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Harvard University
Physics PHYS S - 1ab
Alan Aspuru Guzik

PHYSICAL SCIENCES 1 – SPRING 2012 FINAL STUDY GUIDE THINGS TO NOTE AS PITFALLS OF STUDENTS! - UNITS: be sure of what temperature ( C vs K) and energy (J vs kJ) - MECHANISMS: be careful of K vs k; give n K, set K = [prods]/[reacts]; given k, set R = k[reac1][react2] - CELL TYPE: be careful of whether the question is asking about a voltaic or electrolytic cell; just flip which specie would be what electrode - LEWIS STRUCTURE BOND TYPE TO REDUCE FORMAL CHARGES: always check for a better structure if the single bond structure has formal charges. See if creating double/triple bonds will alleviate formal charges. Think about the practice final problem, XeO . Doubl3 bonds on all 3 O arms removes F.C.s - - - - - - - - - - - - GENERAL INFORMATION SI Unit Prefixes Tera Giga Mega Kilo Deci Centi Milli Micro Nano Pico Femto 12 9 6 3 -1 -2 -3 -6 -9 -12 -15 10 10 10 10 10 10 10 10 10 10 10 Trillion Billion Million Thousand Tenth Hundredth Thousandth Millionth Billionth TrillionthQuadril-th QUANTUM THEORY Light Electron/Particle As a WAVE λ = = λ = = = As a PARTICLE Momentum: p = Momentum: p = mv Energy of light (photon): Energy (classical formula) E = hf = E = mv =2 Eigenfunctions (1-D Particle in a Box) - Energy = E =n ΔE n Hydrogen Atom special case: ΔE = 2.179x10 J( - ) -18 ATOMIC STRUCTURE AND PERIODIC TABLE Periodic Trends - Quantum Numbers - Orbitals and Energy For Hydrogen (and other nuclei stripped down to just one electron), the energy is dependent only the principle quantum number, n, a.k.a. degenerate/degeneracy Degenerate vs Non-Degenerate – refers to how hydrogen atoms (and other nuclei stripped down) have all energy levels of the same principal quantum number at the same energy level (imagine the electron dashes diagram) Pauli Exclusion Principle – no two electrons in the same atom can have the same four quantum numbers Stern-Gerlach Experiment – (uhh…) proved angular momentum in particles(?) Electron Configuration - 2 10 5 i.e. Br: [Ar] 4s 3d 4p - Exceptions: Cr and Cu (which have 4s and 3d filled evenly with 1 or 2 e , respectively) Rules: - Ions removed e from largest n, while add e to smallest n Diamagnetic vs Paramagnetic: - all orbitals filled vs at least one orbital is unpaired - implies ability to interact/bind (got free electron or all of them paired?) Lewis Structures - 1 – Count valence electrons and charge = total electrons 2 – Set up possible, realistic layout of atoms (the order of elements in compound name is usually hint, i.e.: SCN ) - least electronegative atom is usually central atom - H and F are always terminal atoms - in oxyacids (H 2O , 4NO , e3c.), Os are bonded to central atom and H(s) is/are bonded to Os 3 – Connect each adjacent atom with a SINGLE BOND 4 – Dot the structure, using all electrons from step 1, to complete octect - try not to expand octet if possible; if must, add to central atom - the central atom can be an octect or expanded, but never less than an octet - try multiple bonds if central atom lacks an octet 5 – Figure out the charges on each atom (standard # of electrons – # of dots and lines around specie) 6 – Choose Lewis Structure with no separated charges, has total overall charge equal to species’ nomenclature charge, and has the lowest charges on each atom (example |-1|+1|-1| > |-2|+1|0|) 7 – Check for resonance structures (must have just as good of formal charges to be counted as resonance, and usually, resonance structures are simply a switch on where a double/triple bond is) 8 – If the overall molecule is charged, then the final lewis structure MUST HAVE brackets and charge sign BONDING AND MOLECULAR GEOMETRY Bond Energies and Ground State Energies – an atom with more protons and less electrons (ion) will have a lower ground state energy than an atom with less protons. One needs more energy + + to rip an electron from He than H (since the ground state energy of He is lower than H). All the way to the top of that canyon graph is ionization; the deeper, the more energy required to rip electron Valence Bond Theory, Valence-Shell Orbital Diagrams, and Hybridization - covalent bond is formed when two atomic orbitals overlap (electrons are shared) - hybridization is when orbitals mix; example: VSEPR (Valence-Shell Electron-Pair Repulsion) Theory and Molecular Shape Types of Covalent Bonds – (think of midterm 1 example) - σ (sigma) bond: end-to-end overlap, highest electron density between nuclei (Can be s, sp, 2 3 sp , sp ) - π (pi) bond: side-to-side overlap, electron density above and below σ–bond axis (ALWAYS P- orbitals) - # of bonds: single bond: 1 σ-bond double bond: 1 σ-bond and 1 π-bond triple bond: 1 σ-bond and 2 π-bonds How to Designate Labels for Orbitals involved in Forming Bonds - - Count the # of bond pairs (1 = lone pair, single bond, double, or triple bond) - Example: ICN 3 - I has three lone pairs and one single bond to C  4 bond pairs  sp automatically - C has single bond to I and triple bond to N  2 bond pairs  sp automatically - N has one lone pair and one triple bond to C  2 bond pairs  sp automatically Electronegativity – relative ability of a bonded atom to attract shared electrons - Most EN atoms are F, O, N, Cl Bond Energies ΔH orxn ΣBE bonds to be broken in reacbonds formed to make products * Note that I count everything. If some equation is x + y  2NH 3 I have 6 N-H bonds created! Conjugation –structure with alternating single and double bonds due to resonance; when an electron jumps from one atom to the next, a double bond is formed; the electron is free to run across the molecule, back and forth, like the resonation of an eigenwave ; a conjugated molecule such as butadiene contains conjugated pi bonds that allow the pi electrons to delocalize along the length of the molecule. These are comparable to the electrons that move freely along the 1-D line in the particle-in-a-box model. MOLECULAR ORBITAL THEORY Molecular Orbital Diagram / Energy-level Diagram Molecular Orbitals - * Note: this table shows the order for the exceptions B , C , and N2. U2ually, th2 order has the third and fourth lines switched. Molecular Orbital Bond Order – the higher the bond order, the stronger the bond - - equation = x [(# e in bonding MOs) – (#e in anti-bonding MOs)] i.e. looking at diagram above of H : 0.5 x2(2 – 0) = 1 i.e. Paramagnetism – means that an electron is in an orbital that is not full; therefore, the species is more reactive in a magnetic field. B , C , a2d N2are excep2ions to the MO pattern because they must be paramagnetic. The usual MO ordering of 2p electrons would mean require diagmagnetism, which is unacceptable for B , C , and N 2 2 2 INTERMOLECULAR FORCES, PROPERTIES, AND CONCENTRATION Intermolecular Forces - Disperion Forces – exist between all molecules; are generally weak; are stronger with more surface area; also increase as molec.mass increases because electron cloud becomes more polarizable Dipole-Dipole Forces – polarity of a molecule causes dipole charges in certain atoms of a molecule; a polar molecule next to other polar molecules will align according to their dipole charges; a polar molecule next to neutral molecules, it will induce a dipole in the neutral molecule Hydrogen Bonding – is a STRONG type of Dipole-Dipole force; exists for H to O and to F (i.e. bunch of H-F together, or H-O-H (H2O) together) Ion-Dipole Forces - ….????? Ion-Ion Forces – have the strongest interaction energy **Hydrogen Bonding TRICKERY example: The molecule to the right does not have hydrogen bonding, despite seeing the O and Hs. First, the O is most electronegative; therefore, it induces a positive dipole moment in C. However, that doesn’t mean the individual Hs in the CH3tails have positive dipole moments. If anything, they’re slightly negative to make up for the fact the Cs have negative dipole moments because of O. A negative O and negative-neutral H do nothing… there is no hydrogen bonding! (Think CH 3ails) Phase Diagram – Colligative Properties* – when working with solutions, the freezing and boiling points change with respect to the molality(m) of the solution. The freezing point decreases while the boiling point increases ΔT freezingKfreezing ΔT boilingboiling *These properties are only dependent on the solvent (denominator of m) Intermolecular Forces, Properties and Concentration and Related Equations ln () = – ( – )  When using Clausis-Clapeyron, keep pressure in mm Hg/torr; T in Kelvin Molarity (M) = = Molality (m) = = Mass percent (%mass) = Volume percent (%vol) = Mole fraction (X) = CHEMICAL KINETICS Reaction Rates – given: aA + bB  cC + dD the rate of reaction, R, is given by: *Note the negative sign of reactant rates and positive sign of product rates *[A], [B], [C], etc. are concentrations of that species in MOLARITY m n Reaction Order – given: R = k[A] [B] The exponent m is the order of the reaction with respect to reactant A, and n is the order with respect to reactant B. The overall order is given by p = m + n - To find k, usually R is given as the “Initial Rate” in some table. Therefore, manipulate to isolate k. k = Zeroth, First, and Second Orders - First column (order) = differential - Second column (equation) = integrated – used to find a certain concentration at time t I.E. R = k [A] m “rate” “rate constant” “order” Arrhenius Equation (and manipulations) - k = Aexp(–) & ln k = – + A & ln () = – ( – ) * k is extremely dependent on temperature; independent of concentrations appearing in the rate law Arrhenius Equation Problem Solving (Example Problem) The hydrolysis of urea has a first order rate constant of 4.2x10 s. The reaction is catalyzed by urease, which at 21 C has a rate constant of 3x10 s . If the activation energies for the uncatalyzed -1 -1 and catalyzed reactions are 134 kJmol and 43.9 kJmol , respectively, calculate the temperature at which the non-enzymatic hydrolysis of urea would proceed at the same rate as the enzymatic hydrolysis at 21 C. - The “same rate” implies that k catalyzedkuncatalyzed  Think: k = Aexp(–)  Aexp(–) = Aexp(–)  Plug and chug  Answer! Reaction Mechanisms – theoretical detailed description of a chemical rxn** - elementary processes are in the individual steps/events - slowest step is the rate determining step (if none, last step is determining step) - steady state condition takes place when a species is generated and consumed by steps at equal rates Reaction Mechanisms (Example Problem)** Given what’s on the left: what are the rate expressions that can characterize this mechanism for each species? ANSWER: Rate of change of [A] = -k [A]; [B1 = k [A] – 1 [B]; [C2 = k [B] 2 **For future users of this study guide: this study guide does not fully or adequately explain reaction mechanisms. No definition for R (rate) included, no definition of k (rate coefficient), no (1/a)(d[a]/dt) [individual species rates], etc. THERMOCHEMISTRY Law of Conservation of Energy – energy can neither be created nor destroyed First Law of Thermodynamics – heat and work are the means by which a system exchanges energy with its surroundings (U is any state function) ΔU = U final Uinitialq + w q: heat supplied to the system (q > 0) or taken away from the system (q < 0) w: work done on the system (w > 0) or by the system (w < 0) Work (w) and Pistons w = – P ext work equals the opposite sign of external pressure times the change in volume Example Work and Piston Problem: *This problem is tricky. It incorporates the concept of limiting reagents. Don’t forget to consider limiting reagents when working with problems with chemical equations and starting quantities. Latent heat refers to the amount of heat released or absorbed during the phase change at constant temperature. Specific heat of a substance is the amount of heat required to raise the temperature of 1 g of the substance by 1 K. Heat (Q) Equations - q = mcΔT q = nΔH (n is proportionality constant: mol /coefficient ) substance substance q = CΔT (for bombs with capacities without mass factor) Enthalpy (Change only, ΔH) – is equal to the heat measured during a constant-pressure thermochemical process; total enthalpy of a system cannot be measured; only the change can be measured Exothermic: ΔH < 0 - the system releases heat to the surrounding Endothermic: ΔH > 0 - the system gains heat from the surrounding o o ΔH of Formation (Δ H) Usagf in Problems – whenever given Δ H in problem, it’f something to solve Example: [Part of a problem] CH is put into a bomb… Δ H(CH o ) = –94.81kJ/mol. Therefore, 4 f 4(g) think: CH +42O  CO2+ 2H O 2 2  general combustion format: add O ; produce2CO and 2 water ΔH(CH )= 4(n mol)ΔHf° products Σ(m mol)ΔHf° reactants ΔH(CH ) =4(Δ H(CO ) f 2Δ 2(H O)) – (f H(2H ) + 2Δ H(Of))  4ll speciesfΔ 2s will be o f given ΔH(CH ) =4some value to be used for the bigger part of whatever problem this is o *All elemental species have enthalpies of formation = 0 (i.e. Δ H(O ) = 0) f 2 Enthalphy Problem Solving Method: 1) Always, always, always set up 1st law of thermodynamics equation, i.e.: q rxn1+ q rxn2+ q bombcal 0 qrxn+ q soln= 0 qsys= – q surr q rxn1+ q rxn2= – q bombcal qrxn = – q soln qsys + qsurr= 0 * Be sure to give a q for everything in problem (the cal, bomb, species, second species, etc) 2) Determine if I need to use q = mcΔT or q = nΔH for system(s) and surrounding in problem - systems can be phase change (nΔH) or temperature change for same phase (mcΔT) - surroundings are usually exclusively mcΔT, such as bombs and other calorimeters Example Enthalpy Problem: When 2.0 grams of Na CO are dissolved i2 1003 of water, the temperature of the solution increases from 25 C to 26.2 C. Calculate ΔH for the dissolving of Na C2 : 3 Answer: since energy must be conserved, the sum of the heat of the system must be equal and opposite to the heat in the surroundings. I can think q rxn+ q soln 0 -or- q rxn= -q soln nΔH = – mcΔT  (2.0g /106gmol )(ΔH) = – (100g+2g)(4.2J/g C)(26.2-25.0 C)  o ((Answer)) * I always forget that! Everytime I set some q systo qsurr I always set them equal to each other without changing the sign on one side. Don’t forget to think! Switch the sign on one side! Otherwise, some endothermic process might give me a negative ΔH while some exothermic process might give me a positive ΔH. Another Example Enthalpy Problem: Entropy (S) - is the thermodynamic state function that provides a measure of how energy is distributed. An increase in entropy is associated with an increase in the freedom of molecular motions. Qualitative examples of processes where ΔS > 0 1) Increasing temperature 2) Expansion of gas into a vacuum 3) Change in physical state, solid liquid gas 4) Chemical reaction where moles of products > moles of reactants 5) Mixing/dissolving Macroscopic Entropy Definition S = = Second Law of Thermodynamics - whenever a spontaneous event takes place in the universe, the total entropy of the universe must increase ΔSuniverse = ΔSsurroundings + ΔSsystem > 0 - Gibbs Free Energy (G) - relates the entropy of the universe from the perspective of the system. ΔG = ΔH – TΔS (at constant temperature and pressure) ΔG = ΔH – TΔS o ΔG = G final G initialf: < 0: spontaneous process = 0: reversible/equilibrium > 0: nonspontaneous process - Restatement of 2nd Law: At constant T and P, a change is spontaneous if it is accompanied by a decrease in the free energy of a system.
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