* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download File - Get Involved!
Bent's rule wikipedia , lookup
Chemistry: A Volatile History wikipedia , lookup
Rate equation wikipedia , lookup
Condensed matter physics wikipedia , lookup
Low-energy electron diffraction wikipedia , lookup
Electrochemistry wikipedia , lookup
History of chemistry wikipedia , lookup
Photoelectric effect wikipedia , lookup
Metastable inner-shell molecular state wikipedia , lookup
Computational chemistry wikipedia , lookup
Chemical reaction wikipedia , lookup
X-ray fluorescence wikipedia , lookup
Chemical equilibrium wikipedia , lookup
Chemical thermodynamics wikipedia , lookup
Photoredox catalysis wikipedia , lookup
X-ray photoelectron spectroscopy wikipedia , lookup
Electronegativity wikipedia , lookup
Electron scattering wikipedia , lookup
Resonance (chemistry) wikipedia , lookup
George S. Hammond wikipedia , lookup
Atomic nucleus wikipedia , lookup
Rutherford backscattering spectrometry wikipedia , lookup
Marcus theory wikipedia , lookup
History of molecular theory wikipedia , lookup
Physical organic chemistry wikipedia , lookup
Gaseous detection device wikipedia , lookup
Atomic orbital wikipedia , lookup
Molecular orbital diagram wikipedia , lookup
Stoichiometry wikipedia , lookup
Light-dependent reactions wikipedia , lookup
Bioorthogonal chemistry wikipedia , lookup
Hypervalent molecule wikipedia , lookup
Chemical bond wikipedia , lookup
Heat transfer physics wikipedia , lookup
Transition state theory wikipedia , lookup
Metallic bonding wikipedia , lookup
Electron configuration wikipedia , lookup
PCAT Prep Chemistry Section Laura Blue Angie Proctor Introduction to the Chemistry Section Details of the Chemistry Section • Format: 48 Multiple Choice Questions • Time: 30 minutes • Time/Question…37.5 seconds Test Taking Strategies • Eliminate and Simplify – Since you are unable to use a calculator, try to eliminate choices by estimating a probable order of magnitude for the answer or simplifying the math involved • Put it on Paper – Although you are limited on time, do not be afraid to write down equations, periodic trends, etc at the start of the section that may help you work through some of the problems • Know the Trends – Chemistry is all about knowing a little and applying it to a lot of applications. • Don’t know the Answer-GUESS – There is NO penalty for wrong answers • • Stay Calm and Confident Keep Track of Time Atomic Structure Atomic Structure Mass Number A X Z Element Symbol 23 11 Na Atomic Number Mass Number = # of protons + # of neutrons Atomic Number = # of protons # electrons = Atomic Number ± charge # neutrons = Mass Number - Atomic Number Atomic Structure and Ions 23 11 35 Na 17 Cl Neutral Atom Neutral Atom # protons = 11 # protons = 17 # electrons = 11 # electrons = 17 # neutrons = 12 # neutrons = 18 Na Na+ + e- Cl + e- Cl- Charged Atom-Cation Charged Atom-Anion # protons = 11 # protons = 17 # electrons = 10 (Minus 1 electron) # neutrons =12 # electrons = 18 Na+ Cl- (Plus 1 electron) # neutrons =18 Isotopes Isotopes - atoms of the same element with the same number of protons but a different numbers of neutrons 23 Example: 11 Na and 24 11 Na Atomic Mass Unit (amu) – relative mass of element compared to mass of the C12; each isotope has a different amu Molecular weight (MW) – weighted average of relative abundances of different isotopes of element MW of Element Z = xA(amu A) + xB(amu B) +… where: x = fraction of composition or relative abundance A,B, etc. = atomic mass of each isotope of Z Bohr Model An electron can exist only in certain fixed energy states e- can be excited to a higher energy level Excited state return to ground state n=5 n=4 n=3 n=2 n=1 Absorbs Energy/ Photons release energy Emits Energy/Photons eˉ Energy (hc/) 434.1nm (Violet) 656.3 nm (Red) eˉ 486.1 nm (Bluegreen) Wavelength (nm) Quantum Mechanical Model of Atoms • Heisenberg Uncertainty Principle-Unable to calculate both the momentum and location of an electron in an atom simultaneously • Electrons travel in diffuse clouds or orbitals around the nucleus described by probability distributions • Quantum Numbers are used to describe the location of each electron in an atom – – – – n Principle Quantum Number ɭ Angular Momentum Quantum Number mɭ Magnetic Quantum Number ms Spin Quantum Number • Pauli exclusion principle – no 2 e- in a given atom can have the same set of quantum numbers Principal Quantum Number (n) •Integer Value (n = 1, 2, 3….) •Related to the size and energy of the orbital or shell -Larger n: higher the energy of electron and larger radius of electron •Maximum number of electrons in an energy level = 2n2 Energy n=1 n=2 n=3 Angular Momentum Quantum Number (ɭ ) • Integral value from 0 to n-1 for each value of n ɭ Subshell • Corresponds to the shape of the orbital 0 s 1 p 2 d 3 f • Maximum number of electrons in subshell = 4 ɭ + 2 p-orbitals s-orbitals d-orbitals f-orbitals Magnetic Quantum Number (mɭ ) • Values range from – ɭ to + ɭ including 0 • Describes the orientation of an orbital in space Subshell ɭ mɭ s 0 0 p 1 -1, 0, +1 d 2 -2, -1, 0, 1, 2 f 3 -3, -2, -1, 0, 1, 2, 3 Spin Quantum Number (ms) • Denoted as + ½ or - ½ electron spin • Electron spin is based on the behavior of electrons as tiny magnets. +½ e- -½ + +½ -½ e- Electron Configuration-Key Concepts • Aufbau Principle – Electrons enter orbitals of lowest energy first – s subshell is always the lowest in energy 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p Increasing Energy • Hund’s Rule – When electrons occupy orbitals of equal energy one electron enters each orbital until all the orbitals contain one electron with spins parallel, then electrons are paired 2px 2py 2pz Quantum Numbers-Practice Problem What are the four quantum numbers for the 28th electron in Nickel? 1. Write the Electron Configuration 1s2 2s2 2p6 3s2 3p6 4s2 3d8 2. Determine Quantum Numbers n =3 mɭ = 0 ɭ =2 ms = -1/2 S 0 0 d -2 -1 0 1 2 p -2 -1 0 1 2 Quantum Numbers - Practice Question Which of the following combinations of quantum numbers is allowed for a single electron in an element? A. n = 2, ɭ = 2, mɭ = 1, ms = ½ Remember: B. n = 3, ɭ = 1, mɭ = 0, ms = -½ ɭ = (0 n-1) mɭ = (- ɭ ɭ ) C. n = 5, ɭ = 1, mɭ = 2, ms = ½ D. n = 4, ɭ = -1, mɭ = 0, ms = ½ ms = +½ or -½ Quantum Numbers - Practice Question The maximum number of electrons in a shell with the principal quantum number equal to 4 is: A. 2 B. 10 Remember: Max # e- = 2n2 C. 16 D. 32 Electron Configuration - Practice Question Within one principal quantum level of a many electron atom, which orbital has the lowest energy? A. B. C. D. s p d f 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6, 5s2, 4d10, 5p6, 6s2, 4f14, 5d10, 6p6, 7s2, 5f14, 6d10, 7p6 Increasing Energy What is the number of half-filled orbitals in one ground-state atom of sulfur (atomic number 16)? A. B. C. D. 2 3 4 6 1s2 2s2 2p6 3s2 3p4 Sum Superscripts = 16 = # electrons 3px 3py 3pz The Periodic Table Atomic Properties and Electronic Structure • Positively charged nucleus attracts negatively charged electrons • Negatively charged electrons repel other electrons Atomic Properties and Electronic Structure Effective Nuclear Charge (Zeff) Net positive charge experienced by an electron in a multielectron atom Calculating zeff: Zeff = Z – S Z = Number of Protons S = Number of Electrons in inner shells Generally Zeff increases when… -Atomic Radii Decreases -Electronegativity Increases -Ionization Energy Increases Atomic Radius Atomic Radius Increasing Atomic Radius Increasing Atomic Radius Moving Across a Period: • Increasing number of protons • Same size electronic shell • Greater Zeff • Size gets smaller Moving Down a Group: • Larger n means a larger orbital • More shielding of nuclear charge • Overall size is larger Trends in Ion Size • Adding Electrons Forms Anions – Negative ions are larger than the atoms from which they are formed – More electrons means increased repulsions – Occupy a larger volume ee- e- e- e- Cl e e- e- e- + e- e Cl ee- e- e- e- *Electron Structure Similar to Argon Trends in Ion Size • Losing Electrons Forms Cations – Positive ions are smaller than the atoms from which they are formed – Electrons “feel” higher nuclear charge – Remaining electrons are pulled closer to nucleus e- Na e- + Na+ *Electron Structure Similar to Neon Ionization Energy • Ionization Energy (IE) – Energy required to remove an electron from: – An isolated, gaseous atom – An ion in its ground state – Reflects how tightly the electron is held by the atom – Successive ionization energies always increase – Typically endothermic Example: Magnesium Mg(g) 738 kJ/mol Mg+(g) + 1e- Mg+(g) 1450 kJ/mol Mg2+(g) + 1e- Ionization Energy Increasing Ionization Energy Ionization Energy Increasing Electron Affinity • Electron Affinity (EA) – Potential energy change associated with addition of electron to: – A gaseous atom – An ion in its ground state – Relative ease by which atoms gain electrons – Typically exothermic Example: Fluorine F(g) + 1e- F-(g) + 328 kJ Electron Affinity Increasing Electron Affinity Electron Affinity Increasing Electronegativity • Electronegativity (EN) – Attraction of electrons to atoms of certain elements within a compound – Pauling scale from 1-4 used to help predict bonding – Noble gases omitted Element EN F 4.0 O 3.5 Cl 3.0 N 3.0 C 2.5 H 2.1 Ag 1.9 Cs 0.79 Electronegativity Increasing Electronegativity Electronegativity Increasing Periodic Trends - Practice Question • Arrange the following calcium species in terms of increasing size: Ca, Ca+, Ca2+, Ca3+, Ca-, Ca2Positive Ions = Smaller Ionic Radius Negative Ions = Larger Ionic Radius Ca+3 < Ca+2 < Ca+ < Ca < Ca- < Ca2- • Elements in a given period have the same A. B. C. D. E. Atomic weight Maximum angular momentum quantum number Maximum principal quantum number Valence electron structure Atomic number C. Bonding and Chemical Reactions Intermolecular Forces Force Ion-Induced Dipole Strength Players (10-50 kJ/mol) Ions and Polar Solvents Dipole-Dipole (3-4 kJ/mol) Polar Molecules London Dispersion (1-10 kJ/mol) All Molecules Hydrogen Bonding (10-40 kJ/mol) H with O,N,F Figure Bonding (Intramolecular Forces) • Ionic Bond – Transfer of electrons – Often metals (cations) and nonmetals (anions) – The potential energy or lattice energy of the system is lowered (exothermic) when ion complexes form • Covalent Bond – Sharing of electrons – Shift of electron density – Between two or more nonmetals – Molecules are electrically neutral combinations of atoms Bonding (Intramolecular Forces) EN Difference Bond > 1.7 Ionic <1.7 Covalent EN = Electronegativity + Na e + e e - - Cl e - e e - EN 0.9 - e C + H - e - - 3.0 EN Difference = 2.1 Ionic Bond EN 2.5 2.1 EN Difference = 0.4 Covalent Bond Types of Covalent Bonding • Polar Covalent −Unbalanced electron density within a bond −Each atom has partial charge −Partial negative charge (d-) −More electronegative atom −Partial positive charge (d+) −Less electronegative atom d+ d- O=C=O H F No Net Dipole Moment Types of Covalent Bonding • Non-Polar Covalent • Atoms with same electronegativity – equal sharing • Example: Cl2 • Coordinate Covalent • Shared electron pair comes from lone pair of one of the atoms in the molecule H H I I DEN Bond • Example: F B N H > 1.7 Ionic I I H H 1.7 < x < 0.5 Polar Covalent < 0.5 Non-Polar Covalent Bonding • Octet Rule (General) – Atoms gain or lose e- until 8 electrons are in outer shell – Exceptions: Transition metals (>8e-), boron (stable with 6e-), beryllium (stable with 4e-) • Lewis Dot Structure Describes Bonding Between Atoms 1.Decide on Central Atom Least Electronegative 2.Count Total Valence Electrons 3.Add Bonds to Central Atom 4.Complete Octets of More Electronegative Atoms 5.Place Remaining e- in Pairs on Central Atom 6.Consider Multiple Bonding Covalent Bonding • Formal Charge (FC) on Elements FC = [# Valence e-] – [½ of bonding e-+ non-bonding e-] − Electronegative elements will have 0 or negative FC − Electropositive elements will have 0 or positive FC • Multiple Bonding − Higher bond order (double or triple bond) − Shorter bond length − Higher bond energy - .. N O O • Resonance – Hybrid between two or more equivalent Lewis structures – Lowers overall energy – More stable Covalent Bonding: Example Draw Lewis Structure of NCO1. Central Atom 2. Count Total Valence Electrons 3. Add Bond to Central Atom 4. Complete Octets 5. Remaining e- in Pairs on Central Atom 6. Consider Multiple Bonding and Resonance N-C-O : : 1- : : :N-C-O: : : 1- : : N=C=O N=C=O *NOTE: Atoms beyond the third period may be exceptions to the octet rule since valence electrons can occupy the d orbitals 1- : : 1- : : : N C O: 1- : : : : 1- C=4; O=6; N=5 :N C O : • EN C = 2.5 EN O = 3.5 EN N = 3.0 Covalent Bonding: Example • : • Dipole Moment : : : 1- : N=C=O No Net Dipole : : : Calculate Formal Charge FC = [# Valence e-] – [½ of bonding e-+ non-bonding e-] FCOxygen = 6 – [(½) 4 + 4] = 0 1FCCarbon = 4 – [(½) 8 + 0] = 0 N=C=O FCNitrogen = 5 – [(½) 4 + 4) = -1 Total Formal Charge = -1 VSEPR Theory • Valence Shell Electron Pair Repulsion Theory – Electron pairs spread out to minimize repulsion – Gives molecules its geometry • Determining Geometry Using VSEPR Theory – Draw the Lewis structure – Arrange electron pairs around central atom so that they are as far apart from each other as possible – Lone pairs – Bonding groups VSEPR: Molecular Geometry Name Example Hybrid Lone Pair Bond Angle Linear Molecule CO2 sp 0 180o Planar Triangle BCl3 sp2 0 120o Tetrahedral CH4 sp3 0 109.5o NH3 sp3 1 107o Bent H2O sp3 2 104.5o Trigonal Bipyramidal PCl5 sp3d 0 120o/90o Unsymmetrical Tetrahedron SF4 sp3d 1 120o/90o T-Shaped ClF3 sp3d 2 90o Linear I3- sp3d 3 180o SF6 sp3d2 0 90o BrF5 sp3d2 1 90o XeF4 sp3d2 2 90o Pyramidal Structure .. .. Octahedral Square Pyramidal Square Planar .. .. .. Chemical Bonding - Practice Questions • Predict the geometry of ClO2 A. B. C. D. bent linear trigonal planar pyramidal + 1. Count total electrons = 18 1Cl + 2O – 1e- 2. Draw Connectivity 3. Fill in electrons (octet rule) 4. Predict Geometry • Arrange the following compounds in order of increasing boiling point HINT: Consider Bonding Properties A. B. C. D. C2H6 CH3OH LiF HCl bp C2H6 HCl CH3OH LiF Dispersion Dipole-Dipole Hydrogen Bonding Ionic Compounds and Stoichiometry What is a Mole? • Mole – SI Unit for the amount of chemical substance • Avogadro’s Number – 1 mole = 6.022 x 1023 particles – 1 mole Na atoms = 6.022 x 1023 Na atoms – 1 mole He atoms = 6.022 x 1023 He atoms – 1 mole cookies = 6.022 x 1023 cookies The mass of 1 mole of a substance varies depending on the substance!! What is a Mole? • Stochiometric coefficients – Used to indicate the number of moles of a given species involved in the reaction – Two moles of H2 gas must be reacted with one mole of O2 gas to form two moles of water 2 H2 (g) + O2 (g) 2 H2O (g) MW (g/mol) e.g. 1 mole Carbon = 12.011 g Carbon Molar Heat of Reaction (energy/mol) 1 mole = 6.022 x 1023 particles MOLE 1 mole = 22.4 L of a gas at STP Percent Composition % Composition = Mass of X in Formula x 100 Molar Mass of Compound Example: Calculate percent composition of chromium in K2Cr2O7 MW: K = 39 g/mol, Cr = 52 g/mol, O = 16 g/mol Molar Mass of K2Cr2O7 = 2(39g/mol) + 2(52g/mol) + 7(16g/mol) Molar Mass of K2Cr2O7 = 294g/mol %Cr = 2(52g/mol) x 100 294g/mol %Cr = 35.4% Molecular and Empirical Formula • Molecular formula • The exact number of atoms of each element in the compound C 4H 6O 2 • Empirical formula • The simplest whole number ratio of the elements in the compound C 2H 3O 1 Percent Composition and Empirical Formulas Determine the Empirical Formula According to the Following Percent Compositions 2.1 % Hydrogen 29.8% Nitrogen 68.1% Oxygen Step 1: Change Percent Composition to Grams Assuming a 100 g Sample 2.1 grams H 29.8 grams N 68.1 grams O Step 2: Divide by the Molecular Weight of the Element 2.1 grams H 29.8 grams N 68.1 grams O 1.0 g/mol H 14.0 g/mol N 16.0 g/mol O 2.1 mol H 2.13 mol N 4.26 mol O Step 3: Divide by the Smallest Value 2.10 mol H 2.13 mol N 2.10 mol 2.10 mol 1H : 1N 4.26 mol O 2.10 mol : Empirical Formula: HNO2 2O Types of Chemical Reactions 1. Combination Reaction – When 2 or more reactants combine to form one product – Exothermic reaction S (s) + O2 (g) SO2 (g) 2. Decomposition Reaction – When a reactant decomposes into two or more products (at least one gas) – Endothermic reaction 2 H2O (g) 2 H2 (g) + O2 (g) Types of Chemical Reactions 3. Single Displacement Reaction – When a single atom (or ion) of one reactant is displaced by an atom of another element based on activity to form a product Zn (s) + 2 HCl (aq) ZnCl2 (aq) + H2 (g) 4. Double Displacement Reaction – When elements from two different reactants displace each other to form two new products – Exothermic reaction CaCl2(aq) + Na2CO3(aq) CaCO3(s) + 2NaCl(aq) **Always removes one compound from solution as a precipitate (solid) or a gas** Types of Chemical Reactions 5. Combustion Reaction – When a hydrocarbon in the presence of O2 oxidizes or burns to form a gas and H2O – Exothermic reactions 2C4H10 + 13O2(g) 18CO2(g) + 10H2O(aq) 6. Neutralization Reaction − An acid and base react to form a salt in water HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) **More about these in the Acid/Base section** Writing and Balancing Chemical Equations Solubility Rules • • • SOLUBLE Group 1 Metals, NH4+, NO3-, ClO3-, ClO4-, C2H3O2Cl-, Br-, I- ; Except when combined with Pb2+, Ag+, Hg22+ SO42- ; Except when combined with Pb2+, Ag+, Hg22+, Ca2+, Sr2+, Ba2+ • • INSOLUBLE Metal Oxides and Hydroxides Insoluble Except Group 1, NH4+, Ca2+, Ba2+, Sr2+ PO43-, CO32-, SO32-, S2- Insoluble Except Group 1 and NH4+ GASES • • • • • • • H+ + Sulfide H2S (g) H+ + CN HCN (g) H+ + HCO3- CO2 (g) + H2O H+ + CO3-2 CO2 (g) + H2O H+ + HSO3 SO2 (g) + H2O H+ + SO3- SO2 (g) + H2O OH- + NH4+ NH3 (g) + H2O Writing and Balancing Chemical Equations Balance the following reaction: C4H10 (l) + O2 (g) CO2 (g) + H2O (l) Combustion Reaction 1. Balance the carbon in reactants and products C4H10 (l) + O2 (g) 4 CO2 (g) + H2O (l) 2.Balance the hydrogen in reactants and products C4H10 (l) + O2 (g) 4 CO2 (g) + 5 H2O (l) 3. Balance the oxygen in reactants and products C4H10 (l) + 6.5 O2 (g) 4 CO2 (g) + 5 H2O (l) 4. Make stoichiometric coefficients whole numbers 2 C4H10 (l) + 13 O2 (g) 8 CO2 (g) + 10 H2O (l) Coefficients, not subscripts, are used to balance chemical equations Smallest whole-number coefficients used to write balanced equations Applications of Stoichiometry How many grams of calcium chloride are needed to prepare 72 g of silver chloride according to the following reaction: CaCl2 (aq) + 2 AgNO3 (aq) Ca(NO3)2 (aq) + 2 AgCl (s) Calculate Molar Mass of the two compounds: •CaCl2 = 110 g/mol •AgCl = 144 g/mol 72 g AgCl x 1 mol AgCl x 1 mol CaCl2 x 110g CaCl2 144 g AgCl 2 mol AgCl 1 mol CaCl2 = 27.5 g CaCl2 Chemical Kinetics and Equilibrium Reaction Mechanisms A2 + 2 B 2 AB Overall Reaction: rate = -1 D[A2] = Dt -1 D[B] = 2 Dt 1 D[AB] 2 Dt • B consumed twice as fast as A2 • AB produced twice as fast as A2 is consumed • AB produced at same rate as B is consumed Two-step reaction: Step 1: A2 + B A2B Step 2: A2B + B 2 AB (Fast) (Slow) A2B is an intermediate rate-determining step - The slowest step (step 1) in a proposed mechanism Kinetics • Definition of Rate: decrease in [reactants] increase in [products] = rate = time time aA + bB rate = -1 D[A] a Dt = cC + dD -1 D[B] b Dt 1 D[C] = c Dt = 1 D[D] d Dt rate = k [A]x [B]y rate constant reaction order *Note: Do not assume reaction order = coefficients in chemical equation Determining the Exponents of a Rate Law aA + bB cC + dD rate = k[A]x [B]y Experimental determination of x and y: Trial [A]initial(M) [B]initial(M) rinitial(M/sec) 1 1.0 1.0 2.0 2 1.0 2.0 8.1 3 2.0 2.0 16 Find the rate law for the given reaction at 300K. Determining the Exponents of a Rate Law aA + bB cC + dD rate = k[A]x [B]y Experimental determination of x and y: Trial [A]initial(M) [B]initial(M) rinitial(M/sec) 1 1.0 1.0 2.0 2 1.0 2.0 8.1 3 2.0 2.0 16 Trial 1: r1 = k[A]x [B]y = k(1.00)x (1.00)y Trial 2: r2 = k[A]x [B]y = k(1.00)x (2.00)y r2 r1 k(1.00)x (2.00)y 8.1 = 2.0 = k(1.00)x (1.00)y = 4 = (2.00)y y = 2 Determining the Exponents of a Rate Law aA + bB cC + dD rate = k[A]x [B]y Experimental determination of x and y: Trial [A]initial(M) [B]initial(M) rinitial(M/sec) 1 1.0 1.0 2.0 2 1.0 2.0 8.1 3 2.0 2.0 16 Trial 2: r2 = k[A]x [B]y = k(1.00)x (2.00)y Trial 3: r3 = k[A]x [B]y = k(2.00)x (2.00)y r2 r1 k(2.00)x (2.00)y 16 = 8.1 = k(1.00)x (2.00)y = 2 = (2.00)x x = 1 Determining the Exponents of a Rate Law aA + bB cC + dD rate = k[A]x [B]y Experimental determination of x and y: Trial [A]initial(M) 1 1.0 1.0 2.0 2 1.0 2.0 8.1 3 2.0 2.0 16 x=1 y=2 rate = k[A] [B]2 [B]initial(M) rinitial(M/sec) Solving for k: k= Rate [A]1 [B]2 = 2.0 M s-1 = (1.0 M)1 (1.0 M)2 2.0 M-2 sec-1 Reaction Orders Zero-Order – constant rate (M *sec-1); independent of reactants’ concentrations rate = k [A] Differential Equation: slope = -k rate = -d[A]/dt = k Integrate t [A] = [A]0 - kt Half-life – time needed for the concentration of the substance to decrease to one-half the initial value t1/2 = 0.5 A0/k Reaction Orders First-Order – rate proportional to the concentration of one reactant (sec-1) rate = k[A] ln[A] Differential Equation: slope = -k rate = -1 * d[A] = k[A] a dt ln[A] = - kt + ln[A]0 Half-life t1/2 = 0.693/k ** Math Tip: ln x = (log x) (2.3) ** Reaction Orders 1 / [A] Second-Order – rate proportional to the concentration of two reactants, or the square of one reactant (M-1sec-1) rate = k[A]2 or (k[A][B]) slope = k t Energy Diagram H2 + Cl2 2 HCl H H H H H H Cl Cl Cl Cl Cl Cl Energy transition state Ea forward Reactants Ea reverse DHrxn Products Progress of Rxn (Reaction Coordinate) Energy Diagram Energy Exothermic DHreactants > DHproducts DHrxn is negative Temperaturesystem Endothermic DHreactants < DHproducts DHrxn is positive Temperaturesysem Progress of Rxn (Reaction Coordinate) Factors Affecting Reaction Rate • Nature of reactants and differences in chemical reactivity • Ability of reactants to collide • Concentration of reactants • Temperature • Activation energy (Ea) • Presence of catalysts Equilibrium Constant aA+bB cC+dD c d [C] [D] K eq a b [A] [B] Properties of Equilibrium Constant (Keq): • Do not include pure solids and liquids in Keq expression • If Keq > 1, Products Favored If Keq < 1, Reactants Favored If Keq = 1, Products = Reactants • Keq is dependent on temperature • If reaction is more than one step, multiply together Keq from each step in the reaction • When the system is not at equilibrium, use Q instead of Keq Reaction Quotient (Q) – measures degree of reaction completion Le Chatelier and Chemical Equilibria Le Chateliers Principle - if an external stress if applied to the system that is at equilibrium, the system will attempt to adjust itself to offset the stress CH3OH (l) + H2 (g) CH4 (g) + H2O (l) DH = -30 kcal 1. Adding Removing a Reactant or Product -Add Reactant Shift Right; Add Product Shift Left -Remove Reactant Shift Left; Remove Product Shift Right 2. Changing Volume of Gaseous Reactions -Decrease volume of gas increases pressure. Shift towards lower # molecules Ex: 1M H2 vs 1M CH4, since same number moles no effect on change in volume 3. Change in Temperature -Treat as reactant (endothermic) or product (exothermic); shift accordingly 4. Effect of a Catalyst -Affect forward and reverse reactions equally No net change Chemical Kinetics - Practice Question • Which would NOT result in a formation of more XeF6(g) based on Le Chatelier’s Principle? A. Increasing the concentration of F2(g) B. Decreasing the concentration of XeF6(g) C. Decreasing the volume of the container D. Decreasing the pressure of the container Thermochemistry Some Basic Concepts of Thermochemistry Thermodynamics - energy changes determine whether a reaction can occur and how easily reaction will proceed Types of Systems: • Isolated: Cannot exchange energy or matter with surroundings • Closed: Can exchange energy but not matter with surroundings • Open: Can exchange energy and matter with surroundings Types of Processes: • Isothermal: No change in temperature • Adiabatic: No heat exchange • Isobaric: No change in pressure • Isochoric: No change in volume Calorimetry Heat Capacity (C = J g-1 K-1) - The amount of heat required to raise the temperature of 1 g of a substance 1 oC •General Equation q = mCΔT q = the heat absorbed or released (J = kg m2 s-2) m = the mass (g) C = the specific heat of the compound (J g-1 K-1) ΔT = the change in temperature (K) •Constant Pressure (Cp) Measures directly the enthalpy change during the reaction q = mCpDT •Constant Volume (Cv) Measures the internal energy change between reactants and products q = mCvDT •The difference between CP and CV is the capacity of the gas to expand and do work and is equal to Cp = CV + R for an ideal gas where R is the universal gas constant State Functions State function – property whose magnitude depends only on initial and final states of system (not on path) 1) Enthalpy (H)- reflects the heat exchange of a reaction at constant pressure 2) Entropy (S) – is a measure of disorder, or randomness of system 3) Gibbs Free Energy (G) – maximum energy released by the process Enthalpy DHrxn = DHproducts – DHreactants +DHrxn = endothermic process -DHrxn = exothermic process • Standard heat of formation (DHof) DH if one mole of a compound were formed directly from its elements in their standard state • Standard conditions (o) T = 25oC or 298 K, P = 1 atm Hess’s Law Hess’s law - Enthalpies of reaction are additive Overall Reaction (1) 2C (s) + 2O2 2CO2 DHrxn = -787 kJ Multi-Step Reaction: (2) 2C (s) + O2 2CO DHrxn = ? (3) 2CO + O2 2CO2 DHrxn = -566 kJ What is the Heat of Reaction for one mole of CO? + 2 C (s) + 2 O2 2 CO2 DHrxn = -787 kJ 2 CO2 2 CO + O2 DHrxn = 566 kJ DHrxn = -221 kJ 2 C (s) + O2 2 CO The product has a coefficient of 2, so 1 mole = -221 kJ 2 mol = -110 kJ/mol Entropy DSrxn = DSproducts – DSreactants DS = qrev/T +DSrxn = Increase in Disorder -DSrxn = More Ordered • Predicting the Sign of DS -Volume: For gases, entropy increases with increasing volume -Temperature: Temperature = S -Physical State: Ssolid < Sliquid < Sgas Gibbs Free Energy DG = DH – TDS DG < 0 spontaneous reaction DG > 0 non-spontaneous reaction DG = 0 equilibrium (DH = TDS) DH DS Outcome - + Spontaneous at All Temperatures + - Non-Spontaneous at All Temperatures + + Spontaneous only at High Temperatures - - Spontaneous only at Low Temperatures Gibbs Free Energy Equilibrium Constants from Thermodynamics DG = DGo + RT ln Q At Equilibrium… DGo = -RT ln K R = Gas Constant (8.314 J mol-1 K-1) T = Temperature (K) Thermochemistry - Practice Question • Consider the exothermic reaction given above. Predict the sign of DH, DS, and DG A. .DH = neg, DS = neg, DG = neg B. .DH = neg, DS = pos, DG = neg C. .DH = neg, DS = pos, DG = pos D. .DH = pos, DS = pos, DG = pos Remember: Exothermic = -DH Disordered = +DS DG = DH - TDS Phase Changes and the Gas Laws The Phase Diagram Pressure Freezing Critical Point Liquid Melting Condensation Solid Vaporization Deposition Triple Point Gas Sublimation Temperature •Triple Point: -T and P in which all three phases are in equilibrium -For H2O, Triple Point = 0.01oC and 4.58 torr •Critical Point: -T and P in which no distinction between gas and liquid is possible -Substance above critical point = supercritical fluid (e.g. CO2) Kinetic Molecular Theory for Ideal Gases Basic Assumptions • • A gas consists of objects with a defined mass and zero volume A gas consists of a large number of very tiny particles that are in constant random motion. KEavg Temperature • • • Particles collide in perfectly elastic collisions (no energy gained or lost) Molecules travel in rapid, random, straight line motion without attraction or repulsions States of matter are determined by kinetic energy and attraction The Gas Laws Boyle’s Law: the volume of a fixed quantity of gas at constant temperature (isothermal) is inversely proportional to the pressure V1/P P1V1 = P2V2 Charles’s Law: the volume of a fixed amount of gas at constant pressure (isobaric) is directly proportional to its absolute temperature VT V1/T1 = V2/T2 Gay-Lussac’s Law: the pressure of a fixed amount of gas held at constant volume is directly proportional to the absolute temperature PT P1/T1 = P2/T2 Avogadro’s Law: the volume of a gas at constant temperature and pressure is directly proportional to the number of moles Vn n1/V1 = n2/V2 The Ideal Gas Law Ideal gas law – represents hypothetical gas whose molecules have no intermolecular attraction and occupy no volume PV=nRT STP: standard conditions of temperature and pressure 1 atm 101.3 kPa 29.96 inches Hg 760 torr 273.15 K (0oC) R = 0.0821 L atm mol-1 K-1 or 8.314 J K-1 mol-1 At constant T and P, all gases have the same number of moles in the same volume 1 mol gas = 22.4 L Gas Laws - Practice Problem A sample of argon occupies 50 L at standard temperature. Assuming constant pressure, what volume will the gas occupy if the temperature is doubled? (A) (B) (C) (D) (E) 25 L 50 L 100 L 200 L 2500 L Solution: Step1: V1= V2 T1 T2 Step 2: 50L 273K = “x” = 100 L “x” 546K Dalton’s Law of Partial Pressure • Dalton’s Law of Partial Pressures: the total pressure (Ptotal) of a mixture of gases is equal to the sum of the partial pressures (Px) of the individual components Ptotal = PA+ PB + PC + … PA = Ptotal xA • Mole Fraction: XA = nA nTOTAL Colligative Properties Colligative properties – physical properties derived solely from the number of particles present 1) Freezing Point Depression - change in freezing point due to presence of another solute DTf = Kfm 2) Boiling Point Elevation – change in boiling point due to presence of another solute DTb = Kbm 3) Osmotic Pressure – tendency of solute to move from a region of higher concentration/pressure to a region of lower concentration/pressure = MRT molality (m) = mol solute/ kg solvent