{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}


che106_lecture11 - Chemistry 106 Lecture 11 Topics...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chemistry 106 Lecture 11 Topics: Molarity (end of chap 4) Start of Chap 5:Thermochemistry Chapter 4.6, 5.1- 5.4 Working with SoluGons •  The majority of chemical reacGons we have discussed occur in aqueous solu)on. –  When you run reacGons in liquid soluGons, it is convenient to dispense the amounts of reactants by measuring out volumes of reactant soluGons. Molar ConcentraGon •  When we dissolve a substance in a liquid, we call the substance the solute and the liquid the solvent. –  The general term concentra(on refers to the quanGty of solute in a standard quanGty of soluGon. Molar ConcentraGon •  Molar concentra)on, or molarity (M), is defined as the moles of solute dissolved in one liter (cubic decimeter) of soluGon. moles of solute Molarity (M) = liters of solution Molar ConcentraGon •  Let’s try an example. –  A sample of 0.0341 mol iron(III) chloride, FeCl3, was dissolved in water to give 25.0 mL of soluGon. What is the molarity of the soluGon? moles of FeCl 3 Since molarity = liters of solution 0.0341 mole of FeCl 3 then M = = 1.36 M FeCl 3 0.0250 liter of solution Molar ConcentraGon •  The molarity of a soluGon and its volume are inversely proporGonal. Therefore, adding water makes the soluGon less concentrated (dilu)on). –  This inverse relaGonship takes the form of: M i × Vi = M f × V f –  So, as water is added, increasing the final volume, Vf, the final molarity, Mf, decreases. Volumetric Analysis •  Consider the reacGon of sulfuric acid, H2SO4, with sodium hydroxide, NaOH: H 2SO4 (aq) + 2NaOH(aq) → 2H 2O(l ) + Na2SO4 (aq) –  Suppose a beaker contains 35.0 mL of 0.175 M H2SO4. How many milliliters of 0.250 M NaOH must be added to completely react with the sulfuric acid? Volumetric Analysis –  First we must convert the 0.0350 L (35.0 mL) to moles of H2SO4 (using the molarity of the H2SO4). –  Then, convert to moles of NaOH (from the balanced chemical equaGon). –  Finally, convert to volume of NaOH soluGon (using the molarity of NaOH). 0.175 mole H 2SO 4 2 mol NaOH 1 L NaOH soln. (0.0350L ) × × × = 1 L H 2SO 4 solution 1 mol H 2SO 4 0.250 mol NaOH 0.0490 L NaOH solution (or 49.0 mL of NaOH solution) Summary of SoluGons •  Molarity is defined as the number of moles of solute per liter of soluGon. Knowing the molarity allows you to calculate the amount of solute in a given volume of soluGon. •  In volumetric analysis, you determine the amount of a species in a sample by measuring the volumes of reactants and products. This marks the end of Chapter 4 and the beginning of Chapter 5 Thermochemistry •  Thermodynamics is the science of the relaGonship between heat and other forms of energy. •  Thermochemistry is the study of the quanGty of heat absorbed or evolved by chemical reacGons. Energy •  Energy is defined as the capacity to move ma^er. Energy can be in many forms: – Radiant Energy - ElectromagneGc radiaGon. – Thermal Energy - Associated with random moGon of a molecule or atom. – Chemical Energy - Energy stored within the structural limits of a molecule or atom. Energy •  There are three broad concepts of energy: –  Kinetic Energy is the energy associated with an object by virtue of its motion. –  Potential Energy is the energy an object has by virtue of its position in a field of force. –  Internal Energy is the sum of the kinetic and potential energies of the particles making up a substance. Energy •  Kine)c Energy: An object of mass m and speed or velocity v has kineGc energy Ek equal to Ek = 1 mv 2 2 –  This shows that the kineGc energy of an object depends on both its mass and its speed. KineGc Energy Example •  Consider the kineGc energy of a person whose mass is 130 lb (59.0 kg) traveling in a car at 60 mph (26.8 m/s). Ek = 1 ( 59.0 kg ) × ( 26.8 m / s ) 2 2 4 2 2 Ek = 2.12 × 10 kg ⋅ m / s 4 Ek = 2.12 × 10 J –  The SI unit of energy, kg.m2/ s2, is given the name Joule. 1 calorie (cal) = 4.184 J Energy •  Poten)al Energy: This energy depends on the “posiGon” (such as height) in a “field of force” (such as gravity). •  For example, water of a given mass m at the top of a dam is at a relaGvely high “posiGon” h in the “gravitaGonal field” g of the earth. Ep = mgh PotenGal Energy Example •  Consider the potenGal energy of 1000 lb of water (453.6 kg) at the top of a 300 foot dam (91.44 m). 2 Ep = (453.6 kg) × (9.80 m / s ) × (91.44 m ) 5 2 Ep = 4.06 × 10 kg ⋅ m / s 5 Ep = 4.06 × 10 J 2 What to do with PotenGal Energy •  The potenGal energy of 1000 lb of water at the top of a 300 foot dam is 4.06 x 10- 5 J . •  1 Joule = 2.78 x 10- 4 wa^- hour •  4.06 x 10- 5 J = 112 wa^- hour •  A 100- wa^ light bulb uses 100 wa^- hours in an hour. •  The potenGal energy stored in the water could run a 100- wa^ light bulb for more than 1 hour! Energy •  Internal Energy (U) is the energy of the parGcles making up a substance (Ep and Ek of the electrons and nuclei). •  The total energy of a system is the sum of its kineGc energy, potenGal energy, and internal energy, U. Etotal = Ek + E p + U Note: The textbook use “E” to represent internal energy, but it notes (see SecGon 5.2) that many textbooks use “U” as we have done here. There is no difference between the two systems. Energy •  The Law of Conserva)on of Energy: Energy may be converted from one form to another, but the total quanGGes of energy (Etotal) remain constant. •  This is also known as the First Law of Thermodynamics. Conversion of Energy •  Energy can be converted from one type to another. •  For example, the cyclist above has potenGal energy as she sits on top of the hill. Conversion of Energy •  As she coasts down the hill, her potenGal energy is converted to kineGc energy. •  At the bo^om, all the potenGal energy she had at the top of the hill is now kineGc energy. Heat of ReacGon •  In chemical reacGons, heat is oken transferred from the “system” to its “surroundings,” or vice versa. •  The substance or mixture of substances under study in which a change occurs is called the thermodynamic system (or simply system.) •  The surroundings are everything in the vicinity of the thermodynamic system. Heat of ReacGon •  Heat is defined as the energy that flows into or out of a system because of a difference in temperature between the system and its surroundings. •  Heat flows from a region of higher temperature to one of lower temperature. •  Once the temperatures become equal, heat flow stops and thermal equilibrium is achieved. Heat of ReacGon •  Heat is denoted by the symbol q. – The sign of q is posiGve if heat is absorbed by the system. – The sign of q is negaGve if heat is evolved by the system. – Heat of Reac)on is the value of q required to return a system to the original temperature at the compleGon of the reacGon. Heat of ReacGon •  An exothermic process is a chemical reacGon or physical change in which heat is evolved (q is negaGve). •  An endothermic process is a chemical reacGon or physical change in which heat is absorbed (q is posiGve). Heat of ReacGon •  Exothermic – “out of” a system •  Endothermic – “into” a system An Exothermic Process Test tube contains anhydrous copper (II) sulfate, graduated cylinder contains water. 79.0° F An Exothermic Process Water from a graduated cylinder is added to a test tube where copper (II) sulfate forms the hydrated copper (II) ion. 194° F Enthalpy and Enthalpy Change •  The heat absorbed or evolved by a reacGon depends on the condiGons under which it occurs. •  Usually, a reacGon takes place in an open vessel, and therefore at the constant pressure of the atmosphere. •  The heat of this type of reacGon is denoted qp, the heat at constant pressure. Enthalpy and Enthalpy Change •  Enthalpy, denoted H, is an extensive property of a substance that can be used to obtain the heat absorbed or evolved in a chemical reacGon. – An extensive property is one that depends on the quanGty of substance (mass, volume, etc.). – Enthalpy is a state funcGon, a property of a system that depends only on its present state and is independent of any previous history of the system. An Analogy to Illustrate a State FuncGon The difference in alGtude of A and B is 3600 k. It does not ma^er how one travels, the difference is the same. Enthalpy and Enthalpy Change •  The change in enthalpy (!H) for a reacGon at a given temperature and pressure (called the enthalpy of reac)on) is obtained by subtracGng the enthalpy of the reactants from the enthalpy of the products. ΔH = H (products) − H ( reactants ) Enthalpy Diagram 2 Na(s) + 2H2O(l) → 2NaOH(aq) + H 2 (g) The diagram tells us: 2 moles of Na react with 2 moles of water to produce 368.6 kJ of heat. Enthalpy and Enthalpy Change •  The change in enthalpy (!H) is equal to the heat of reacGon at constant pressure. •  This represents the enGre change in internal energy (!U) minus any expansion “work” done by the system. ΔH = q p Enthalpy and Enthalpy Change •  Enthalpy and Internal Energy   The internal energy of a system, U, is – precisely defined as the heat at constant pressure plus the work done by the system: U = qp + w –  In chemical systems, work is defined as a change in volume at a given pressure, that is: w = −PΔV Pressure- Volume Work 2 Na(s) + 2H2O(l) → 2NaOH(aq) + H 2 (g) Enthalpy and Enthalpy Change •  Since the heat at constant pressure, qp, represents ΔH, then ΔU = ΔH − PΔV –  This means that the internal energy (U) of the system changes by: – energy entering or leaving (!H) – or the system increase or decreases in volume ( - P!V) Thermochemical EquaGons •  A thermochemical equa)on is the chemical equaGon for a reacGon (including phase labels) in which the equaGon is given a molar interpretaGon, and the enthalpy of reacGon for these molar amounts is wri^en directly aker the equaGon. N 2 (g ) + 3H 2 (g ) → 2NH 3 (g ); ΔH = -91.8 kJ Thermochemical EquaGons •  In a thermochemical equa)on it is important to note phase labels because the enthalpy change, DH, depends on the phase of the substances. 2H 2 (g ) + O 2 (g ) → 2H 2O(g ) ; ΔH o = - 483.7 kJ o 2H 2 (g ) + O 2 (g ) → 2H 2O(l ) ; ΔH = - 571.7 kJ Thermochemical EquaGons •  The following are two important rules for manipulaGng thermochemical equaGons: 1.  When a thermochemical equaGon is mul(plied by any factor, the value of !H for the new equaGon is obtained by mulGplying the !H in the original equaGon by that same factor. 2.  When a chemical equaGon is reversed, the value of !H is reversed in sign. Applying Stoichiometry and Heats of ReacGons •  Consider the reacGon of methane, CH4, burning in the presence of oxygen at constant pressure. Given the following equaGon, how much heat could be obtained by the combusGon of 10.0 grams CH4? CH 4 (g ) + 2O 2 (g ) → CO 2 (g ) + 2H 2O(l );ΔH o = -890.3 kJ Thermochemical EquaGons How much heat could be obtained by the combusGon of 10.0 grams CH4? CH 4 (g ) + 2O 2 (g ) → CO 2 (g ) + 2H 2O(l );ΔH o = -890.3 kJ 1 mol CH 4 10.0 g CH 4 × 16.0 g − 890 3 kJ × 1 mol.CH = −556 kJ 4 Next Lecture •  Topics: Thermochemistry •  Text Reading: 5.5- 5.8 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online