Lecture 10 Professor Hicks Inorganic Chemistry (CHE151) Ability to do work What is energy? Work means moving something against a force Energy thought of as an imaginary liquid that gets moved from one container to another when processes occur 1
Types of energy Kinetic energy- energy of motion. KE = ½ mv 2 Potential energy- associated with position. different types PE gravity = mass x gravity x height PE (opposite charges) = increases with separation (like gravity) + - Heat SI unit Joule Units of energy 1 Joule = 1 kg m 2 s 2 Other common units 1) 1 calorie = 4.18 Joules 2) 1 dietary Calorie = 1 kilocalorie (energy to warm 1 kg H 2 O 1 C) dietary and scientific calorie are not the same dietary calorie written uppercase Calorie James Joule 2
First law of thermodynamics Energy cannot be created or destroyed Applied when energy changes forms Energy like a liquid kinetic energy potential energy heat Rudolf Clausius When things go downhill A 10 kg bowling ball at a height of 100 m has potential energy due to gravity Potential energy = mgh = 10 kg x 9.8 m/s2 x10 m = 9800 J As it falls it speeds up converting potential energy into kinetic energy Right before it hits ground all the potential energy has become kinetic energy = 9800 J Heat 9800 J After it hits the ground it is not moving so kinetic energy is zero- all the energy has become heat = 9800 J 3
energy changes forms (but does not get destroyed) 1) Potential energy = 9800 J 9800 J 9800 J on 9800 J as ball falls impact PE KE KE heat potential energy kinetic energy heat 2) Kinetic energy = 9800 J 3) Heat 9800 J First Law requires heat is like a reactant or product in every chemical equation Reactants + Heat Products Reactions that absorb heat are said to be endothermic Reactions that release heat are said to be exothermic reactants products + heat 4
Second Law Second law states that matter and energy tend to spread out spontaneously A famous statement for energy is Heat only flows from hotter to cooler Explains why processes that release heat tend to not reverse themselves Sadi Carnot Second Law: Energy Spreads Out When the ball hit the ground and the energy becomes heat the ball warms up a little. The second law of thermodynamics says the bowling ball is trapped on the earth because the energy it would need to go back to the top of building has left as heat Heat 9800 J Hotter Same Cooler temperature 5
Second law: Energy and matter spread out Fuels heat CO 2 (g) 2H 2 O (g) reaction products are trapped in the lowest energy state after heat is released, like the bowling ball Heat and matter spread out prevents products from turning back into reactants!!! System and Surroundings System region of interest Surroundings everything else energy system surroundings Energy can be either heat = q work = w 6
Internal Energy (U) (ALL the energy that is in there) Change in internal energy ( U) U = q + w q heat released (negative in sign) w work done by system - (negative in sign) system surroundings q heat absorbed + (positive in sign) work done on system + (positive in sign) Work Movement performed against a force (resistance) requires energy Amount of energy=work=force distance Examples of work Lifting an object against gravity Stretching a spring Gas forming and blowing up a balloon (PV work) P V = work Force Force V = Area Length Area Area 7
Enthalpy (H) Heat released under constant pressure = Enthalpy change ( H) Constant pressure PV work can be done - Conditions for reactions of life - Conditions of most chemical reaction unless they are run in rigid containers Only reflects part of the internal energy change since it is only the heat-not the work The alternative to constant pressure is constant volume like the aging of wine in a glass bottle Enthalpy is heat released under constant pressure change in internal energy E = q + w change in enthalpy H = q work heat C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O (g) system surroundings 8
A sample of nitrogen gas expands in volume from 1.6 L to 5.4 L at constant temperature. Calculate the work done in joules if the gas expands (a) against a vacuum (b) against a constant pressure of 0.80 atm A gas expands and does P-V work on the surroundings equal to 325 J. At the same time, it absorbs 127 J of heat from the surroundings. Calculate the change in the internal energy of the gas. 9
Calculate the work done when 255 g of tin dissolves in excess acid at 1.10 atm and 22 C: Assume ideal gas behavior. Sn (s) + 2H + (aq) Sn 2+ (aq) + H 2 (g) Thermochemical equations Chemical equation with H or E included 6 CO 2 + 6 H 2 O C 6 H 12 O 6 + 6 O 2 H = 2830 kj Endothermic = heat as a reactant = positive Exothermic = heat as a product = negative The heat is associated with the whole reaction Heat + moles of any substance conversion factors 2830 kj 6 moles CO 2 2830 kj 6 moles O 2 1 mole C 6 H 12 O 6 2830 kj etc. 10
First law and thermochemical equations If a process is reversed an equal amount of energy must flow in the opposite direction H and E for reverse process will just have the sign reversed H and E depend upon the amount of material reacted so doubling, tripling reaction doubles, triples the H and E Hess Law states that if a series of chemical reactions occurs the overall H or E will just be the sum of each step s H and E values 11
Standard Conditions To be able to compare reactions Standard Conditions are defined All substances at a concentration of 1.0 M & gases at a partial pressure of 1.0 atm Heat released under constant P and standard conditions is the Standard Enthalpy Change ( H o ) First law and thermochemical equations Example: Determine the H o for 2CO 2 (g) 2C (s) + 2O 2 (g) given that H o = 2 x 393 kj = 786 kj C (s) + O 2 (g) CO 2 (g) H o = - 393 kj 12
Hess law If a process happens in steps the H o is the sum of the H o for the steps A + B C H o = +100 kj C + D E H o = -150 kj A + B + C + D C + E H o = +100 150 kj A + B + D E H o = 50 kj overall 13
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Standard Formation Reactions Form 1 mole of a substance from elements in most stable state most elements just element solids some elements exist as diatomic gases F 2, Cl 2, N 2, O 2, H 2 Br 2 and Hg liquid and I 2 is a solid Standard Enthalpy for this reaction is called the Standard Enthalpy of Formation ( H o f) 15
Standard Formation Reactions Example: a) Write standard formation reactions for CO 2 (g), H 2 O (l), C 8 H 18 (l), and CHCl 3 (l). H o f Use the Standard Formation Reaction Data Tables to determine H o for these reactions See tables full size appended to the end of the lecture packet 16
Elements in the standard state junction home station standard state is like a train junction where you transfer for your final destination destination -enthalpy formation for A - H o f (A) elements in standard state +enthalpy formation for B H o f (B) substance A substance B H o = -? H o f (A) + H o f (B) Hess Law H o = n p H o f - n r H o f reaction elements in standard state - H o f (CH 4 ) + - H o f (O 2 ) x 2 + H o f (CO 2 ) + + H o f (H 2 O) x 2 CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) H o =? 17
See tables full size appended to the end of the lecture packet Hess Law H o = n p H o f - n r H o f reaction elements in standard state - H o f (CH 4 ) - H o f (O 2 ) x 2 H o f (CO 2 ) H o f (H 2 O) x 2 CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) H o =? H o = 1 x (-393.5) + 2 x (-241.8) - (1 x - 74.6 + 2 x 0) = products - reactants n p H o f - n r H o f -802.5 kj 18
Heat Capacity and Specific Heat Properties that describe how much temperature changes when heat is absorbed Heat capacity (C) - extensive property -units are Joule C Specific heat capacity (s) - intensive property -units are Joule g* C Specific heat capacity is sometimes referred to as just specific heat 19
Heat capacity (C) Imaginary container that holds copper add 100 J the imaginary liquid heat same mass of both water add 100 J T Temperature 100 J heat capacity copper T water has a larger HC than copper so its temperature changes less when same amount of heat is added heat capacity water 100 J Conversion factor Heat capacity (C) change temperature heat T x C = q (heat) ( C) x (Joule) = (Joule) ( C) old unit conversion factor new unit changes in quantities are given symbol always calculated final initial T = T final T initial = T f - T i answers the question if I have a certain change in temperature for an object how much heat must have flowed to cause it 20
Specific heat capacity (s) Conversion factor mass heat capacity mass x specific heat = heat capacity m x s = C (grams) x (Joule) = (Joule) (gram C) ( C) old unit conversion factor new unit specific heat capacity is heat capacity per gram multiplying a mass by the specific heat capacity answers the question if I have a certain mass of a substance what is its heat capacity? How much heat would be required to raise the temperature of 50.0 grams of water from 12.0 C to 22.0 C? (s (water) = 4.18 J/g* C) Given Information T i = 12.0 T f = 22.0 calculate T = 22.0 12.0 = 10.0 C mass = 50.0 grams water s (water) = 4.18 J/g* C s x mass = HC (use this equation to calculate heat capacity of 50 g water) 4.18 J/g* C x 50.00 grams = 209 J/ C (heat capacity) HC x T = q (use this equation to calculate amount of heat to raise temperature of 50 g water 10 C ) 209 J/ C x 10.0 C = 2.09 x10 3 J these two equations can be combined: q = mass x specific heat x change temperature 21
Calorimetry measuring heat flow calorimetry measure T calculate heat add room temperature water add hot metal sample at about 100 C First law - energy cannot be created or destroyed heat lost by metal was gained by the water (and the cup which we neglect here) q water = - q metal q water = mass water s (water) (T f -T i ) q metal = mass metal s (metal) (T f -T i )? q heat surroundings and system 10 20 C coffee cup calorimeter no heat can escape perfectly insulated q metal = -q water q water = mass water s (water) (T f -T i ) q metal = mass metal s (metal) (T f -T i ) mass metal s (metal) (T f -T i ) = - mass water s (water) (T f -T i ) s (metal) = - mass water s (water) (T f -T i ) mass metal (T f -T i ) metal and water have same final temperature metal and water have different initial temperatures 22
A sample of a pure substance with a mass 125 g has a heat capacity of 307.5 J/ C. What is the specific heat of this substance? 23
A sheet of gold weighing 5.0 g and at a temperature of 99.9 C is placed flat on a sheet of iron weighing 25 g and at a temperature of 16.5 C. What is the final temperature of the combined metals? Assume that no heat is lost to the surroundings. (Hint: The heat gained by the gold must be equal to the heat lost by the iron. The specific heats of the metals are given in Table 6.2.) 24
A 44.0-g sample of an unknown metal at 199.0 C was placed in a constant-pressure calorimeter containing 155 g of water at 37.0 C. The final temperature of the system was found to be 48.4 C. Calculate the specific heat of the metal. (The heat capacity of the calorimeter is 12.4 J/ C.) 25