As long as the relative ratios are constant the amounts are correct. Using these ratios to determine quantities is called Stoichiometry.


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1 The Meaning of the Balanced Equation Tuesday, October 11, :05 PM The Balanced Equation is a measure of the relative amounts of a compounds that participate in or are produced by a reaction. Since it is a relative amount, as long as the ratios are kept consistent the values are valid. EXAMPLE 2 SO 2(g) + O 2(g) > 2 SO 3(g) This is a balanced equation, and as with all balanced equations the lowest whole number ratios are used. However if we were talking about actual molecules, it is just as valid to say 100 molecules of Sulfur Dioxide react with 50 molecules of Oxygen gas to produce 100 molecules of Sulfur Trioxide. Likewise we could say, 2 moles of Sulfur Dioxide reacts with 1 mole of Oxygen gas to form 2 moles of Sulfur Trioxide. As long as the relative ratios are constant the amounts are correct. Using these ratios to determine quantities is called Stoichiometry. Chapter 6 Page 1
2 MoleMole Conversions Tuesday, October 11, :05 PM Since balanced equations are relative ratios, we can use the relationships in a balanced equation as conversion factors. EXAMPLE Given the following equation 2 SO 2(g) + O 2(g) > 2 SO 3(g) We know the following ratio exist Moles SO 2(g) 2 1 Moles O 2(g) =  = =  Moles SO 3(g) 2 1 Moles SO 3(g) 2 With these ratios we can calculate how much of any reactant is used to produce a known amount of product, or vice versa. Given 3.4 moles of Sulfur Trioxide, how many moles of Oxygen gas must be consumed?? Moles O 2(g) = 3.4 moles SO 3(g) (1 mole O 2(g) / 2 moles SO 3(g) ) = 1.7 mole O 2(g) PROBLEMS Given the following balanced equation for the combustion of Methanol, calculate the number of moles of oxygen gas that will need to react with 1.00 gallon of methanol (1.00 gallon contains 94.5 moles of Methanol). 2 CH 3 OH (l) + 3 O 2(g) > 2 CO 2(g) + 4 H 2 O (g) Chapter 6 Page 2
3 Given the reaction of Al (s) (produces a 3+ ions) and Cl 2(g), answer the following. 1. What is the most likely type of reaction here? 2. Write the balanced chemical equation for this reaction? 3. If you react mol of Al (s), how many mol of Cl 2(g) will be needed? Chapter 6 Page 3
4 MassMass Conversions Tuesday, October 11, :06 PM We an also use these ratios to determine the mass or volume of a substance needed (or produced) in a reaction. However, since the masses (or volumes) are not directly related to each other in the equation we need to convert them to a ratiovalid unit. The unit will use is the mole, since all masses (and volumes, once converted to mass through the density formula) can be converted to moles through the molar mass value of the compound. To solve a problem of this type follow these steps. 1. Convert any nonmass values to mass (i.e. volume, through density) 2. Convert given mass values to number of moles through the use of the molar mass value for the compound. 3. Use the ratios from the balanced equation (as described in the last section) to figure out the quantity needed in moles. 4. Using molar mass again, convert the newly determined number ofmoles to a mass. EXAMPLE 2 SO 2(g) + O 2(g) > 2 SO 3(g) Given g of sulfur dioxide, how many grams of oxygen gas would be needed to react? SO 2(g) = g/mole O 2(g) = g/mole Ratio of SO 2(g) :O 2(g) = 2:1?moles SO 2(g) = g (1 mole/64.07 g) = moles?g O 2(g) = moles (1/2) (32.00 g/1 mole) = g The combined mass of all the reactants must equal the combined mass of all the products (Law of Conservation of Mass). PROBLEM Given the following equation, answer the questions that follow. 2 H 2(g) + O 2(g) > 2 H 2 O (l) A. What type of reaction is this? B. If you use 1.80 X 10 5 kg of hydrogen, what quantity of oxygen will be needed? Chapter 6 Page 4
5 C. What mass of water will be produced? 2. The oxidation of Al(s) with oxygen gas produces Al2O3(s). Answer the following questions. A. Write the balanced equation for the reaction. B. Suppose a sheet of aluminum gains g of mass, what mass oxygen reacted with the metal? C. What mass of Aluminum reacted with the oxygen? D. How much Aluminum (III) Oxide was produced? Chapter 6 Page 5
6 Limiting Reactants Tuesday, October 11, :06 PM While a reaction will proceed according to the wholenumber ratios set out in the chemical reaction, it will always be limited by a component. The limiting reagent is the reagent compound/element present in the least amount needed for the reaction. For example 2 H2(g) + O2(g) > 2H2O(l) is the balanced equation for the combination of oxygen and hydrogen to make water. If we react 4 moles of hydrogen gas with 1 mole of oxygen gas, what would the product be? According to the balanced equation we react 2 moles of hydrogen gas for every mole oxygen, producing 2 moles of water But we have four moles of hydrogen, what happens to the extra 2 moles? Nothing, they simply do not react, therefore oxygen is our limiting reagent PROBLEMS 1. Recall the reaction to produce sulfur trioxide from sulfur dioxide and oxygen. If we react 3 moles of oxygen gas with 2 moles of sulfur dioxide, what will be limiting reagent and the product of the reaction? 2. The same type of problem can be done with mass conversions. Assume the same reaction as above, but we use g sulfur dioxide and 156g oxygen gas. What is the limiting reagent and what is the product mass? Chapter 6 Page 6
7 Chapter 6 Page 7
8 Percent Yield Tuesday, October 11, :06 PM In the last Problem we calculated that the reaction should produce 292 grams of Sulfur Trioxide. This is the maximum amount that can be produced in a reaction, called the theoretical yield. However, life rarely goes as planned and inefficiencies and other factors lower the amount a reaction produces from the maximum. The amount we actually recover is called the actual yield. We use the percent yield to describe the relationship between the two. actual yield Percent Yield = 100% X theoretical yield Example If the sulfur trioxide reaction produced 200 g of SO 3, what would the percent yield be? actual yield 200 g Percent Yield = 100% X = X 100% = 68.5% theoretical yield 292 g PROBLEM What is the percent yield of a the following reaction: 2Ag (s) + Cl 2(g) > 2AgCl (s) A. If you use 3.5 moles of silver and 1.8 moles of chlorine, but get 0.85 moles of Silver Chloride? Chapter 6 Page 8
9 B. If you use g Silver (where silver is limiting) and recover g Silver Chloride? Chapter 6 Page 9
10 Energy Changes Tuesday, October 11, :06 PM Okay, we are going a little off the reservation here. The text will come back to this topic later and treat it much more accurately, so we are going to only gloss over some of these concepts right now. Thermodynamic Laws (though your book skirts around this issue here) Law of Conservation of Energy  In a closed system (such as a chemical reaction) the amount of energy in the system must remain constant, though it may change forms. The book discusses efficiency and avoids that it is discussing the second law of Thermodynamics, Law of Increasing Entropy  all closed systems will progress towards increasing entropy. Reactions can be classified on the basis of the energy change that occurs during them. Exothermic (more appropriately Exergonic) is a reaction that releases energy. Endothermic (more appropriately Endergonic) is a reaction that absorbs energy Quantities of Heat Heat is the energy transferred between objects due to a difference in temperature. What Unit do We Use for Heat? Calorie (cal) = The amount of heat required to raise 1 g of water 1 C. 1 Calorie = Joules (J) *the preferred unit for chemistry However, there is a different Calorie (Cal) that is the unit you are used to from food. 1 Cal = 1000 cal Example (from text) If an American soda contains 180 Cal and an Australian soda contains 900 J, which is the diet soda? Chapter 6 Page 10
11 First convert the values to the same units.?joules = 180 Cal (1000 cal/1 Cal) (4.184 J/cal) = J Therefore the American soda has significantly more energy content than the Australian soda, and the Australian soda must be the diet soda. Specific Heat  The amount of heat required to raise the temperature of 1 g of a given substance 1 C. It is a unique value for any substance and must be determined empirically. Example Aluminum has a specific heat of J/(g C) and copper has a specific heat of cal/(g C), which would be a better material for cookware, if your sole criteria is how fast it heats up? Since specific heat is a measure of how much energy is needed to heat a material, a lower specific heat would mean less energy is needed to raise the temperature of a substance. Therefore since the specific heat of copper is almost an order of magnitude lower than that of aluminum, copper would be a better choice. Using specific heat we are able to calculate the heat absorbed or released by a substance with the following formula. where the following is true. q = m C ΔT q= heat m = mass of material C = specific heat of material ΔT = change in temperature (positive means temperature increases, negative temperature decreases) EXAMPLE If a 1.00 kg block of aluminum (specific heat J/(g C)) is heated from 34 C to 48 C, how much heat is absorbed by the block of aluminum? Chapter 6 Page 11
12 q = m C ΔT PROBLEM What is heat change when 55.0 g of water cools from 60.0 C to 25.0 C (use your text to determine the appropriate specific heat to use)? Calorimetry Using the Law of Conservation Energy we can determine the heat lost or gained in a reaction by detecting the heat lost or gained from/to the surroundings. The simplest setup for this is a calorimeter, as described in your text (Figure 6.18) Example A brick is placed in a water calorimeter containing kg of water. The temperature of the water decreases from 25.0 C to 19.4 C when equilibrium was reached. A) Was the brick initially warmer or colder than the water? B) What was the heat change q of the brick? A) The brick was initially colder than the water since the water loses heat when the brick is added. B) Chapter 6 Page 12
13 q= unknown m = kg = 5000 g C = J/(g C) ΔT = 19.4 C C = 5.6 C q = m C ΔT q = (5000 g) (4.184 J/(g C)) 5.6 C = kj for water q water + q brick = 0 q water =  q brick Therefore q brick = 11.7 kj PROBLEM A sample of a metal alloy is heated and then placed in 125.0g water held in a calorimeter at 22.5 C. The final temperature is 29.0 C. Assume heat exchange only between the alloy and the water. What is the heat change q for the alloy. Chapter 6 Page 13
14 Chapter 6 Page 14
15 Heat Changes in a Reaction Tuesday, October 11, :06 PM Bomb Calorimetry We can expand on the system in the last setup to determine the amount of heat released in a chemical reaction, such as a combustion reaction. The results are reported in either kilojoules per gram (kj/g) or kilojoules per mole (kj/mol). PROBLEM Isopropanol (C 3 H 7 OH) has a heat change of 54 kj/g, what would be the heat change in kj/mol? What would be the final temperature of a bomb calorimeter containing kg of water at an initial temperature of 25.0 C if you combusted 4.00 g of isopropanol? Chapter 6 Page 15
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