Slide 40 / 118 # # State Functions Understanding thermodynamics enables us to harness energy flow for useful purposes. For instance, whether the battery is shorted out or is discharged by running the fan, its # E is the same. A B But q and w are different in the two cases. If the battery shorts Charged battery out, all of the energy is lost as heat, whereas if it is used to Heat run the fan, some energy is Energy Heat used to do work. Q and w are lost by # E Work A B NOT state functions. battery Discharged battery
Slide 41 / 118 # # Work Done by a Gas When a gas expands it does Initial state Final state work on its surroundings: P= F/A P= F/A W = Fd. In this case: F = PA d = ∆ h. and ∆ V ∆ h W = Fd W = (PA) ∆ h h W = P ∆ V A = cross sectional area
Slide 42 / 118 Work Done By a Gas # # Initial state Final state Since the gas expands, and does work on its P= F/A P= F/A surroundings, the energy of the gas decreases, this is considered negative work. (-) work is done by system ∆ h ∆ V W = - P ∆ V h A = cross sectional area
Slide 43 / 118 Work Done By a Gas # # Initial state Final state P= F/A P= F/A If the surroundings compress the gas, decreasing its volume, this increases the energy of the gas, it is considered ∆ V ∆ h positive work. (+) work is done on system h W = + P ∆ V A = cross sectional area
Slide 44 / 118 # # Work Done by a Gas We can measure the work done by the gas if the reaction is done in a vessel fitted with a piston. HCl Solution + Zinc HCl Solution
Slide 45 / 118 Enthalpy Return to Table of Contents
Slide 46 / 118 Enthalpy The word "enthalpy" is derived from the Greek noun "enthalpos" which means heating. The enthalpy of a system (H) is a combination of the internal energy of a system (E) plus the work that needs to be done to create the space for the substance to occupy. H = E + W It is impossible to measure enthalpy, H, directly. Only the change ( ∆ H) can be measured. ∆H = ∆E + W
Slide 47 / 118 Enthalpy At constant pressure, the change in enthalpy is the heat gained or lost by the system. ∆H = (q+W) - W ∆H = q Note: This is only true at constant pressure. See ** for more information ( W = P∆V)
Slide 48 / 118 ** Enthalpy The enthalpy of a system (H) is a combination of the internal energy of a system (E) plus the work that needs to be done to create the space for the substance (PV) to occupy. H = E + PV It is impossible to measure enthalpy, H, directly. Only the change ( ∆ H) can be measured. ∆H = ∆E + PV
Slide 49 / 118 ** Enthalpy If a process takes place at constant pressure (as most processes we study do), we can account for heat flow during the process by measuring the enthalpy of the system. ∆H = ∆E + ∆(PV) ∆H = ∆E + P∆V ∆H = (q+w) - w ∆H = q p (at constant pressure) At constant pressure, the change in enthalpy is the heat gained or lost by the system.
Slide 50 / 118 Enthalpy Enthalpy is an extensive property; its value depends on the quantity of the substance present. Combustion of 16 grams of CH 4 ∆ H = -891 kJ Combustion of 32 grams of CH 4 ∆ H = -1782 kJ ∆ H for a reaction in the forward direction is equal in size, but opposite in sign, to ∆ H for the reverse reaction. CH 4 (g) + 2O 2 (g) --> CO 2 (g) + 2H 2 O(g) ∆ H = -891 kJ 2H 2 O(g) + CO 2 (g) --> CH 4 (g) + 2O 2 (g) ∆ H = +891 kJ ∆ H for a reaction depends on the state of the products and the state of the reactants. CH 4 (g) + 2O 2 (g) --> CO 2 (g) + 2H 2 O(g) ∆ H = -891 kJ CH 4 (g) + 2O 2 (g) --> CO 2 (g) + 2H 2 O(l) ∆ H = -979 kJ
Slide 51 / 118 20 When 114 grams of gasoline combust, the ∆H is equal to -5,330 kJ. What is the ∆H for combustion of 57 grams of gasoline?
Slide 52 / 118 Endothermic and Exothermic Processes A process is exothermic A process is endothermic when ∆ H is negative. when ∆ H is positive. Surroundings Surroundings System System +q -q Heat Heat ∆H>0 ∆H<0 Endothermic Exothermic
Slide 53 / 118 21 The reaction A + B --> C is endothermic. The D H for this reaction is +50 J. What is the D H for the reaction C --> A + B? A Cannot be determined. B 0.02 J C - 50 J D 100 J
Slide 54 / 118 22 The reaction A + B --> C is exothermic. The ∆H for this reaction is -150 J. What is the ∆H for the reaction C --> A + B? A +150 B zero C -150 D this reaction will not happen
Slide 55 / 118 23 Dissolving NaOH in water will increase the temperature of the solution. This reaction is _____ exothermic A endothermic B adiabatic C isothermal D
Slide 56 / 118 24 NH 3 NO 3 + H 2 O --> NH 4+ + NO 3- + OH - ∆H = +25.69 kJ/mol. This reaction is exothermic. Yes No
Slide 57 / 118 Measuring Enthalpy Changes: Calorimetry Return to Table of Contents
Slide 58 / 118 Measuring Enthalpy Changes: Calorimetry Since we cannot know the exact enthalpy of the reactants and products, we measure ∆ H through calorimetry, the measurement of heat flow by making use of the expression: ∆ H = q p The subscript p on q means the process is occurring at constant pressure. No work is being done only heat is being exchanged. You will not always see the subscript but it is implied when we are speaking of Enthalpy.
Slide 59 / 118 Heat Capacity and Specific Heat The amount of energy required to raise the temperature of a substance by 1 K (1°C) is its heat capacity. The amount of energy required to raise the temperature of one gram of a substance by 1 K (1°C) is its specific heat (c).
Slide 60 / 118 25 Heat capacity is an example of an A Intensive property B Extensive property
Slide 61 / 118 26 Specific heat is an example of an A Intensive property B Extensive property
Slide 62 / 118 Heat Capacity and Specific Heat heat transferred Specific heat = mass x temperature change q c = m∆T
Slide 63 / 118 27 The specific heat of marble is 0.858 J/g-K. How much heat (in J) is required to raise the temperature of 20g of marble from 22 °C to 45 °C? q = m c∆ T
Slide 64 / 118 28 An 26 g sample of wood (c = 1.674 J/(g-K)) absorbs 200 J of heat, upon which the temperature of the sample increases from 20.0 °C to ______°C. q = m c∆ T
Slide 65 / 118 29 A sample of silver (c = 0.236 J/g-K) absorbs 800 J of heat, upon which the temperature of the sample increases from 50.0 °C to 80°C. What is the mass of the sample? q = m c∆ T
Slide 66 / 118 Constant Pressure Calorimetry thermometer Many chemical reactions happen stirrer in aqueous solutions. The apparatus to the left is a insulated cover calorimeter. How could you use it to measure the heat change for a chemical reaction in an aqueous solution? styrofoam cup sparknotes.com
Slide 67 / 118 Constant Pressure Calorimetry thermometer Because the specific heat for water stirrer is well known (4.184 J/g-K), we can measure ∆ H for the reaction with this equation: insulated cover D ∆ H = q = mc ∆ T (at constant pressure) styrofoam cup sparknotes.com
Slide 68 / 118 Constant Pressure Calorimetry Example: A student wishes to determine the enthalpy change when ammonium chloride (NH 4 Cl) dissolves in water. The student masses out 20.00 grams of ammonium chloride and adds it to 500 grams of water in a styrofoam cup at a temperature of 16.1 Celsius. The student observes the temperature to decrease to 13.2 Celsius. What is the enthalpy change for the dissolution of ammonium chloride? 1. Find enthalpy change of solution using q = mc∆T = 520 g x -2.9 C x 4.2 J = -6,334 J g ℃ 2. Since heat released by surroundings (solution) is equal to the heat gained by system (ammonium chloride). ∆H for dissolving of NH 4 Cl = +6334 J
Slide 69 / 118 Bomb Calorimetry Thermometer Ignition wires Reactions can be carried out in a sealed “bomb” such as this one. The heat absorbed (or released) by the water is a very good Water Oxygen approximation of atmosphere the enthalpy change for the reaction. Sample
Slide 70 / 118 Bomb Calorimetry (Constant Volume) Thermometer Ignition wires Because the volume in the bomb calorimeter is constant, what is measured is really the change in internal energy, ∆ E , not ∆ H . Water For most reactions, the Oxygen difference is very small. atmosphere Sample
Slide 71 / 118 30 The reaction below takes place in a bomb calorimeter. If the student found that the temperature of the water was 12.2 Celsius to start and the heat capacity of the calorimeter was 34 J/C, what must be the enthalpy change of the reaction if the temperature of the water increased to 15.6 Celsius? 4Fe(s) + 3O 2 (g) --> 2Fe 2 O 3 (s)
Slide 72 / 118 31 A mysterious meteorite is discovered in your backyard. To determine its identity, you determine its specific heat. The 164.6 gram sample of metal is heated to 90 C and then dumped in 300 grams of water in a styrofoam cup at an initial temperature of 10 C. After the metal is added, the temperature rises to 11.3 C. Identify the metal. Metal Specific Heat (J/gC) Cu 0.386 A B Au 0.126 C Al 0.900
Slide 73 / 118 Energy Changes Associated with Changes of State Return to Table of Contents
Slide 74 / 118 Energy Changes Associated with Changes of State Chemical and physical changes are usually accompanied by changes in energy. Recall the following terms: When energy is put into the system, the process is called _____________. When energy is released by the system, the process is called _____________.
Slide 75 / 118 Energy Changes Associated with Changes of State Fill in the blanks Endothermic processes Exothermic processes Energy is released from the Energy is taken into the system from the surroundings (∆H > 0) system to surroundings ( ∆ H < 0) Answer boiling or evaporating a liquid freezing a liquid condensing a gas melting a solid deposition of a gas sublimation of a solid
Slide 76 / 118 Phase Changes Gas Condensation Vaporization Energy of system Sublimation Deposition Liquid Melting Freezing Solid
Slide 77 / 118 32 Which of the following is/are exothermic? I. boiling II. melting III. freezing I only A I and II only B III only C I, II and III D
Slide 78 / 118 Energy Changes Associated with Changes of State The heat of fusion (∆H fus ) is the energy required to change a solid at its melting point to a liquid. Heat of fusion(H 2 O) = 6.0 kJ/mol The heat of vaporization (∆H vap ) is defined as the energy required to change a liquid at its boiling point to a gas. Heat of vaporization(H 2 O) = 41.0 kJ/mol Class Question: Why is the heat of vaporization much higher than the heat of fusion for a substance? In order to change phase from a solid to liquid, the particle attractions need only be strained somewhat. When a move for answer material changes from a liquid to a gas, the particle attractions must be essentially broken.
Slide 79 / 118 Energy Changes Associated with Changes of State Heat of fusion and vaporization kJ/mol 80 Heat of fusion Heat of vaporization Note that these quantities are 40 usually per 1.00 mole, whereas 58 q = mc ∆ T involves mass in grams. 29 41 10 24 5 7 6 23 Butane Diethyl Water Mercury ether
Slide 80 / 118 33 The heat of vaporization for butane is 24 kJ/mol. How much energy is required to vaporize 2 mol of butane? 2kJ A 12 kJ B 22 kJ C 48 kJ D
Slide 81 / 118 34 The heat of vaporization for butane is 24 kJ/ mol. How much energy is required to vaporize 0.33 mol of butane? 8kJ A 12 kJ B 16 kJ C 72 kJ D
Slide 82 / 118 35 How much energy is required to melt 0.5 mol of water? Heat of Heat of 2kJ A fusion vaporization 3 kJ B for H 2 O (s) for H 2 O (l) 6 kJ C Answer 12 kJ 6 kJ/mol 41 kJ/mol D
Slide 83 / 118 36 How much energy is released when 3.0 mol of water freezes? Heat of 2kJ A Heat of fusion vaporization 3 kJ B for H 2 O (s) for H 2 O (l) 18 kJ C 123 kJ Answer D 6 kJ/mol 41 kJ/mol
Slide 84 / 118 37 How much energy is needed to melt 10.0 mol solid Hg? 2.3 kJ A Heat of Heat of fusion 5.8 kJ B vaporization for Hg (s) 23 kJ for Hg (l) C 230 kJ D 23 kJ/mol 58 kJ/mol Answer E 580 kJ
Slide 85 / 118 How much energy is needed to vaporize 38 2.0 mol Hg (l) ? Heat of Heat of fusion vaporization for Hg (s) for Hg (l) 23 kJ/mol 58 kJ/mol Answer
Slide 86 / 118 Energy Changes Associated with Changes of State F 125 Water vapor This graph is called D 100 a heating curve. E 0 C) Liquid water and vapor It illustrates 75 Temperature ( (vaporization) how temperature changes over time 50 Liquid water as constant heat is applied to a pure 25 solid substance. B C Ice and liquid water (melting) 0 Ice -25 A Heat added (each division corresponds to 4 kJ)
Slide 87 / 118 Energy Changes Associated with Changes of State F From A to B, ice is 125 Water vapor heating up from D -25 o C to 0 o C. 100 E 0 C) Liquid water and vapor 75 Since Temperature ( (vaporization) Q = mc∆T 50 Liquid water ∆T = Q 25 (mc) B C Ice and liquid water (melting) So the slope is 0 1 Ice (mc) -25 A Heat added (each division corresponds to 4 kJ)
Slide 88 / 118 Energy Changes Associated with Changes of State F 125 From B to C, Water vapor ice is melting. D 100 E 0 C) The added heat is Liquid water and vapor 75 Temperature ( breaking the hydrogen (vaporization) bonds of the solid, 50 Liquid water so the temperature is constant. 25 B C That's why the slope 0 Ice and liquid water (melting) is zero. Ice -25 A Heat added (each division corresponds to 4 kJ)
Slide 89 / 118 Energy Changes Associated with Changes of State From C to D, F liquid water is 125 heating up from Water vapor D 0 C to 100 C. 100 E 0 C) Liquid water and vapor Once again, the 75 Temperature ( (vaporization) slope is 1/(mc). 50 Liquid water But "c" is different for all the phases 25 of a substance, so B C the slope is 0 Ice and liquid water (melting) Ice different for solid, -25 liquid and A Heat added gaseous H 2 O. (each division corresponds to 4 kJ)
Slide 90 / 118 Energy Changes Associated with Changes of State F From D to E, liquid 125 water is boiling into Water vapor D vapor. 100 E 0 C) Liquid water and vapor The added heat is 75 Temperature ( (vaporization) breaking the IM forces of the liquid, so the 50 Liquid water temperature is constant. 25 B C That's why the 0 Ice and liqui d water (melting) Ice slope is zero. -25 A Heat added (each division corresponds to 4 kJ)
Slide 91 / 118 Energy Changes Associated with Changes of State From E to F water F 125 vapor, steam, is Water vapor heating up from D 100 100 C to 125 C. E 0 C) Liquid water and vapor 75 Temperature ( Once again, the (vaporization) slope is 1/(mc). 50 Liquid water But "c" is different for all the phases 25 of a substance, so the slope is B C 0 Ice and liquid water (melting) different for for Ice solid, liquid -25 and gaseous H 2 O. A Heat added (each division corresponds to 4 kJ)
Slide 92 / 118 Energy Changes Associated with Changes of State 1 Recall that slope = mC where m = mass and C = specific heat of the substance A B This graph shows heat (DT) Temperature (C) C transfer versus change in temperature for 1 D gram of 4 different substances. (q) Energy Added (Joules)
Slide 93 / 118 39 Which substance has the lowest specific heat? A A A B B B (DT) Temperature (C) C C C D D D (q) Energy Added (Joules)
Slide 94 / 118 40 Which substance requires the highest amount of heat added to raise the temperature? A A A B B B (DT) Temperature (C) C C C D D D (q) Energy Added (Joules)
Slide 95 / 118 41 Which segment(s) contain solid sodium chloride? C AB, BC and CD A AB only BC, CD and DE D B AB and BC
Slide 96 / 118 42 Which segment(s) contain liquid sodium chloride? C AB, BC and CD A AB only D BC, CD and DE B AB and BC
Slide 97 / 118 43 What is the melting point (in o C) of sodium chloride?
Slide 98 / 118 44 What is the freezing point (in o C) of sodium chloride?
Slide 99 / 118 45 Which is greater: the specific heat of solid NaCl, or the specific heat of molten (liquid) NaCl? A C Cannot be determined. solid D B They are equal. liquid
Slide 100 / 118 Features of a Heating Curve F 125 Segments Segments Water vapor D AB, CD, EF BC & DE 100 E 0 C) Liquid water and vapor 75 slope = 0 Temperature ( slope = nonzero (vaporization) ∆T = 0 T increasing 50 Liquid water KE constant KE increasing PE increasing 25 PE constant B C 0 Ice and liquid water (melting) apply apply Ice ∆H fus or -25 q = mc∆T ∆H vap A Heat added (each division corresponds to 4 kJ)
Recommend
More recommend
