Chem 1211 1st EditionExam #3 Study Guide Lectures: 14-21Chapter 5.3-5.8Explain heat of fusion and heat of vaporization.Heat of fusion is the energy that is required to convert a substance from a solid at its melting point to a liquid; heat of vaporization is the energy transferred as heat to convert a liquid at its boiling point to a vapor.-The total energy transferred as heat is the sum of the energies of the individual stepsQtotal= q1+q2+q3…For example, to convert 500 grams of ice at -50 degrees C to water vapor at 200 degrees celcius…1. Warm ice to 0 degrees celcius 2. Converti to liquid water at 0 degrees celcius3. Warm liquid water to 100 degrees celcius4. Evaporate at 100 degrees celcius5. Warm the water vapor to 200 degrees celcius- Then add all of the q’s togetherFirst law of Thermodynamics- ΔU= q + w- Aka, change in energy content= energy transferred as heat to/from system + energy transferred as work to/from the system- SIGN IS VERY IMPORTANTo For an endothermic reaction, q is positive; for an exothermic reaction, q is negativeo When work is done on the system, w is positive; when work is done by the system, w is negative- W= -P x ΔVo Work (at constant pressure) = -pressure x change in volume - State function: changes in quantities depend only on the initial and final state (ex. –enthalpy, temperature, volume, energy)o You can determine enthalpy change using two other reactionso Example: What is the enthalpy change for the first reaction? 2 C (s) + O2 (g)  2 CO (g) delta Hrxn=? 2 CO (g) + O2 (g)  2 CO2 (g) delta Hrxn= -566kJ C (s) + O2 (g) CO2 (g) delta Hrxn= -393kJ- ANSWER: -221 kJ- Reasoning: o Double everything in the third equation (gives you -786 kJ) o Delta H1 + Delta H2 = Delta H3o ? + -566 = -786o ? = -220- Hess’s Law- if a reaction is the sum of two or more other reactions, the enthalpy change ΔrH for the overall process is the sum of the ΔrH values of those reactionso Example: Calculate the enthalpy change for the reaction in which 30.0 g of aluminum reacts with oxygen to form Al2O3 at 25 degrees Celsius and one atmosphere:o 4 Al (s) + 3 O (g)  2 Al2O3 (s) delta Hf (Al2O3) = -1676 kJ/mol ANSWER: Use stoichiometry. Convert 30 grams of Al to 1.11 moles Al. It takes two moles of Al to get one mole of Al2O3 (according to the equation), so divide 1.11 by 2 to get .555 moles Al2O3. Multiply .555 by -1676 kJ/moles to get your answer (approx. -930.2 kJ)- ΔrH= ΣnΔfH (products) – ΣnΔfH (reactants)o To find the enthalpy change, add up the molar enthalpies of formation of the products, each multiplied by its stoichiometric coefficient n, and subtract from this the sum of the molar enthalpies of formation of the reactants, each multiplied by its stoichiometric coefficient- Product-favored reaction at equilibrium- reactions in which reactants are largely converted to products when equilibrium is reached- Reactant-favored reaction at equilibrium- reactions in which only small amounts of product are present at equilibrium- In most cases, reactions with a positive enthalpy change are reactant-favored, and reactions with a negative enthalpy are product-favoredChapter 6- Wavelength- distance between successive crests/high points of a wave; measured in meters or nanometers- Frequency- the number of waves that pass a given point in some unit of time (usually per second); the frequency that stands for 1 wave per second is a hertz- Wavelength and frequency are inverseso As wavelength goes down, frequency goes up. Red has the greatest wavelength, so it has the lowest frequency. (Follow ROY G BIV)- C (speed of light- 3 x 108 mps) = w (wavelength in meters) x f (frequency in Hertz)- E (energy)=h(Planck’s constant) x fo H= 6.626 x 10-34 Jso Example: What is the energy of .122 m radiation from a microwave oven? Answer: 1.63 x 10-24 J Reasoning: To find the energy, use the equation e=hf, but first, you need to find the frequency. You know the wavelength (.122m), so plug it into the equation c=wf to get a frequency of 2.45x109 Hz. Then multiply that frequency times Planck’s constant to get the energy. o Cosmic rays – x-rays - UV - visible - infrared - microwave - radiowave energy and frequency wavelength  increase this way increases this way-total energy of an electron in the nth level = -(Rhc)/n2-R= Rydberg constant (1.0974 x 107 m-1), P= Planck’s constant, c= speed of light, n= principle quantum number-Rydberg Equation: (1/wavelength) = R [(1/n12) – (1/n22)]- n1 < n2- the quantum number n defines the energies of the allowed orbits in the H atoms-the electron gap decreases as n increases-deBroglie wavelength: w= (h)/(mv)-m=mass-v=velocity-paramagnetic- elements and compounds that have unpaired electrons are attracted to a magnet-diamagnetic-substances in which all electrons are paired and experience a repulsion from a magnetic fieldChapter 7Quantum numbers:a. n= periodic row, or shellb. l= shape: s, p, d, f, g, hc. m= (-l … +l)d. spin number +1/2 or -1/2Aufbau Principlea. Rule 1- fill starting with lowest order of (n+l)b. Rule 2- if (n+l) is the same, start with lower n-Example: Which two orbitals have the same value for n+l?i. 2s, 2pii. 2s, 3siii. 2p, 3piv. 2p, 3s1. ANSWER: iv (for 2p, add 2+1, and for 3s, add 3+0)-In the periodic table,- Rows 1A and 2A are s orbitals (start with 1)-Rows 3A and 8A are p orbitals (start with 2)-B rows are d orbitals (start with 3)Example: What is the electron configuration for chlorine? I. 1s22s22p5II. [He]2s22p5III. 3s23p5IV. [Ne]3s23p5V. [Ne]3s13p61. ANSWER: [Ne]3s23p52. Reasoning: noble gases are completely filled. [He]= 1s2, [Ne]= 1s22s22p6. Chlorine is all of Neon plus 3s2 and 3p5-cations have a positive charge, so they have fewer electrons-anions have a negative charge, so they have more electrons-cations shrink and anions expand i. + charge: smaller than corresponding atomii. - charge: larger than their corresponding atomTrends:-Atomic radius increases as you move from the top to the bottom of the periodic table-Atomic radius decreases as you move left to right across the periodic table-Atomic energy is has the opposite trendsChapter 8Definition: isoelectronic- has the same number of electrons Example: With what atoms are K+ and Br- isoelectronic?i. He, Neii. Ne, Ariii. Ar, Kriv. Kr, Xe 1. ANSWER: Ar, Kra. REASONING: Because K is +, it has one less electron than normal, and because Br is -, it has one more electron than normalOctet rule: element