liquid state so we will use the (l) symbol after H 2 O to show that. Illustration reaction #1: AB + CD AD + CB Example reaction #10: 2NaOH(aq) + H 2 SO 4 (aq) Na 2 SO 4 (aq) + 2H 2 O(l) Recall the formation of water provides a driving force for a double replacement reaction. Recall that a small amount of water splits into 2 parts: H + and OH. Sometimes it s easier to write water as if it s a combination of these 2 ions: HOH. It is perfectly acceptable and sometimes it s even preferred to write water s formula in that fashion (as seen here). I ve included the complete and net ionic equations here. Note that water is a liquid, so it doesn t get broken up into separate ions. Example reaction #10: 2NaOH(aq) + H 2 SO 4 (aq) Na 2 SO 4 (aq) + 2HOH(l) Complete ionic 2Na + (aq) + OH (aq) + 2H + (aq) + SO 2 4 (aq) 2Na + (aq) + SO 2 4 (aq) + 2HOH(l) equation 2Na + (aq) + OH (aq) + 2H + (aq) + SO 2 4 (aq) 2Na + (aq) + SO 2 4 (aq) + 2HOH(l) Net ionic equation OH (aq) + 2H + (aq) 2HOH(l) Take note that the ions OH (aq) and H + (aq) are not the same thing as HOH(l), so they cannot cancel each other out. Predicting the products of a synthesis reaction Synthesis Recall that a synthesis reaction can be thought of as 2 or more reactants combining into a single or at the very least a smaller number of products. Here is an illustration form of a synthesis reaction: Illustration reaction #1: A + B AB

25 Unit 5 Notes Chemistry I CP Page 25 Any synthesis reaction for which you will be given reactants and for which you will be expected to predict products will form a with which that you should already be familiar. Illustration reaction #1: A + B AB Example reaction #11: N 2 (g) + 3H 2 (g) 2NH 3 (g) Example #1: Illustration reaction #1: A + B AB Example reaction #12: 2CO(g) + O 2 (g) 2CO 2 (g) Note: Again I have balanced this equation for you in this example, but you will have to learn to balance equations on your own and you will always be required to use an atom inventory. Predicting the products of a hydrocarbon combustion reaction Combustion of hydrocarbons Recall that in this class, we will only deal with the simplest of combustion reactions; those between a hydrocarbon or organic fuel and oxygen from the air and which produce carbon dioxide gas, water (usually as a gas), light, and heat. So, predicting the products of combustion reactions of the type that we will work with in this class will be very easy. The products will ALWAYS be carbon dioxide, gaseous water, heat, and light. Here is an illustration form of a combustion reaction: Hydrocarbon + O 2 (g) CO 2 (g) + H 2 O(g) + heat + light or organic fuel Contains carbon, hydrogen, and sometimes oxygen From the air or from the atmosphere A spark to get things started Carbon dioxide (water) Usually this is a gas because of the heat Sometimes symbolized Δ Sometimes symbolized hν As with the earlier reactions, the order of the s in the equation is not important (as long as you don t switch the reactants and products) but the order of the parts of the compound are. Oxygen, carbon dioxide, and water have standard formulas. The hydrocarbon can be written in a condensed structural formula or a molecular formula (C x H y O z ). Example: C 2 H 4 (g) + 3O 2 (g) 2CO 2 (g) + 2H 2 O(g) + heat + light Note: Again I have balanced this equation for you in this example, but you will have to learn to balance equations on your own and you will always be required to use an atom inventory.