First Ionization Energy Review Key

AP Chemistry Fall Semester Review 1. 1972 Consider the following melting points in degrees Celsius: Alkali metals Halogens Li 181° F2 -119° Na 98° Cl2 -101° K 63° Br2 -7° Rb 39° I2 +104° Cs 29° (a) Account for the trend in the melting points of the alkali metals. (b) Account for the trend in the melting points of the halogens. (Make sure that your discussion clarifies the difference between the two trends.) Answer: (a) Because of their large sizes and limited numbers of valence electrons, bonding between alkali metal atoms is not as strong as in most metals. Since the atoms increase in size down the family, those near the bottom (Rb and Cs) have the greatest internuclear distances. (b) Attractive forces between halogen molecules are rather weak, they are of the instantaneous dipole-induced dipole type (London forces), and increase in strength with increasing molecular weight (and increasing numbers of electrons). 2. 1973 D First ionization Energy (kilocalories/mole) Covalent Radii, Å Li 124 1.34 Be 215 0.90 B 191 0.82 C 260 0.77 N 336 0.75 O 314 0.73 F 402 0.72 The covalent radii decrease regularly from Li to F, whereas the first ionization energies do not. For the ionization energies, show how currently accepted theoretical concepts can be used to explain the general trend and the two discontinuities. Answer: AP Chemistry Fall Semester Review The trend in moving across a period is that the first ionization energy, I1, increases from group 1 (Li) to group 7 (F) because of an increase in effective nuclear charge, atoms get smaller (decrease covalent radii) and less metallic through the period. The I1 is less for B than Be because the electron to be ionized in B is in a higher energy orbital (2p) than is the electron (2s) to be ionized in Be. The I1 is less for O than N because the electron to be ionized in O is a paired electron in the 2p orbitals. At N, the outer sublevel of its atom is half-filled, resulting in a symmetrical spherical electron cloud. The extra electron in O reduces this symmetry and so less energy is required to remove this electron. 3. 2001 B Answer the following questions about acetylsalicylic acid, the active ingredient in aspirin. (a) The amount of acetylsalicylic acid in a single aspirin tablet is 325 mg, yet the tablet has a mass of 2.00 g. Calculate the mass percent of acetylsalicylic acid in the tablet. (b) The elements contained in acetylsalicylic acid are hydrogen, carbon, and oxygen. The combustion of 3.000 g of the pure compound yields 1.200 g of water and 3.72 L of dry carbon dioxide, measured at 750. mm Hg and 25°C. Calculate the mass, in g, of each element in the 3.000 g sample. (c) A student dissolved 1.625 g of pure acetylsalicylic acid in distilled water and titrated the resulting solution to the equivalence point using 88.43 mL of 0.102 M NaOH(aq). Assuming that acetylsalicylic acid has only one ionizable hydrogen, calculate the molar mass of the acid. (d) A 2.00 × 10-3 mole sample of pure acetylsalicylic acid was dissolved in 15.00 mL of water and then titrated with 0.100 M NaOH(aq). The equivalence point was reached after 20.00 mL of the NaOH solution had been added. Using the data from the titration, shown in the table below, determine (i) (ii) the value of the acid dissociation constant, Ka, for acetylsalicylic acid and the pH of the solution after a total volume of 25.00 mL of the NaOH solution had been added (assume that volumes are additive). Volume of 0.100M NaOH Added (mL) pH 0.00 5.00 10.00 15.00 20.00 25.00 2.22 2.97 3.44 3.92 8.13 ? Answer: (a) × 100% = 16.3% (b) 1.200 g H2O × + 16 g H2O) = 0.134 g H n = = = 0.150 mol CO2 0.150 mol CO2 × = 1.801 g C 3.000 g ASA – (1.801 g C + 0.134 g H) = 1.065 g O (c) 0.08843 L × = 0.00902 mol base 1 mol base = 1 mol acid = 180 g/mol (d) (i) HAsa ↔ Asa– + H+ × = 0.133 M pH = –log[H+]; 2.22 = –log[H+] [H+] = M = [Asa–] [HAsa] = 0.133 M – 6.03 × 10-3 M = 0.127 M AP Chemistry K = = × = 2.85× 10-4 Fall Semester Review OR when the solution is half-neutralized, pH = pKa at 10.00 mL, pH = 3.44; K = 10–pH = 10–3.44 = 3.63×10-4 (ii) 0.025 L × 0.100 mol/L = 2.50 × 10-3 mol OH2.50 × 10-3 mol OH- - 2.00 × 10-3 mol neutralized = 5.0 × 10-4 mol OH- remaining in (25 + 15 mL) of solution; [OH-] = 5.0×10-4 mol/0.040 L = 0.0125 M pH = 14 – pOH = 14 + log[OH-] = 14 – 1.9 = 12.1 4. 1995 B A sample of dolomitic limestone containing only CaCO3 and MgCO3 was analyzed. (a) When a 0.2800 gram sample of this limestone was decomposed by heating, 75.0 milliliters of CO2 at 750 mm Hg and 20°C were evolved. How many grams of CO2 were produced. (b) Write equations for the decomposition of both carbonates described above. (c) It was also determined that the initial sample contained 0.0448 gram of calcium. What percent of the limestone by mass was CaCO3? (d) How many grams of the magnesium-containing product were present in the sample in (a) after it had been heated? Answer: (a) = 3.08×10-3 mol 3.08×10-3 mol × (44.0 g CO2/1 mol) = 0.135 g CO2 (b) CaCO3 → CaO + CO2 MgCO3 → MgO + CO2 (c) = 40.0% CaCO3 (d) 60.0% of 0.2800 g sample = 0.168 g of MgCO3 5. 1988 B An electrochemical cell consists of a tin electrode in an acidic solution of 1.00 molar Sn2+ connected by a salt bridge to a second compartment with a silver electrode in an acidic solution of 1.00 molar Ag+. (a) Write the equation for the half–cell reaction occurring at each electrode. Indicate which half–reaction occurs at the anode. AP Chemistry Fall Semester Review (b) Write the balanced chemical equation for the overall spontaneous cell reaction that occurs when the circuit is complete. Calculate the standard voltage, E°, for this cell reaction. (c) Calculate the equilibrium constant for this cell reaction at 298K. Hint: n= number of moles of electrons being transferred; E= E° cell; (d) A cell similar to the one described above is constructed with solutions that have initial concentrations of 1.00 molar Sn2+ and 0.0200 molar Ag+. Calculate the initial voltage, E°, of this cell. Answer: (a) Sn → Sn2+ + 2e- anode reaction Ag+ + e- → Ag (b) 2 Ag+ + Sn → 2 Ag + Sn2+ E° = [0.80v - (-0.14v)] = 0.94v (c) (d) 6. 1993 D A galvanic cell is constructed using a chromium electrode in a 1.00-molar solution of Cr(NO3)3 and a copper electrode in a 1.00-molar solution of Cu(NO3)2. Both solutions are at 25°C. (a) Write a balanced net ionic equation for the spontaneous reaction that occurs as the cell operates. Identify the oxidizing agent and the reducing agent. (b) A partial diagram of the cell is shown below. (i) Which metal is the cathode? (ii) What additional component is necessary to make the cell operate? (iii) What function does the component in (ii) serve? (c) How does the potential of this cell change if the concentration of Cr(NO3)3 is changed to 3.00-molar at 25C? Explain. AP Chemistry Fall Semester Review Answer: (a) 2 Cr + 3 Cu2+ → 2 Cr3+ + 3 Cu Cr = reducing agent; Cu2+ = oxidizing agent (b) (i) Cu is cathode (ii) salt bridge (iii) transfer of ions or charge but not electrons (c) Ecell decreases. use the Nernst equation to explain answer 7. Suppose the substances in the reaction below are at equilibrium at 600K in volume V and at pressure P. State whether the partial pressure of NH3(g) will have increased, decreased, or remained the same when equilibrium is reestablished after each of the following disturbances of the original system. Some solid NH4Cl remains in the flask at all times. Justify each answer with a one-or-two sentence explanation. NH4Cl(s) ↔ NH3(g) + HCl(g) ΔH = +42.1 kilocalories (a) A small quantity of NH4Cl is added. (b) The temperature of the system is increased. (c) The volume of the system is increased. (d) A quantity of gaseous HCl is added. (e) A quantity of gaseous NH3 is added. Answers: (a) PNH3 does not change. Since NH4Cl(s) has constant concentration (a = 1), equilibrium does not shift. (b) PNH3 increases. Since the reaction is endothermic, increasing the temperature shifts the equilibrium to the right and more NH3 is present. (c) PNH3 does not change. As V increases, some solid NH4Cl decomposes to produce more NH3. But as the volume increases, PNH3 remains constant due to the additional decomposition. (d) PNH3 decreases. Some NH3 reacts with the added HCl to relieve the stress from the HCl addition. (e) PNH3 increases. Some of the added NH3 reacts with HCl to relieve the stress, but only a part of the added NH3 reacts, so PNH3 increases. 8. Ammonium hydrogen sulfide is a crystalline solid that decomposes as follows: NH4HS(s) ↔ NH3(g) + H2S(g) (a) Some solid NH4HS is placed in an evacuated vessel at 25ºC. After equilibrium is attained, the total pressure inside the vessel is found to be 0.659 atmosphere. Some solid NH4HS remains in the vessel at equilibrium. For this decomposition, write the expression for KP and calculate its numerical value at 25ºC. (b) Some extra NH3 gas is injected into the vessel containing the sample described in part (a). When equilibrium is reestablished at 25ºC, the partial pressure of NH3 in the vessel is twice the partial pressure of H2S. Calculate the numerical value of the partial pressure of NH3 and the partial pressure of H2S in the vessel after the NH3 has been added and the equilibrium has been reestablished. AP Chemistry Fall Semester Review (c) In a different experiment, NH3 gas and H2S gas are introduced into an empty 1.00 liter vessel at 25ºC. The initial partial pressure of each gas is 0.500 atmospheres. Calculate the number of moles of solid NH4HS that is present when equilibrium is established. Answers: (a) KP = (PNH3)(PH2S) PNH3 = PH2S = 0.659/2 atm = 0.330 atm KP = (0.330)2 = 0.109 (b) PNH3 = 2 PH2S (2x)(x) = 0.109 ; x = 0.233 atm = PH2S PNH3 = 0.466 atm (c) Equilibrium pressures of NH3 and H2S are each 0.330 atm. Amounts of each NH3 and H2S that have reacted correspond to (0.500 - 0.330) = 0.170 atm. n = mol of each reactant = mol of solid product n = PV/RT = 0.170 atm x 1L / 0.0821L atm/mol K x 298K = 6.95 x 10-3 mol 9. When H2(g) is mixed with CO2(g) at 2,000 K, equilibrium is achieved according to the equation below. In one experiment, the following equilibrium concentrations were measured. CO2(g) + H2(g) ↔ H2O(g) + CO(g) [H2] = 0.20 mol/L [CO2] = 0.30 mol/L [H2O] = [CO] = 0.55 mol/L (a) What is the mole fraction of CO(g) in the equilibrium mixture? (b) Using the equilibrium concentrations given above, calculate the value of Kc, the equilibrium constant for the reaction. (c) Determine Kp in terms of Kc for this system. Hint: Δn= (total moles of gas on products side)-(total moles of gas on reactants side); ; Kp=Kc(RT)Δn (d) When the system is cooled from 2,000 K to a lower temperature, 30.0 percent of the CO(g) is converted back to CO2(g). Calculate the value of Kc at this lower temperature. (e) In a different experiment, 0.50 mole of H2(g) is mixed with 0.50 mole of CO2(g) in a 3.0-liter reaction vessel at 2,000 K. Calculate the equilibrium concentration, in moles per liter, of CO(g) at this temperature. Answer: (a) mole fractionCO = (0.55 mol /1.6 mol) = 0.34 (b) Kc = ([H2O][CO]) / ([H2][CO2]) = (0.55x0.55)/(0.20x0.30) = 5.04 (c) since Δn = 0, Kc = Kp (d) (e) [CO] = 0.55 - 30.0% = 0.55 - 0.165 = 0.385 M [H2O] = 0.55 - 0.165 = 0.385 M [H2] = 0.20 + 0.165 = 0.365 M [CO2] = 0.30 + 0.165 = 0.465 M K = (0.385)2 / (0.365x0.465) = 0.87 let x = Δ[H2] to reach equilibrium [H2] = 0.50 mol/3.0L - X = 0.167 - X [CO2] = 0.50 mol/3.0L - X = 0.167 - X AP Chemistry Fall Semester Review [CO] = +X ; [H2O] = +X K = X2/(0.167 - X)2 = 5.04 ; X = [CO] = 0.12 M 10. A 0.682 g sample of an unknown weak monoprotic organic acid, HA, was dissolved in sufficient water to make 50 milliliters of solution. It was titrated with 0.135 molar NaOH solution. After the addition of 10.6 milliliters of base, a pH of 5.65 was recorded. The equivalence point (end point) was reached after addition of 27.4 milliliters of the 0.135 molar NaOH. a) Calculate the number of moles of acid in the original sample. b) Calculate the molar mass of the acid, HA. c) Calculate the number of moles of unreacted HA remaining in solution when the pH was 5.65. d) Calculate the hydronium ion concentration at pH = 5.65 e) Calculate the value of the ionization constant, Ka, of the acid HA. Answer: a) 0.135M * 0.0274L= 0.003699 mols or NaOH @ equivalence point… moles of acid is also equal to this b) 0.682g/ 0.003699 mol= 184.4 g/mol c) WA= weak acid 0.682g WA * 1 mol/ 184.4 g= 0.003698 mol WA/ 0.050 L= 0.07397 M for WA @ start MaVa=MbVb X(.05L)= 0.135M(.0106L) X=0.0286M 0.07397M-0.0286M= 0.04537M remains 0.04537M* (0.05L + 0.0106L)= 0.00275 mols remain d) [H3O+]= 10-pH [H3O+]= 10-5.65 [H3O+]= 2.24*10-6M 11. + – CH3NH2 + H2O ↔ CH3NH3 + OH Methylamine, CH3NH2, is a weak base that reacts according to the equation above. The value of the ionization constant, –4 + – Kb, is 5.25 X 10 . Methylamine forms salts such as methylammonium nitrate, (CH3NH3 )(NO3 ). – (a) Calculate the hydroxide ion concentration, [OH ], of a 0.225–molar aqueous solution of methylamine. (b) Calculate the pH of a solution made by adding 0.0100 mole of solid methylammonium nitrate to 120.0 milliliters of a 0.225–molar solution of methylamine. Assume no volume change occurs. (c) How many moles of either NaOH or HCl (state clearly which you choose) should be added to the solution in (b) to produce a solution that has a pH of 11.00? Assume that no volume change occurs. (d) A volume of 100. milliliters of distilled water is added to the solution in (c). How is the pH of the solution affected? Explain. Answer: AP Chemistry Fall Semester Review (a) + – CH3NH2 + H2O ⁄ CH3NH3 + OH [ ]i ∆[ ] [ ]eq 0.225 –X 0.225–X 0 +X X 0 +X X – –2 X = [OH ] = 1.09¥10 M – –2 solved using quadratic: X = [OH ] = 1.06¥10 M (b) + [CH3NH3 ] = 0.0100 mol / 0.1200 L = 0.0833 M or CH3NH2 = 0.120 L ¥ 0.225 mol/L = 0.0270 mol – –3 X = [OH ] = 1.42¥10 M; pOH = 2.85; pH = 11.15 OR OR (c) HCl must be added. X = 0.0228 M –3 0.0228 mol/L ¥ 0.120 L = 2.74¥10 mol HCl OR –3 0.0227 M ¥ 0.120 L = 2.73¥10 mol HCl (d) The ratio does not change in this buffer solution with dilution, therefore, no effect on pH.