Openstax: Chemistry Chapter Notes
Openstax: Chemistry Chapter Notes CHEM 102
Popular in General Descriptive Chemistry II
Popular in Chemistry
This 34 page Bundle was uploaded by Joy Furigay on Thursday January 28, 2016. The Bundle belongs to CHEM 102 at University of North Carolina - Chapel Hill taught by Dr. Jennifer Krumper in Spring 2016. Since its upload, it has received 103 views. For similar materials see General Descriptive Chemistry II in Chemistry at University of North Carolina - Chapel Hill.
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Date Created: 01/28/16
9.5 The KineticMolecular Theory Sunday, January 3, 2016 10:10 PM The kinetic molecular theory (KMT) is a simple microscopic model that effectively explains the gas laws. Based on the five postulates: o Gases are composed of molecules that are in continuous motion, travelling in straight lines and changing direction only when they collide with other molecules or with the walls of a container o The molecules composing the gas are negligibly small compared to the distances between them. o The pressure exerted by a gas in a container results from collisions between the gas molecules and the container walls. o Gas molecules exert no attractive or repulsive forces on each other or the container walls; therefore, their collisions are elastic (do not involve a loss of energy) o The average kinetic energy of the gas molecules is proportional to the kelvin temperature of the gas. The KineticMolecular Theory Explains the Behavior of Gases, Part 1 Amonton's Law If the temperature is increased, the average speed and kinetic energy of the gas molecules increase. o If the volume is held constant, the increased speed of the gas molecules results in more frequent and more forceful collisions with the walls of the container, therefore increasing the pressure. Charles' Law If the temperature of a gas is increased, a constant pressure may be maintained only if the volume occupied by the gas increases. o This will result in greater average distances traveled by the molecules to reach the container walls, as well as increased wall surface area. o These conditions will decrease both the frequency of moleculewall collisions and the number of collisions per unit area, the combined effects of which outweigh those of increased collision forces due to the greater kinetic energy at the higher temperature. o The net result is a decrease in gas pressure. Boyle's Law If the gas volume is decreased, the container wall area decreases and the molecule wall collision frequency increases, both of which increase the pressure exerted by the gas. Avogadro's Law At constant pressure and temperature, the frequency and force of moleculewall collisions are constant. o Under such conditions, increasing the number of gaseous molecules will require a proportional increase in the container volume in order to yield a decrease in the number of collisions per unit area to compensate for the increased frequency of collisions. Dalton's Law Because of the large distances between them, the molecules of one gas in a mixture bombard the container walls with the same frequency whether other gases are present or not, and the total pressure of a gas mixture equals the sum of the partial pressures of the individual gases. Molecular Velocities and Kinetic Energy In a gas sample, the molecular speed distribution and average speed are constant. o This molecular speed distribution is known as a MaxwellBoltzmann distribution 2 KE = .5mu In the KMT, the root mean square velocity of a particle, u ,rms defined as the square root of the average of the squares of the velocities with n = the number of particles. 2 o Therefore, KE avg5mu rms Ke avg3/2 R ( Gas Constant) T (Kelvin) Gas Constant = 8.314 J/K 2 .5mu rms = 3/2 RT o U rms (√RT / m) If the temperature of a gas increases, its average KE increases, more molecules have higher speeds and fewer molecules have lower speeds, and the distribution shifts toward higher speeds overall to the right. At a given temperature, all gases have the same average KE for their molecules. Gases composed of lighter molecules have more highspeed particles and a higher u , withrms speed distribution that peaks at relatively higher velocities. Example 9.23 Calculation of u rms Calculate the rootmeansquare velocity for a nitrogen molecule at 30 degrees C. Convert the temperature into Kelvin 30 + 273 = 303 K Determine the mass of a nitrogen molecule in kilograms: 28 g / 1 mol x 1 kg / 1000g = .028 kg/mol Replace the variables and constants into the rootmeansquare velocity equation U rms√( 3RT / m) = 519 m/s Check Your Learning Calculate the rootmeansquare velocity for an oxygen molecule at 23 degrees C. Answer = 441 m/s The KineticMolecular Theory Explains the Behavior of Gases, Part II According to Graham/s law, the molecules of a gas are in rapid motion and the molecules themselves are small. The average distance between the molecules of a gas is large compared to the size of the molecules. o As a consequence, gas molecules can move past each other easily and diffuse at relatively fast rates. The rate of effusion of a gas depends directly on the average speed of its molecules. Effusion A / Effusion B =√ (m / m ) B A 10.1 Intermolecular Forces Tuesday, January 12, 2016 6:34 PM Particles in a solid are tightly packed together and often arranged in a regular pattern; in a liquid, they are close together with no regular arrangement; in a gas, they are far apart with no regular arrangement. Particles in a solid vibrate about fixed positions and do not generally move in relation to one another; in a liquid, they move past each other, but remain in essentially constant contact; in a gas, they move independently of one another except when they collide. The phase in which a substance exists depends on the relative extents of it intermolecular forces and the kinetic energies of its molecules. o IMFs are the various forces of attraction that may exist between the atoms and molecules of a substance due to electrostatic phenomena. They hold particles together, while KE provides the energy required to overcome the attractive forces and this increase the distance between particles. The increased pressure will bring the molecules of gas closer together, such that the attractions between the molecules become strong relative to their KE. o Consequently, they form liquids. If the temperature of a liquid becomes sufficiently low, or the pressure on the liquid becomes sufficiently high, the molecules of the liquid no longer have enough KE to overcome the IMF between them, and a solid forms. Forces Between Molecules Intramolecular forces are those within the molecule that keep the molecule together. Intermolecular forces are the attractions between molecules, which determine many of the physical properties of a substance. o They are weak compared to intramolecular forces. All of the attractive forces between neutral atoms and molecules are known as van der Waals forces. Dispersion Forces Because the electrons of an atom or molecule are in constant motion, at any moment in time, an atom or molecule can develop a temporary, instantaneous dipole if its electrons are distributed asymmetrically. o The presence of this dipole can distort the electrons of a neighboring atom or molecule, producing an induced dipole. o These two rapidly fluctuating, temporary dipoles result in a relatively weak electrostatic attraction between the species, called a dispersion force. Larger and heavier atoms and molecules exhibit stronger dispersion forces than so smaller and lighter atoms and molecules. In a larger atom, the valence electrons are farther from the nuclei than in a smaller atom. o Therefore, they are less tightly held and can more easily form the temporary dipoles that produce the attraction The measure of how easy or difficult it is for another electrostatic charge to distort a molecule's charge distribution is known as polarizability. o A molecule that has a charge cloud that is easily distorted is said to be very polarizable and will have large dispersion forces o One with a charge cloud that is difficult to distort is not very polarizable and will have small dispersion forces. The shapes of molecules also affect the magnitudes of the dispersion forces between them. o The larger the surface area, the stronger the force. Example 10.1 London Forces and Their Effects Order the following compounds of a group 14 element and hydrogen from lowest to highest boiling point: CH4, SiH4, GeH4, and SnH4. Ch4 SiH4 GeH4 SnH4 (Look at the molar masses) Check Your Learning Order the following hydrocarbons from lowest to highest boiling point: C2H6, C3H8, and C4H10 Answer = C2H6 C3H8 C4H10 Dipole Dipole Attractions Polar molecules have a partial positive charge on one side and a partial negative charge on the other side of the molecule a separation of charge called a dipole. A dipole dipole attraction is the electrostatic force between the partially positive end of one polar molecule and the partially negative end of another. The higher the boiling point, the stronger the dipole dipole attraction. Example 10.2 Predict which will have the higher boiling point: N2 or CO. Explain your reasoning. CO because it is polar and can experience dipoledipole attraction, while N2 is nonpolar so it cannot exhibit dipoledipole attractions. Check Your Learning Predict which will have the high boiling point: ICl or Br2. Explain your reasoning. Answer = ICl because it is a polar molecule, while Br2 is not. Hydrogen Bonding Hydrogen atoms bonded with F, O, or N atoms lead to highly concentrated partial charges with these atoms. o Molecules with FH, OH, or NH moieties are strongly attracted to similar moieties in nearby molecules, a particularly strong type of dipoledipole attraction called hydrogen bonding. Hydrogen bonds are intermolecular attractive forces and are much weaker than covalent bonds, but are much stronger than other dipoledipole attractions and dispersion forces. Example 10.3 Effect of Hydrogen Bonding on Boiling Points Consider the compounds CH3OCH3, CH3CH2OH, and CH3CH2CH3. Their boiling points are 42.1 degrees C, 24.8 degrees C, and 78.4 degrees C. Match each compound with its boiling point. Explain your reasoning. The VSEPR predicted shapes and molar masses of all three are similar so they will exhibit similar dispersion forces. CH3CH2CH3 is nonpolar so it may exhibit only dispersion forces. CH3OCH3 is polar so it will experience dipoledipole attractions. CH3CH2OH has an OH groups so it will experience hydrogen bonding. Therefore CH3CH2CH3 has the lowest boiling point while CH3CH2OH has the highest boiling point. Check Your Learning CH3CH3 has a melting point of 183 degrees C and a boiling point of 89 degrees C. Predict the melting and boiling points of CH3NH2. Explain your reasoning. CH3NH2 will have a higher melting and boiling point due to its NH group. 12.1 Chemical Reaction Rates Thursday, January 14, 2016 11:52 PM The rate of reaction is the change in the amount of a reactant or product per unit time. o Determined by measuring the time dependence of some property that can be related to reactant or product amounts. The mathematical representation of the change in species concentration over time is the rate expression for the reaction. Relative Rates of Reaction The rate of reaction may be expressed in terms of the change in the amount of any reactant or product, and may be simply derived from the stoichiometry of the reaction. Example 12.2 Expressions for Relative Reaction Rates 4NH3(g) + 5O2(g) > 4NO(g) + 6H2O(g) Write the equations that relate the rates of consumption of the reactants and the rates of formation of the products. .25(Δ[NH3] / Δ t) = .2(Δ[O2] / Δt) = .25(Δ[NO] / Δ t} = 1/6(Δ[H2O] / Δt) Check Your Learning The rate of formation of Br2 is 6 x 10^6 mol/L/s in a reaction described by the following net ionic equation: 5Br + BrO3 + 6H+ > 3Br2 + 3H2O Write out the equations that relate the rates of consumption of the reactants and the rates of formation of the products. Answer: − 1/5(Δ[Br−]/Δt) = − (Δ[BrO3−] /Δt) = − 1/6(Δ[H+]/Δt) = 1/3(Δ[Br2]/ Δt )= 1/3(Δ[H2O]/Δt) Example 12.2 Reaction Rate Expressions for Decomposition of H2O2 2H2O2 > 2H2O + O2 Based on this data, the instantaneous rate of decomposition of H2O3 at t=11.1 is determined to be 3.2 x 10^2 mol/L/h. What is the instantaneous rate of production of H2O and O2? .5(Δ[H2O2] / Δt) = .5 (Δ[H2O] / Δt) = ΔO2 / Δt .5(3.2 x 10^2) = ΔO2 / Δt ΔO2 / Δt = 1.6 x 10^2 mol L^1 h^1 Check Your Learning If the rate of decomposition of ammonia, NH3, at 1150 K is 2.1 x 10^6 mol/L/s, what is the rate of production of nitrogen and hydrogen? Answer = 1.05 x 10^6 for N2 and 3.15 x 10^6 for H2 12.2 Factors Affecting Reaction Rates Friday, January 15, 2016 8:13 PM The Chemical Nature of the Reacting Substances Reactions that appear similar may have different rates under the same conditions, depending on the identity of the reactants. The State of Subdivision of the Reactants Except for substances in the gaseous state or in solution, reactions occur at the boundary, or interface, between two phases. o The rate of reaction between two phases depends to a great extent on the surface contact between them. Temperature of the Reactants Chemical reactions typically occur faster at higher temperatures. An increase in temperature of only 10 degrees C will approximately double the rate of a reaction in a homogenous system. Concentration of the Reactants Rates usually increase when the concentration of one or more of the reactants increases. The Presence of a Catalyst A catalyst is a substance that increases the rate of a chemical reaction by lowering the activation energy without itself being consumed by the reaction. o Activation energy is the minimum amount of energy required for a chemical reaction to proceed in the forward direction. o It increases the reaction rate by providing an alternative pathway or mechanism for the reaction to follow. 12.3 Rate Laws Sunday, January 17, 2016 9:56 PM Rate laws or rate equations are mathematical expressions that describe the relationship between the rate of a chemical reaction and the concentration of its reactions. o Rate = k [A]^m [B]^n [C]^p  represents the molar concentrations of reactants and k is the rate constant The exponents m, n, and p are usually positive integers The rate constant k is independent of the concentration of A, B, or C, but it does vary with temperature and surface area. The exponents in a rate law describe the effects of the reactant concentrations on the reaction rate and define the reaction order. o The overall reaction order is the sum of the orders with respect to each reactant. The method of initial rates helps determine the orders in rate laws. o To use this method, we select two sets of rate data that differ in the concentration of only one reactant and set up a ratio of the two rates and the two rate laws. o After canceling terms that are equal, we are left with an equation that contains only one unknown, the coefficient of the concentration that varies. Example 12.3 Writing Rate Laws from Reaction Orders NO2 (g) + CO (g) > NO (g) + CO2 (g) NO2 is second order and CO is zero order at 100 degrees C. What is the rate law for the reaction? Rate = k [NO2]^2 [CO]^0 = k [NO2]^2 Check Your Learning H2 (g) + 2NO (g) > N2O (g) + H2O (g) Rate = k [NO]^2[H2]. What are the orders with respect to each reactant, and what is the overall order of the reaction? Answer: Order in NO = 2; order in H2 = 1; overall order = 3 Check Your Learning CH3OH + CH3OCOCH3 > CH3OCOCH3 + CH3CH2OH Rate = k [CH3OH]. What is the order of reaction with respect to methanol and ethyl acetate, and what is the overall order of the reaction? Answer: Order in CH3OH = 1; order in CH3CH2OCOCH3 = 0; overall order = 1 Example 12.4 Determining a Rate Law from Initial Rates NO (g) + O3 (g) > NO2 (g) + O2 (g) Determine the rate law and the rate constant for the reaction. Step 1. Determine the value of m from the data in which NO varies and O3 is constant. Step 2. Determine the value of n from data in which O3 varies and NO is constant. Rate = k[NO]^1[O3]^1 = k[NO][O3] Step 3. Determine the value of k from one set of concentrations and the corresponding rate. K = rate / ( [NO][O3]) = .66 x 10^4 / [( 1 x 10^6) (3 x 10^6)] = 2.2 x 10^7 mol L^1 s^1 Check Your Learning CH3CHO (g) > CH4 (g) + CO (g) Reaction Order and Rate Constant Units Rate laws are determined by experiment only and are not reliably predicted by reaction stoichiometry. 12.4 Integrated Rate Laws Wednesday, January 20, 2016 10:06 PM Integrated Rate Laws related the concentration of the reactants and time. o Helps determine the amount of reactant or product present after a period of time or to estimate the time required for a reaction to proceed to a certain extent. First Order Reactions Ln([A] t [A] )0= kt Ln([A] / [A]) = kt o 0 t Ln[A] = kt + ln[A] 0 The ln[A] plot is a straight line with a slope of k and an intercept of ln[A] .0 Example 12.6 The Integrated Rate Law for a FirstOrder Reaction Example 12.7 Determination of Reaction Order by Graphing SecondOrder Reactions Rate = k[A] 2 1/[A] = kt + 1/[A] 0 This 1/[A] plot is a straight line with a slope of k and an intercept of 1/[A] 0 Example 12.8 The Integrated Rate Law for a SecondOrder Reaction Example 12.9 Determination of Reaction Order by Graphing ZeroOrder Reactions Rate = k[A] = k [A] = kt +[A] 0 The HalfLife of a Reaction The halflife of a reaction (t ) is the time required for onehalf of a given amount of reactant to 1/2 be consumed. o The halflife of a first order reaction is independent of the concentration of the reactant. o Halflives of reactions with other orders depend on the concentrations of the reactants. FirstOrder Reactions T 1/2.693 / k Exampled 12.10 Calculation of a FirstOrder Rate Constant Using HalfLife Calculate the rate constant for the firstorder decomposition of hydrogen peroxide in water at 40 degrees C, using the data given in Figure 12.13. SecondOrder Reactions T1/2 1/ k[A] 0 For a second order reaction, t is1/2versely proportional to the concentration of the reactant, and the halflife increases as the reaction because the concentration of reactant decreases. Unlike with the firstorder reactions, the rate constant of a secondorder reaction cannot be calculated directly from the halflife unless the initial concentration is known. ZeroOrder Reactions T1/2 [A] /02k The halflife of a zero order reaction increases as the initial concentration increases. 12.5 Collision Theory Tuesday, January 26, 2016 6:20 PM Collision Theory is based on the following postulates o The rate of a reaction is proportional to the rate of reactant collisions o The reacting species must collide in an orientation that allows contact between the atoms that will become bonded together in the product. o The collision must occur with adequate energy to permit mutual penetration of the reacting species' valence shells so that the electrons can rearrange and form new bonds (and new chemical species). Every reaction requires a certain amount of activation energy for it to proceed in the forward direction, yielding an appropriate activated complex along the way. The arrangements of the atoms is called a proposed activated complex or transition state. o In most cases, it is impossible to isolate or identify a transition state or activated complex. With an increase in concentration of any reacting substance, the chances for collisions between molecules are increased because there are more molecules per unit of volume. o More collision mean a faster reaction rate, assuming the energy of the collisions is adequate. Activation Energy and the Arrhenius Equation The minimum energy necessary to form a product during a collision between reactants is called the activation energy (Ea). o In a single reactant reaction, activation energy may be provided by a collision of the reactant molecule with the wall of the reaction vessel or with molecules of an inert contaminant. o If the activation energy is much large then the average kinetic energy of the molecules, the reaction will occur slowly. Only a few fastmoving molecules will have enough energy to react o If the activation energy is much smaller than the average kinetic energy of the molecules, the fraction of molecules possessing the necessary kinetic energy will be large Most collisions between molecules will result in reaction, and the reaction will occur rapidly. Arrhenius Equation is k = Ae^(Ea / RT) o R = 8.314 J/mol/K o T is in the Kelvin scale o Ea is the activation energy in joules per mole o E is the constant 2.71183 o A is a constant called the frequency factor, which is related to the frequency of collisions and the orientation of the reacting molecules. Related to the rate at which collisions having the correct orientation occur o Exponential term, e^(Ea/RT), is related to the fraction of collisions providing adequate energy to overcome the activation barrier of the reaction. For two reactions at the same temperature, the reaction with the higher activation energy has the lower rate constant and the slower rate. The larger value of Ea results in a smaller value for the exponential term, reflecting the smaller fraction of molecules with enough energy to react. o The reaction with the smaller Ea, has a larger fraction of molecules with enough energy to react An increase in temperature has the same effect as a decrease in activation energy. The rate constant is directly proportional to the frequency factor, A. Ln k = (Ea/R)(1/T) + ln A
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