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UNIT A: THERMOCHEMICAL CHANGES CHAPTER 11 (P.476 – 519) CHEM 30.

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Presentation on theme: "UNIT A: THERMOCHEMICAL CHANGES CHAPTER 11 (P.476 – 519) CHEM 30."— Presentation transcript:

1 UNIT A: THERMOCHEMICAL CHANGES CHAPTER 11 (P.476 – 519) CHEM 30

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3 Unit Intro Most of the energy we have and use comes from the sun. Kinetic Energy: Energy of movement. Potential Energy: Energy of an object depending on how HIGH it is above the ground. Phase Change: Refers to a change in state; (l)  (g) or (s)  (l) or (s)  (g)

4 Unit Intro The Kinetic Molecular Theory Kinetic: Means moving Molecular: aka molecules + = The Kinetic Molecular Theory: Everything is made of molecules and they are always moving.

5 Unit Intro Endothermic & Exothermic ____ ___ Endo: Means inside….these reactions absorb (take in) energy when they happen. [COLD PACKS]. Exo: Means outside….these reactions release energy when they happen [Heat Packs]. Ex……..Explosion! Do you feel heat when your next to an explosion?

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7 The process of the photosynthesis converts solar energy from the sun and converts (changes it) into chemical energy (Glucose C 6 H 12 O 6 ) Because this energy is stored in the chemical bonds that connect all the C’s, H’s, and O’s together it then has the Potential to be released at some later time, its called Chemical Potential Energy. Like a barrel full of gasoline…..theres’ always the potential (chance) that the barrel could catch fire and explode….releasing a lot of energy at once.

8 What Kind of Chemical Potential Energy Do We Use…Have A Lot of In Alberta? Where did the energy in things like Oil and Gas come from….originally?

9 From The Sun To Your Gas Tank Photosynthesis Dead Plant Covered By Sediment Millions of Years Pass By Fossil Fuel

10 Major Energy Demands: There are four of them…..what do you think?

11 Questions Pg. 482 #’s: 2, 3, 4 Pg. 483 #’s: 7,8 Pg. 484 #’s: 11, 12, 15

12 Calorimetry is the technological study of energy changes (transfers). [APPLICATION] Works on the principal that all energy lost or gained by whatever is burned or reacted will either be gained by the water (exothermic) or taken from the water (endothermic). Thermochemistry is the study of energy changes (transfers) [THEORY] Can anyone think of a problem with that?

13 Open, Closed, and Isolated Systems Open System: It exchanges matter and energy with its surroundings. (An Open Popcan) Closed System: It exchanged energy ONLY with its surroundings. (A Closed Popcan) ? Isolated System: It doesn’t exchange EITHER energy or matter with its surroundings.

14 How Can We Find Out How Much Energy Something Has? When something is burned it is assumed that all its energy is released into its surroundings…. So the amount of energy the peanut has is EQUAL to the amount of energy gained by the surroundings(The Calorimerer). ****We Can Measure This**** ***The Law of Conservation of Energy*** Hint

15 How Can We Calculate The Amount of Energy Gained By The Surroundings? The more energy something has the higher its temperature. Question: If something is burned and the surrounding (water) gains the energy (exothermic), will the temperature of the water go up or down? So lets brainstorm for a sec. When you put a pot of water on the stove and heat it up (add energy to it), what factors determine how quickly it boils? How much water your boiling Turn up the burner Salt the water?

16 Q=mc∆t Q= Amount of energy in joules or kilojoules (Kilo just means 1000)… m= The mass of substance being heated or cooled. (g or Kg) c= Is the specific heat capacity of the substance being heated or cooled. (J/g°C) ∆t= The change in temperature of the substance. *** The specific heat capacity is unique and different for every substance THE SPECIFIC HEAT CAPACITY WILL BE GIVEN TO YOU OR YOU CAN LOOK IT UP. ***The change in temperature (∆t) is just the difference between how hot something was before and after……does it get hotter?--  +∆t Colder--  -∆t Q= Amount of energy in joules or kilojoules (Kilo just means 1000)… m= The mass of substance being heated or cooled. (g or Kg) c= Is the specific heat capacity of the substance being heated or cooled. (J/g°C) ∆t= The change in temperature of the substance.

17 m= c= ∆t= ******1ml = 1 g***** 115g 4.19 J/g°C ∆t= 79.2°C Q=mc∆t Q= (115g)(4.19 J/g°C)(79.2°C) Q= 38162.52 □ ??? Units ??? Q= 38162.52 J / ~ 38.2 KJ SIGFIGS!

18 Practice

19 Bomb Calorimetry Special type of calorimetry where you don’t have a specific mass Special type of calorimetry where you don’t have a specific mass. Usually the Specific Heat Capacity (c) tells you how many joules per gram to raise the temperature 1°C. Bomb calorimetry has a C (Specific Heat Capacity) but its only energy need to raise environment (ifinit or unknown mass) by 1 °C.

20 Percent Efficiency Percent Efficiency: Also called percent yield. Is a measure of the amount of product you can actually make when you do an experiment, compare to theoretically How much you could of make theoretically when you used calculations. Formula % Efficiency = Actual Amount Theoretical Amount X 100 %

21 So when I read this question the main problem is calculating efficiency…. So lets remember back to Chem 20, how do we calculate percent efficiency? Formula? Percent Efficiency = Experimental Energy x 100% Theoretical Energy Experimental is calculated by using the information about what actually happened. Theoretical is either given to you in the question or is calculated by using theory…Ideal Conditions….what should have happened NOT WHAT ACTUALLY DID. Multiply by 100 so you can put a % sign.

22 Percent Efficiency = Experimental Energy x 100% Theoretical Energy Percent Efficiency = Experimental Energy x 100% 30KJ or 30000J So how do we calculate the amount of energy Actually Gained? Q=mc∆t Q= (150g)(4.19J/g°C)(20.6-52.8°C) Q= (150g)(4.19J/g°C)(32.2°C) Q= 20237.7 Joules or 20.2377Kilojoules

23 Percent Efficiency = Experimental Energy x 100% Theoretical Energy Percent Efficiency = 20237.7 J x 100% 30000 J Percent Efficiency = 67.459% It doesn’t matter if you use Joules or Kilojoules, but you have to use the same unit both times!!!! ~ 67.5%

24 Enthalpy….What is it??? Enthalpy is a measure of BOTH the kinetic (energy of motion/heat) AND potential energy (energy stored in chemical bonds). ∆H Q To make it simpler, a change in the potential energy (energy stored in chemical bonds) is represented by ∆H and a change in kinetic energy(heat) is represented by Q

25 ∆H So energy doesn’t just appear out of nowhere, it has to come from somewhere. And if all the amount of energy lost (exothermic) or gained (endothermic) by a chemical reaction is equal to the amount of energy gained by the surroundings (water in calorimeter)(***which we can calculate using Q=mc∆t***) then Q = -∆H

26 ∆H and Q

27 Summary A change in Potential Energy (chemical bonds) is represented by ∆H A change in Kinetic Energy (motion/heat) is represent by Q (Q can be calculated) Energy lost by a chemical system (∆H) is equal to the amount of energy gained by the surroundings (Q). If the water gains energy (+) then the chemical system must lose energy (-). So ∆H is equal to Q, but it has the OPPOSITE sign (+ -). So if you want to know ∆H, calculate Q using Q=mc∆t and switch the sign when your done. *** + ∆H is endothermic - ∆H is exothermic***

28 Pg 489 has a couple of examples to do for practice (take a look on your own) MY EXAMPLE A scoop of magical heating powder is dissolve in a 150ml cup of water and the cup of water increased in temperature from 23.0 °C to 48.7 °C. What is the enthalpy change(∆H)? So….∆H = Q (with opposite symbol) so ∆H = -(Q) Q= mc∆t so…… ∆H = -(mc∆t) ∆H = -(mc∆t) ∆H = -((150g)(4.19 J/g°C)(25.7°C)) ∆H = -(16152.45) ∆H = -16152.45J m= c= ∆t= 150 g 4.19 J/g°C 23.0 – 48.7°C 25.7°C -16.2 kJ ~

29 Molar Enthalpy So chemists want to be very accurate and precise so even more than the ∆H, they want to know ∆H/mol of whatever is reacted, whether something is: Burned (∆ c H) Reacted (∆ r H) Formation (∆ f H) Decomposition (∆ d H) ***To find Molar Enthalpy you just divide the enthalpy (∆H) you calculated by the number of moles***

30 MY EXAMPLE A scoop of magical heating powder is dissolve in a 150ml cup of water and the cup of water increased in temperature from 23.0 °C to 48.7 °C. What is the enthalpy change(∆H)? ∆H = -16152.45J -16.2 kJ ~ 12.3 g of NH 3 NO 3 is reacted in a 150ml cup of water and the cup of water increased in temperature from 23.0 °C to 48.7 °C. What is the Molar Enthalpy (∆rHm)? ÷ n NH 3 NO 3 nNH 3 NO 3 = m M = 12.3 g 79.05 g/mol = 0.15559772mol So… -16152.45J ÷ 0.15559772mol =-103809.04J~ -104 kJ/mol

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33 Molar Enthalpy can be expressed in a variety of ways. We have already looked at these but lets do a final review just to make sure we understand them all.

34 1. Molar Enthalpies of Reaction ∆ r H m Reaction refers to a whole bunch of things, its just a general term that includes: Combustion (∆ c H m °) Formation (∆ f H m °) Decomposition (∆ d H m °) Molar enthalpies of reaction are J/mol or kJ/mol The molar enthalpy of common reactions are often given to you in a table…or just given to you in the question. To turn a molar enthalpy in total enthalpy you just take the molar enthalpy given and multiply by how many moles. Exothermic ***The – symbol just means when this reaction happens it releases that amount of energy…Exothermic***

35 2. Enthalpy Change ∆ r H Same as 1. but its not energy/mol. Your give the enthalpy change per mole and you have to turn it into the total enthalpy change by multiply the enthalpy/mol by how many moles in a balanced chemical reaction…..so you have to balance it. H 2 + O 2 H 2 0 ∆ c H m = -258.8 kJ/mol Combustion of Hydrogen to form water 21 ∆ c H° = n∆ c H m ∆cH° = (2mol)(-258.8kJ/mol) ∆cH° = -572kJ No written question so no sig figs…so your done. 1/2

36 3. Energy As A Term In A Chemical Equation H 2 + O 2 H 2 0 ∆cHm = -258.8 kJ/mol In order to make the energy a term in the chemical equation we just right in + energy on whichever side of the arrow depending on if the reaction is exothermic (add energy to right side) or endothermic (add energy to left side). H 2 + O 2 H 2 0 + 258.8kJ

37 4. Chemical Potential Energy Diagrams These diagrams show the energy of the reactants and the products. Gains (absorbs) Energy Loses (releases) Energy

38 Summary of 4 Ways (-) (+)

39 Practice -

40 C 6 H 12 O 6(s) + O 2(g)  CO 2(g) + H 2 O (l) 1 6

41 Practice C 6 H 12 O 6(s) + O 2(g)  CO 2(g) + H 2 O (l) 1 6 Exothermic….so add energy to left side or right side? + 2802.5 kJ

42 Practice Exothermic or Endothermic?

43 AKA…The Law of Adding Enthalpy Equations Together To Get The Reactants and Products We Want. So some reactions happens so slowly or so randomly that its impossible to actually use a calorimeter and find out an enthalpy change for them. So to calculate the enthalpy changes for these reactions we use the enthalpy equation of reactions that we know are related. Sort of like Taylor is 1.3 centimeters taller than Jared, Jared is 116.7 centimeters tall. How tall is Taylor?

44 Hess’ Law So Hess’ Law states that the sum of all individual steps in a reaction is equal to the overall (sometimes unknown) enthalpy for the reaction. This is just the capital letter Sigma….its Greek….it just means sum, you’v probably seen it before in Math??

45 Hess’ Law: “Rules” 1. If you take an enthalpy equation of flip it around like a 180 so the reactants become the products and the products become the reactants YOU HAVE TO CHANGE THE SYMBOL ON THE ENTHALPY!!! (+  -) or (-  +). H 2 O (l)  H 2 (g) + ½ O 2 (g) ∆ d H = + 258.8 kJ H 2 (g) + ½ O 2 (g)  H 2 O (l) ∆dH = - 258.8 kJ Your Given This but want the opposite reaction.

46 Hess’ Law: “Rules” 2. you have to multiply the Enthalpy as well. 2. If balance a given reaction equation and multiply it by a number to get it to balance with whole number coefficients for example, you have to multiply the Enthalpy as well. H 2 O (l)  H 2 (g) + ½ O 2 (g) ∆ d H = + 258.8 kJ 2H 2 O (l)  2H 2 (g) + 1O 2 (g) ∆dH = + 517.6 kJ Your given this but you need to balance it with whole numbers…so get rid of the ½ …..so you have to multiply all three terms by 2. X 2

47 Lets Use The Rules and Try One. GIVEN: So we need a carbon (C) on the left side…. Which equation will give us a carbon (C) on the left side??? / ***Little hint, if you leave Oxygens till last they usually work themselves out. + 2CO 2 (g)  2CO (g) + O 2 (g) ∆ c H° = +556.0 kJ 2 CO 2 (g)  CO (g) + ½ O 2 (g) ∆ c H° = +283.0 kJ

48 Next were going to cancel out terms that are the same on both sides of the equation. 1/2 Hess’ Law states that the overall enthalpy for the reaction is the sum of all the individual parts. + So

49 Practice Questions

50 Hess’s Law Summary

51 Works the same as using Hess’ Law to find out overall enthalpy changes. Just remember to balance and because its MOLAR enthalpy, you have to either balance so that whatever your looking at is either balance using a 1 or you divide the final enthalpy by the balancing coefficient. And the final unit is energy/mol.

52 If Your Not Given Enthalpy Reaction Equations. Common enthalpy equations can be found in the data booklet or will be given to you….you don’t have to memorize them. If you want the overall enthalpy of a reaction you just add up the individual energy from the PRODUCT side and subtract the total energy from the REACTANT side. ***When your calculating the individual energy of individual compounds make sure you balance and multiply the energy by the balancing coefficient!!**** ***If your having trouble finding enthalpy equations for sing elements….like O2 or H2, its because all PURE ELEMENTS have an enthalpy of zero.***

53 Thermal Stability :

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