Chapter 19 - Chemical Reactions 1 PS 101 Kim Cohn address

Chapter 19 - Chemical Reactions 2 5 th Grade Standards 2 Students know that images of atoms and molecules show there are often well ordered arrays. Students know differences in chemical and physical properties of substances can be used to separate mixtures. Students know properties of some substances such as sugar, salt, water, helium, oxygen, nitrogen and carbon dioxide.

Chapter 19 - Chemical Reactions 3 Chemical Equation 1 Reactants  Products C(s) + O 2 (g)  CO 2 (g) 1C(s) + 1O 2 (g)  1CO 2 (g) 1 carbon atom reacts with 1 oxygen molecule to form 1 molecule of carbon dioxide.

Chapter 19 - Chemical Reactions 4 Chemical Equation 2 Usually don’t put in ones. C(s) + O 2 (g)  CO 2 (g) Conserve mass. The number of carbon atoms on the left is equal to the number of carbon atoms on the right. The same is true for oxygen atoms.

Chapter 19 - Chemical Reactions 5 Balancing Equations Fe(s) + O 2 (g)  Fe 2 O 3 (s) Focus on one element at a time. One Fe on left, two on right. Put a two in front of the Fe on the left. 2Fe(s) + O 2 (g)  Fe 2 O 3 (s) Now try oxygen

Chapter 19 - Chemical Reactions 6 Balancing Equations 2 2Fe(s) + O 2 (g)  Fe 2 O 3 (s) Now try oxygen. There are 2 on the left and three on the right. You can not change subscripts. 2Fe(s) + 3O 2 (g)  2Fe 2 O 3 (s) Fe is now wrong. 4Fe(s) + 3O 2 (g)  2Fe 2 O 3 (s)

Chapter 19 - Chemical Reactions 7 Balancing Equations 3 Try to balance H 2 (g) + N 2 (g)  NH 3 (g) 3H 2 (g) + N 2 (g)  2NH 3 (g) Try to balance CH 4 (g) + O 2 (g)  CO 2 (g) + H 2 O(l)

Chapter 19 - Chemical Reactions 8 Balancing Equations 4 Remember - conserve mass. There are 12 yellow atoms on right, therefore there must be atoms on the left. The yellow atoms on the right are found in the molecule yellow 2. The yellow atoms on the left are found in the solid (red yellow).

Chapter 19 - Chemical Reactions 9 Energy If the products have less potential energy than the reactants, the reaction will give off energy during the reaction. This type of reaction is called exothermic. The energy is usually given off in the form of heat. Lighting a match, burning food.

Chapter 19 - Chemical Reactions 10 Energy 2 Sometimes you need to put energy in to have a reaction occur. This type of reaction is called endothermic. For example, melting ice, charging a battery.

Chapter 19 - Chemical Reactions 11 Catalyst A compound that speeds up a reaction without being changed is called a catalyst. Enzymes in your mouth change starch into sugar. Gasoline is burned more quickly in the presence of platinum. Your catalytic converter uses platinum to burn all of the gasoline.

Chapter 19 - Chemical Reactions 12 Cultural Enrichment Next time you are sitting at a bar – bet the person next to you they can’t make a sugar cube burn, but you can. If you try to light a sugar cube on fire it goes out. If you rub the cube in the ashes of cigarette it will burn. There are enough metal ions in the ash to act as a catalyst for the reaction.

Chapter 19 - Chemical Reactions 13 Ozone 1 Oxygen atoms are usually found as the oxygen molecule, O 2. It is possible for three oxygen atoms to bind together to form the more energetic ozone molecule, O 3. At sea level ozone is bad for you. It destroys rubber, causes smog.

Chapter 19 - Chemical Reactions 14 Ozone 2 In the upper atmosphere, ozone is good for you. It absorbs harmful ultraviolet radiation. Too much ultraviolet radiation causes skin and plant damage. Molecules that contain the fluorine atom destroy ozone.

Chapter 19 - Chemical Reactions 15 Ozone 3 On the left of the screen is a typical molecule, difluorodichloromethane, found in cooling systems. C-Cl and C-F bonds are very stable and long lasting. The compound is volatile and floats to the upper atmosphere where it will destroy the ozone. Such compounds are no longer being used.

Chapter 19 - Chemical Reactions 16 Ozone 4 We are not sure destruction of the ozone in the atmosphere occurs to any appreciable extent and Substitutes for CF 2 Cl 2 (refrigerants) are not as good and The substitutes are more expensive. The question revolves around risk/reward ratios. Discussion.

Chapter 19 - Chemical Reactions 17 Chemical Equilibrium The object of the game is to get as many balls in your opponents court as possible.

Chapter 19 - Chemical Reactions 18 Chemical Equilibrium 2 At the start of the game B is winning.

Chapter 19 - Chemical Reactions 19 Chemical Equilibrium 3 After a few minutes A has enough balls to start throwing them back.

Chapter 19 - Chemical Reactions 20 Chemical Equilibrium 4 After a few more minutes there are just as many balls in A’s court as B.

Chapter 19 - Chemical Reactions 21 Chemical Equilibrium 5 Now both A and B are throwing balls back and forth without any change in the number of balls on each side.

Chapter 19 - Chemical Reactions 22 Chemical Equilibrium 6 This is an equilibrium condition.

Chapter 19 - Chemical Reactions 23 Chemical Equilibrium 7 If B were bigger and stronger, the equilibrium would lie to the left.

Chapter 19 - Chemical Reactions 24 Chemical Equilibrium 8 The same sort of equilibrium occurs with chemical reactions. The equilibrium can be affected by temperature and pressure. You are most familiar with the physical equilibrium between ice and water.

Chapter 19 - Chemical Reactions 25 Chemical Equilibrium 9 When water turns to ice it gives off heat. Water  Ice + heat. If you add heat to ice, the reaction goes to the ?

Chapter 19 - Chemical Reactions 26 Chemical Equilibrium 10 If the reaction gives off heat, the addition of heat pushes the reaction to the ? Consider opening a can of soda pop. Soda pop  water + gas. This reaction makes more gas molecules on the right of the reaction than on the left, the addition of pressure causes the reaction to go to the ?

Chapter 19 - Chemical Reactions 27 Chemical Equilibrium 11 When you burn gasoline in your car you also burn or oxidize some nitrogen to form NO 2. NO 2 will dimerize: 2NO 2  N 2 O 4 + Energy. NO 2 is dark brown, N 2 O 4 is light brown. Why is the air brown over LA in the summer?

Chapter 19 - Chemical Reactions 28 Relative Masses 1 A recipe calls for an equal number of both grapefruits and lemons. You could count both items in the store or Weigh them. You would not buy equal weights of both because grapefruits weigh more than lemons.

Chapter 19 - Chemical Reactions 29 Relative Masses 2 Consider: 1C(s) + 1O 2 (g)  1CO 2 (g) This recipe calls for 1 carbon atom and 1 oxygen molecule to make 1 molecule of CO 2, carbon dioxide.

Chapter 19 - Chemical Reactions 30 Relative Masses 3 1C(s) + 1O 2 (g)  1CO 2 (g) If you wanted to make CO 2 you could count out 1 carbon atom and 1 oxygen molecule and put them together to make a molecule of CO 2. Because atoms and molecules are too small to see you can’t count them out.

Chapter 19 - Chemical Reactions 31 Relative Masses 3 Because you can’t count atoms and molecules to use the recipe and make CO 2 we need to weigh atoms or molecules. To get the right proportions, we need to know the relative masses. We do! The atomic masses = the relative masses.

Chapter 19 - Chemical Reactions 32 Relative Masses 4 On the periodic table the atomic mass is usually placed under the symbol for the atom.

Chapter 19 - Chemical Reactions 33 Relative Masses 6 C(s) + O 2 (g)  CO 2 (g) The atomic mass of C is The atomic mass of one oxygen atom is The oxygen molecule contains two oxygen atoms and has a relative mass of (2 x 16) or 32. Therefore g of carbon will react with g of O 2.

Chapter 19 - Chemical Reactions 34 Relative Masses g of carbon will react with g of O 2. Or oz. of carbon will react with oz. of O 2. Or pounds of carbon will react with ? pounds of O 2.

Chapter 19 - Chemical Reactions 35 Relative Masses 8 C(s) + O 2 (g)  CO 2 (g) The atomic mass of C is 12.00, and of O 2 is g of carbon will react with g of O 2. g of carbon will react with 8.00 g of O 2 ? 3.00 g of C Here is the way you solved the problem.

Chapter 19 - Chemical Reactions 36 Relative Masses 9 First you realized that 12 g of C react with 32 g of O 2 To determine how many g of C react with 8.00 g of O 2 you set up the proportion -

Chapter 19 - Chemical Reactions 37 Relative Masses 10 And then you solved for X

Chapter 19 - Chemical Reactions 38 Relative Masses 11 C(s) + O 2 (g)  CO 2 (g) The atomic mass of C is 12.00, and of O 2 is g of carbon will react with g of O 2. g of O 2 will react with 6.00 g of C? 16 g of O 2

Chapter 19 - Chemical Reactions 39 Relative Masses 12 C(s) + O 2 (g)  CO 2 (g) 32 g of O 2 will react with 1000 g of C to produce g of CO 2 ? In this case there is more than enough C so that the only limitation on the amount of carbon dioxide produced is on the amount of oxygen. Out of the 1000 g of C you will only use up 12 g of C g of CO 2 will be produced.

Chapter 19 - Chemical Reactions 40 Relative Masses 13A C(s) + O 2 (g)  CO 2 (g) The atomic mass of C is 12.00, O 2 is g of carbon will react with g of O 2. Here is a harder problem – 12 g of O 2 will react with 8.00 g of C to produce g of CO 2 ?

Chapter 19 - Chemical Reactions 41 Relative Masses 13B 12 g of O 2 will react with 8.00 g of C to produce g of CO 2 ? If you had enough C, 12 g of O 2 would make 12/32 the amount of CO 2 or (12/32) x 44 = 16.5 g of CO 2 If you had enough O 2,8 g of C would make 8/12 the amount of CO 2 or 29.3 g of CO 2

Chapter 19 - Chemical Reactions 42 Relative Masses 14 C(s) + O 2 (g)  CO 2 (g) Remember that 6.02 x of anything is a Mole. Therefore, you can say that one mole of carbon reacts with one mole of oxygen molecules to form one mole of carbon dioxide.

Chapter 19 - Chemical Reactions 43 Relative Masses 15 Most chemists will say the atomic mass of a substance is equal to it’s weight in grams and The molecular mass of a substance is equal to it’s weight in grams. Consequently 12 g of C contains 6.02 x atoms of C. Each atom weighs 12/6.02 x g or 2 x g

Chapter 19 - Chemical Reactions 44 Relative Masses 16 What is the weight of 1 atom of C? 6.02 x atoms weigh g Therefore 1 atom weighs divided by 6.02 x or about 2.0 x g