CHEMICAL REACTIONS. represent energy (heat, light, energy) as either a product or a reactant by writing it on the appropriate side of the equation.

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1 Page1 CHEMICAL REACTIONS We have already discussed the fact that sometimes individual atoms (and in the case of polyatomic ions small groups of atoms) can join together to form compounds in order to increase stability and decrease the total amount of chemical potential energy. Sometimes compounds can come together and they can become more stable still by rearranging atoms. These changes are called chemical reactions, and are different from the nuclear reactions we studied earlier, which are when nuclei come together (or split apart) and create new atoms. Chemists represent chemical reactions with equations. Equations consist of a list of formulas for all of the reactants (compounds or elements added together to begin a chemical reaction) on the left side and a list of all of the products (compounds or elements which form during a chemical reaction) on the right side. Each formula may be followed by a subscript (solid (s), liquid (l), gas (g), aqueous (aq)) to show what state each compound is in. We may represent energy (heat, light, energy) as either a product or a reactant by writing it on the appropriate side of the equation. Forming A Compound Lab Objective: To observe and document the formation of a common compound 1. Label the test tubes TREATED and UNTREATED. Place the piece of steel wool into the test tubes. Push them all the way to the end. Check to make sure neither piece will fall out when the test tubes are inverted. Rubber band the two test tubes together and set aside for now. 2. Get 100 ml of water in the glass beaker. Place the test tubes in the water upside down and locate them next to the side of the beaker, NOT in the middle. Place the packing material around the test tubes so that they stay exactly upright. Try not to shake the table during the experiment. 3. Record the STARTING TIME on your lab sheet. Measure the height of the water in the two test tubes (use cm) and record that on the lab sheet. Measure from the bottom of the test tube up to water level in test tube minutes after the starting time, look for any differences in the two test tubes or the steel wool. Measure the water heights again. Record this on your lab sheet minutes after the starting time, record any differences you see between the two pieces of steel wool or the test tubes again. Measure the height of the water again. 6. After 25 minutes, record any differences you see in the two test tubes and measure the height of the water again. 1

2 Page2 7. Remove the test tubes from the beaker and add water from the beaker to each test tube until they are ½ full. Cover tops and shake the test tubes. Record any differences you see between the water in the test tubes on the lab sheet. 8. Remove the two pieces of steel wool from the test tube and place them on a paper towel. Examine them carefully using all your senses (except taste!) Record ALL the differences between the two pieces. 9. CLEAN UP! Start time: Treated H 2 O level Untreated H 2 O level Observations List 3 ways you know that a chemical reaction took place minutes minutes 25 minutes 3. CONCLUSION: The chemical reaction for this experiment is: Iron +2 + water Iron (II) oxide + hydrogen (g) Rewrite the chemical reaction using your formula writing rules. Law of Conservation of Mass: In a chemical reaction the arrangement of atoms will change, but the total number of each type of atom, and therefore the total mass, will remain constant. All chemical reactions must obey this law and therefore all chemical equations must obey it also. In order to ensure that our chemical equations obey the law of cons. of mass, we balance all of our equations. To do this: 1. You must make sure that all of the individual compound s formulas are correct. (Hydrogen, oxygen, nitrogen, and the halogens are all diatomic elements, meaning that whenever they are not found in a compound they form a molecule with themselves and have a formula with a subscript 2). Counting Atoms Problem Solving 2

3 Page3 There are several rules to counting the numbers of atoms in a compound, your job is to look at the clues and figure out what those rules are. DO NOT work out the problems on the calculation sheet until you have all four rules filled in! Here are the clues to the first rule: CaCO calcium = 1 carbon = 1 oxygen =1 NaCl sodium = 1 chlorine = 1 KOH potassium = 1 oxygen = 1 hydrogen = 1 Write the first rule about counting atoms on the next page. Here are the clues to the second rule about subscripts: H 2 O hydrogen = 2 oxygen = 1 NH 3 nitrogen = 1 hydrogen = 3 PbCrO 4 lead = 1 chromium = 1 oxygen = 4 C 14 H 9 C l5 carbon = 14 hydrogen = 9 chlorine = 15 Write the second rule about subscript on the next page. Here are the clues to the third rule about parentheses: Fe(OH) 3 iron = 1 oxygen = 3 hydrogen = 3 Sn 3 (PO) 2 tin = 3 phosphorus = 2 oxygen = 2 Be(F 2 Li) 2 beryllium = 1 fluorine = 4 lithium = 2 Al 2 (SO 4 ) 2 aluminum = 2 sulfur = 2 oxygen = 8 (Sr 2 O) 3 Mg strontium = 6 oxygen =3 magnesium = 1 Write the third rule about parentheses on the next page. Here are the clues to the fourth rule about coefficients: 2Fe(OH) 3 iron = 2, oxygen = 6, hydrogen = 6 3Sn 3 (PO) 2 tin = 9, phosphorus = 6, oxygen = 6 6Be(F 2 Li) 2 beryllium = 6, fluorine = 24, lithium = 12 10Al 2 (SO 4 ) 2 aluminum = 20, sulfur = 20, oxygen = 80 4(Sr 2 O) 3 Mg strontium = 24, oxygen =12, magnesium = 4 Write the fourth rule about coefficients on the next page. Counting Atoms Calculations and Rule Sheet First rule of counting atoms: Yours: Mine: Practice with the first rule: 1. FeS 2. HCl 3

4 Page4 3. AgCl 4. NaOH Second rule of subscripts: Yours: Mine: Practice with the second rule: 1. Fe 2 O 3 2. C 12 H 22 O KNO 3 4. MgCl 2 Third rule of parentheses: Yours: Mine: Practice with the third rule: 1. Mg(OH) 2 2. Ca(CO 3 ) 3 3. Ag 2 (SO 4 ) 2 4. (Cu 3 Os) 2 Hg Fourth rule of coefficients: Yours: Mine: Practice with the fourth rule: 1. 3Mg(OH) Ca(CO 3 ) Ag 2 (SO 4 ) (Cu 3 Os) 2 Hg Now work out the problems on the next page, following the rules you have written and then we will check your answers. Counting Atoms Practice Count the number of atoms in each element represented in the molecule. N=1 H=4 Cl=1 Total = 6 4. CaCO 3 8. Ag 3 PO 4 1. CO 2 Ex. NH 4 Cl 12. Mg(OH) 2 2. HCl 5. NH 3 9. NaOH 13. Al(OH) 3 6. PbCrO KOH 14. Ni(NO 3 ) 2 3. CaCl 2 7. Ag 2 S 11. Fe(OH) Ca(OH) 2 4

5 Page5 16. Ni(NO 3 ) HNC(NH 2 ) C 5 H 2 OH 28. 2(NH 4 ) 2 CO Al 2 (SO 4 ) CaCl Fe(OH) C 7 H 6 O Sn 3 (PO 4 ) HCl 26. 4Ni(OH) C 55 H 72 MgN (NH 2 ) 2 CO 23. 3KNO Hg(NO 3 ) 2 The number of each atom on the reactants side must equal the number of each atom on the products side. This is called BALANCING which is adding coeffscients. Balancing Equations by the Element Inventory Method There are four easy steps that you need to follow to make this work. Here they are: 1. Get yourself an unbalanced equation. I might give this to you, or I might make you figure it out by giving you a word equation or a skeleton equation. Either way, if you don't have an equation with all the correct chemical formulas and the arrow and all that other stuff, then you're out of luck. 2. Draw boxes around all the chemical formulas. Never, ever, change anything inside the boxes. Ever! Really!! If you do, you're guaranteed to get the answer wrong. 3. Make an element inventory. How are you going to know if the equation is balanced if you don't actually make a list of how many of each atom you have? You won't. You have to make an inventory of how many atoms of each element you have, and then you have to keep it current throughout the whole problem. 4. Write numbers in front of each of the boxes until the inventory for each element is the same both before and after the reaction. Whenever you change a number, make sure to update the inventory - otherwise, you run the risk of balancing it incorrectly. When all the numbers in the inventory balance, then the equation can balance, and you can relax. Example #1 Balance the equation that takes place when sodium hydroxide reacts with sulfuric acid to form sodium sulfate and water." How do we solve this using the steps above? 1. Get yourself an unbalanced equation. Here's where you use your knowledge of formulas to help you out. If you know what the formula of sodium hydroxide, sulfuric acid, sodium sulfate, and water are, you'd be able to write the following skeleton (unbalanced) equation: 5

6 Page6 2. Draw boxes around all the chemical formulas. This is the step that people frequently don't do because they feel that it's a stupid thing to do. You're drawing those boxes so that you'll be sure not to mess around with the formulas to balance the equation. Never, ever change subscripts! 3. Make an element inventory. In this inventory, your job is to figure out how many atoms of each element you have on the left and right sides of the equation. Now, if you look at the equation, you should be able to see that on the left side of the equation there is one sodium atom, five oxygen atoms (one from the sodium hydroxide, four from the sulfuric acid), three hydrogen atoms (one from the sodium hydroxide, two from the sulfuric acid), and one sulfur atom. On the right side of the equation, there are two atoms of sodium, one atom of sulfur, five atoms of oxygen (four from the sodium sulfate and one from the water), and two atoms of hydrogen. Thus, your element inventory should look like this: 4. Write numbers in front of each of the boxes until the inventory for each element is the same both before and after the reaction. Now, what happens when we put a number in front of a formula? Basically, anything in that box is multiplied by that number, because we're saying that we have that many of that kind of molecule. So, looking at the inventory, what should we do? Well, we can see that on the left side of the inventory, there is one atom of sodium and on the right there are two. The solution: Stick a "2" in front of the sodium hydroxide on the left side of the equation so that the numbers of sodium atoms are the same on both sides of the equation. When we do this, the new atom inventory should look like this: (I'll let you figure out how this is done) Now what? Well, looking at the new inventory, we can see that we now have two sodium atoms on both the left and the right sides, but the others still don't match up. What to do? You can see from the inventory that on the right side of the equation, there are two hydrogen atoms and on the left there are four. Using your amazing powers of mathematics you can see that two multiplied by the number two becomes four. That's what you need to do. How? Put a "2" in front of the water on the right side of the equation to make the hydrogens balance out. Now that this is done, you should make a new inventory that looks something like this: Since both sides of the inventory match, the equation is now balanced! All other equations will balance in exactly the same way, though it might take a few more steps in some cases. Problems you might encounter: (1) What happens when you do the inventory, and you find that there are two atoms of element X on the left side of the equation and three atoms of element X on the right. How can you make those numbers match? When you run into this problem, find the lowest common denominator (LCD) of those two numbers, and then put the numbers in front of those two boxes which allow the inventory on both sides to match. In the element X example, the lowest common denominator of two and three is six, so you'd put a "3" in front of the molecule on the left, and a "2" in front of the one on the right. Element X will 6

7 Page7 then match up, and you can use a new inventory to see what else needs to be done. This is sometimes called the criss-cross method because you are taking the subscript of the reactant and making it the coefficient of the product, and taking the subscript of the product and making it the coefficient of the reactant. (2) Another common problem: What do you do when the only way you can get a problem to work out is to make one of the numbers a decimal or fraction? This sometimes happens with combustion reactions: C 2 H 6 + O 2 CO 2 + H 2 O When this happens, find the largest molecule in the equation (in the example above: C 2 H 6 ) and stick a "2" in front of it. Then start the problem over. Will this work all the time? Well, no. But it will work sometimes, and give you a new strategy for hard problems. Most importantly: Always remember to keep the inventory of the elements current! If you try to keep it in your head, you'll screw it up. Nobody can keep a bunch of changing numbers in their head for very long. I certainly can't, and you can't either. In more difficult equations, trial and error is not effective so there are some additional rules to simplify the process. Additional Rule #1: Find the elements which appear in the fewest numbers of molecules and balance these first. Continue in sequence until you balance the element which appears in the most molecules last. For example in the following equation: NaOH + H 2 SO 4 Na 2 SO 4 + H 2 O Na and S appear in only 2 molecules, H in 3 molecules, and O in four. This is the order in which they should be balanced. The S is already balanced (1 on each side), and the Na may be balanced as follows: 2 NaOH + H 2 SO 4 Na 2 SO 4 + H 2 O The hydrogen is a little more difficult. There are two H atoms in 2 NaOH and another two in H 2 SO 4 so there are four H atoms on the reactant side of the equation. But there are only two H atoms on the product side of the equation. The H is then balanced by placing a two in front of the water: The equation is now balanced 2 NaOH + H 2 SO 4 Na 2 SO H 2 O Special hints: 1. Never change your subscripts 2. Balance elements that only appear in one formula on each side first 3. After your first free choice of a coefficient, you must only use coefficients that balance the elements 4. For polyatomic ions, balance them as though they are single groups Balance the following chemical equations. 3. Al + Fe 3 N 2 AlN + Fe 1. Cu 2 O + C Cu + CO 2 4. Ag 2 S Ag + S 8 2. H 2 O 2 H 2 O + O 2 5. ZnS + AlP Zn 3 P 2 + Al 2 S 3 7

8 Page8 16) CoBr 3 + CaSO 4 CaBr 2 + Co 2 (SO 4 ) 3 6. Fe(OH) 3 Fe 2 O 3 + H 2 O 17) Na 3 P + CaF 2 NaF + Ca 3 P 2 Given the two chemical equations, circle the one that is balanced. 7. a. 2Na + Cl 2 2NaCl 18) Mn + HI H 2 + MnI 3 b. 2Na + 2Cl 2 2NaCl 19) Li 3 PO 4 + NaBr Na 3 PO 4 + LiBr 8. a. C 3 H 8 + 5O 2 3CO 2 + 4H 2 O 20) CaF 2 + Li 2 SO 4 CaSO 4 + LiF b. 2C 3 H 8 + 5O 2 3CO 2 + 8H 2 O 21) HBr + Mg(OH) 2 MgBr 2 + H 2 O 9. a. 2NH 3 + 5O 2 2NO + 3H 2 O 22) LiNO 3 + CaBr 2 Ca(NO 3 ) 2 + LiBr b. 4NH 3 + 5O 2 4NO + 6H 2 O 23) AgNO 3 + Li LiNO 3 + Ag 10. a. Y(NO 3 ) 2 + GaPO 4 YPO 4 + Ga(NO 3 ) 2 24) Si(OH) 4 + NaBr SiBr 4 + NaOH b. 2Y(NO 3 ) 2 + 2GaPO 4 2YPO 4 + Ga(NO 3 ) 2 25) NaCN + CuCO 3 Na 2 CO 3 + Cu(CN) 2 Balance these equations! 11) AlBr 3 + K KBr + Al 12) FeO + PdF 2 FeF 2 + PdO 13) P 4 + Br 2 PBr 3 14) LiCl + Br 2 LiBr + Cl 2 15) PbBr 2 + HCl HBr + PbCl 2 8

9 Page9 Write the word equations below as chemical equations and balance them. 1) zinc and lead (II) acetate react to form zinc acetate and solid lead. 2) aluminum bromide and chlorine gas react to form aluminum chloride and bromine gas 3) sodium phosphate and calcium chloride combine to form calcium phosphate and sodium chloride 4) potassium metal and chlorine gas react to form potassium chloride 5) calcium hydroxide and hydrogen phosphate react to form calcium phosphate and water 6) water and carbon dioxide combine to yield heptacarbon octahydride and oxygen gas 7) tin (IV) sulfate and potassium phosphate react to form tin(iv) phosphate and potassium sulfate 8) tetraphosphorus solid reacts with bromine liquid to yield phosphorus pentabromide 9) potassium hydroxide reacts with calcium nitrate to produce potassium nitrate and calcium hydroxide 10) When you burn propane gas (aka tricarbon octahydride), it reacts with oxygen gas to form carbon dioxide and water vapor 9