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OB: Introduction to Thermochemistry; energy units and energy unit conversion math, review of Table B with descriptions, review of Table I and descriptions, and the inclusion of heat to Stoichiometry of balanced chemical equations. Get out reference tables, get a calculator
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Thermo = heat, the energy changes that happen in a chemical reaction. Sometimes heat is given off which is exothermic, sometimes energy is absorbed to make the reaction happen, this is endothermic. Energy Units: there are four, some you “know”, some will sound odd, that’s okay. Calories are “food” calories. A loose science word that means 1000 calories (with a small “c”) Calories are really kilo-calories, or 1000 cal. In our class we will use CAPITAL C for the food or kilo calories.
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With a lower case “c”, calories are the “science” calories, a smaller amount of energy. There are 1000 calories in 1 Calorie. A calorie is the amount of energy required to raise the temperature of one gram of water by exactly 1 Kelvin (or 1.0°C) 1000 cal = 1 C = 1 kcal
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1 Hershey bar has about 230 Calories, how many calories is that? This is a one step unit conversion problem (do it now…)
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1 Hershey bar has about 230 Calories, how many calories is that? This is a one step unit conversion problem (do it now…) 230 C 1 x 1000 cal 1 C = 230,000 cal
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Another unit of energy is the JOULE, named after J.P. Joule, the English physicist who studied energy and work. His ideas led to the development of the theory (now a law) of conservation of energy in a chemical reaction. (wowza!) Joules and calories are related with this equality: 4.18 Joules = 1.0 calories The last unit is kilo-Joule or kJ 1000 Joules = 1 kilo-Joule 1.0 Calorie = 1000 cal = 4180 joules = 4.180 kJ Converting back and forth between units will take some practice, but not much.
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An idea that sometimes escapes students is this: Food is the equivalent of energy. You eat food, which has its energy content measured in Calories. Too much food eaten = too many Calories of energy eaten = too much Charlie Eat less food means taking in less energy. You get tired quicker, and begin to use the stored energy in your body (fat becomes energy, you lose mass) That’s called going on a diet. Food is stored by the body (this is basic biology) as fat. Food (energy) is stored to be used when the body needs energy. If you never need more energy because you live now, and there are supermarkets and lunchrooms, you don’t need to worry about your hunting skills, or the weather, then the season of less food never comes. Your body stores the excess energy you eat (in the form of cupcakes or salmon, food is energy), and you grow. Food is energy. Calories turn into usable energy, or stored energy. The more food = the more energy
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Table B (let’s take a lookie-look at what gifts we have there!) The Physical Constants of Water. These 3 constants are for water. Every substance will have these 3 constants, although their numbers will be different. The Heat of Fusion The heat of fusion for water means the amount of energy required to be added to one gram of ice at the melting point, to melt it into one gram of water, without changing the temperature. This means, 273 K solid to 273 K liquid. It’s the cold phase change energy for water. It also means the reverse: how much energy needs to be removed from one gram of liquid water at the freezing point to turn it into a solid. Heat of fusion for water is 334 Joules/gram
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This is ONE GRAM of ICE CUBE. 1 gram of H 2 O (S) To MELT this gram of ice it takes a CERTAIN AMOUNT of energy added in, called the HEAT OF FUSION for water. + 334 Joules of energy Energy must be added
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This is ONE GRAM of WATER. 1 gram of H 2 O (L) To FREEZE this gram of ice it takes a CERTAIN AMOUNT of energy removed, called the HEAT OF FUSION for water. – 334 Joules of energy Energy must be removed
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The Heat of Vaporization The heat of vaporization for water means the amount of energy required to be added to one gram of water at the boiling point, to vaporize it into one gram of steam gas without changing the temperature. This means, 373 K liquid to 373 K gas. It’s the hot phase change energy for water. It also means the reverse: how much energy needs to be removed from one gram of steam at the condensing point to turn it into a liquid. Heat of vaporization for water is 2260 Joules/gram ----------------------------------------------- For water… H F = 334 J/g H V = 2260 J/g
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The last constant for water here is the specific heat capacity. This is the amount of energy required to change the temperature of one gram of water by one Kelvin (or one degree centigrade). This change can be heating, or cooling. Energy added will heat the water, energy removed will cool the water. This constant can only be used when H 2 O is liquid phase, and there is a change in temperature. for water, the specific heat capacity (C) is C = 4.18 Joules/gram x Kelvin C = 4.18 J/g·K ----------------------------------------------- Go to table B now. Under it, in small letters, add: 1 C = 1000 cal 4.18 J = 1 cal 1000 J = 1 kJ
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Table I shows us the HEATS OF REACTION for 25 different balanced chemical reactions. This means how much energy is released (exo) or absorbed (endo) by these reactions when they occur at room temperature and normal pressure. Let’s look over this first reaction, the combustion of methane gas. CH 4 + 2O 2 ---> CO 2 + 2H 2 O we can read this as: 1 mole CH 4 combusts with 2 moles O 2, into 1 mole CO 2 + 2 moles H 2 O
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The ΔH for this reaction is -890.4 kJ/mole which means that when one mole of methane combusts, it releases 890.4 kJ of energy. (the - sign only means exothermic!) There’s NO SUCH THING as negative energy. All energy is energy. So, this becomes... CH 4 + 2O 2 ---> CO 2 + 2H 2 O + 890.4 kJ we can read this as: 1 mole CH 4 combusts with 2 moles oxygen, into 1 mole CO 2 and 2 moles water + 890.4 kJ energy
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CH 4 + 2O 2 ---> CO 2 + 2H 2 O + 890.4 kJ Remember how to do stoich? You can determine the answer to this (right) If 23.4 moles of methane combust, how many molecules of water form? Well, now I’m telling you that ENERGY is also in the mole ratio as well. the mole ratio here is 1:2:1:2:890.4 kJ You could do this problem now too… If 23.4 moles of methane combust, how many kJ of energy are released?
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If 23.4 moles of methane combust, how much energy is released? Or, if I were feeling devilish on a Wednesday: If 23.4 moles of methane combust, how many Calories of energy is released (or joules, or calories)? MR CH 4 energy 1 890.4 kJ 23.4 X kJ X = (890.4)(23.4) = 20,835.36 kJ X = 20,800 kJ with 3 SF
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