CH 301 Exam Review: Atomic Structure, Gases, Intermolecular Forces and Thermodynamics

Terms in this set (30)

1. Rydberg equation for hydrogen emission lines1/λ = R_H(1/n_low^2 − 1/n_high^2), with R_H = 1.097×10^7 m^−1. Transitions down to n = 2 give the visible Balmer series.
2. Energy levels of a particle in a 1-D boxE_n = n^2·h^2/(8mL^2), n = 1, 2, 3, ... Energy is quantized, the lowest energy is not zero, and levels get closer together as the box length L or the mass m increases.
3. How many radial and angular nodes does an orbital have?Angular nodes = l. Radial nodes = n − l − 1. Total nodes = n − 1. Example: a 3p orbital has 1 angular and 1 radial node.
4. Shapes of s, p and d orbitalss orbitals are spherical. p orbitals are dumbbell-shaped along x, y or z. Most d orbitals are four-lobed (cloverleaf); d(z^2) has two lobes and a ring.
5. Why is a 2s orbital lower in energy than a 2p orbital in multi-electron atoms?2s electrons penetrate closer to the nucleus, so they are less shielded and feel a higher effective nuclear charge. In hydrogen, with one electron, 2s and 2p have equal energy.
6. Coulomb's law and what it predicts for ionic attractionPotential energy E ∝ (q1·q2)/r. Larger ion charges and smaller distances between ions give stronger attraction and more energy released.
7. How does bond order relate to bond length and bond strength?Higher bond order means shorter and stronger bonds. For carbon-carbon bonds: triple is shorter and stronger than double, which is shorter and stronger than single.
8. N2 molecular orbital configuration and propertiesFor B2, C2 and N2, the π2p orbitals lie below σ2p. N2 has bond order 3 and all electrons paired, so it is diamagnetic and very strongly bonded.
9. Boyle's, Charles's and Avogadro's lawsBoyle: P1V1 = P2V2 at constant n and T. Charles: V1/T1 = V2/T2 at constant n and P. Avogadro: V/n is constant at constant P and T. All follow from PV = nRT.
10. Density of an ideal gasd = PM/(RT), where M is molar mass. Heavier gases are denser at the same temperature and pressure.
11. Mini-problem: How many moles of gas are in 2.00 L at 1.00 atm and 273 K?n = PV/(RT) = (1.00)(2.00)/(0.08206 × 273) ≈ 0.0893 mol.
12. What do the a and b terms in the van der Waals equation correct for?(P + a·n^2/V^2)(V − n·b) = nRT. The a term corrects for intermolecular attractions, which lower the measured pressure; b corrects for the volume occupied by the molecules themselves.
13. How does the Maxwell-Boltzmann speed distribution change with temperature and molar mass?Higher temperature or lower molar mass flattens and broadens the curve and shifts the most probable speed higher. Lower temperature or heavier gas gives a narrower, taller peak at lower speed.
14. Four main types of intermolecular forces, weakest to strongest (typical)London dispersion forces (all molecules), dipole-dipole (polar molecules), hydrogen bonding (special strong dipole-dipole), and ion-dipole (ions in polar solvents).
15. When can hydrogen bonding occur?When H is covalently bonded to N, O or F and is attracted to a lone pair on another N, O or F atom.
16. Why is F2 a gas, Br2 a liquid and I2 a solid at room temperature?All are nonpolar, so only dispersion forces act. Dispersion forces grow with the number of electrons and polarizability, so they increase from F2 to I2, raising melting and boiling points.
17. Rank by boiling point: H2O, CH4, H2S. Explain.CH4 < H2S < H2O. CH4 has only dispersion forces; H2S adds dipole-dipole forces; H2O forms strong hydrogen bonds, giving it the highest boiling point despite its low mass.
18. How do intermolecular forces affect vapor pressure and boiling point?Stronger IMFs mean fewer molecules escape to the gas phase, so vapor pressure is lower and boiling point is higher. A liquid boils when its vapor pressure equals the external pressure.
19. Phase diagram: what are the triple point and critical point? What is odd about water's?Triple point: the T and P where solid, liquid and gas coexist. Critical point: above it, liquid and gas are indistinguishable (supercritical fluid). Water's solid-liquid line slopes negatively because ice is less dense than liquid water.
20. First law of thermodynamics and the chemistry sign conventionΔU = q + w. q > 0 when heat flows into the system; w > 0 when work is done on the system. For expansion against constant external pressure, w = −P_ext·ΔV.
21. What is a state function? Give examples and non-examples.A property that depends only on the current state, not the path taken. U, H, S, G, P, V and T are state functions; heat q and work w are not.
22. How are ΔH and ΔU related for reactions involving gases?ΔH = ΔU + Δn_gas·RT, where Δn_gas is moles of gaseous products minus gaseous reactants. If Δn_gas = 0, ΔH ≈ ΔU.
23. What is entropy at the molecular level, and what increases it?S = k ln W, where W is the number of microstates. Entropy increases on melting or vaporizing, when gas moles increase, with higher temperature, and when substances mix or (usually) dissolve.
24. Second law of thermodynamics and entropy of the surroundingsFor any spontaneous process, ΔS_universe = ΔS_sys + ΔS_surr > 0. At constant T and P, ΔS_surr = −ΔH_sys/T, so exothermic reactions increase the entropy of the surroundings.
25. Gibbs free energy: predict spontaneity from the signs of ΔH and ΔS.ΔG = ΔH − TΔS. ΔH < 0, ΔS > 0: spontaneous at all T. ΔH > 0, ΔS < 0: never spontaneous. ΔH < 0, ΔS < 0: spontaneous at low T. ΔH > 0, ΔS > 0: spontaneous at high T.
26. Mini-problem: ΔH = +40.0 kJ/mol and ΔS = +100. J/(mol·K). Above what temperature is the process spontaneous?Set ΔG = 0: T = ΔH/ΔS = 40,000 J/mol ÷ 100 J/(mol·K) = 400 K. Above 400 K, ΔG < 0. Remember to convert kJ to J.
27. Relationship between ΔG° and the equilibrium constantΔG° = −RT ln K. If ΔG° < 0, K > 1 (products favored); if ΔG° > 0, K < 1 (reactants favored); if ΔG° = 0, K = 1.
28. Why is temperature constant during a phase change on a heating curve?Added heat goes into overcoming intermolecular forces rather than increasing kinetic energy. The heat needed is q = n·ΔH_fus for melting or q = n·ΔH_vap for boiling.
29. Entropy change for a phase transition at its transition temperatureΔS = ΔH_transition / T_transition, with T in kelvin. At the normal boiling point, liquid and gas are in equilibrium, so ΔG = 0.
30. State the third law of thermodynamics.The entropy of a perfect crystal at absolute zero (0 K) is zero. This allows absolute standard molar entropies S° to be tabulated, unlike enthalpies.