Exam 3 – Solutions to Selected Problems Chemistry 102 [Items from the text are noted by Chang.chapter.number, items from the online worksheet are noted by OW.number, items from Exam homework on Moodle by M.number and supplemental items as S.chapternumber] S.5 In 2.00 min, 29.1 mL of He effuses through a small hole. Under the same conditions of pressure and temperature, 9.74 mL of a mixture of CO and CO2 effuses through the hole in the same amount of time. Using the following mathematical expression relating rate of effusion to molar mass (MM), calculate the average molar mass of the mixture of CO and CO2. rateHe
=
rateCO + CO2
MM CO + CO2
MM He
To solve this, you will first determine the rates (although you could also relate the amounts of helium versus the mixture since the time is the same). Then you will calculate the average (weighted) molar mass of the mixture of CO and CO2. MM CO + CO2
rate He
=
rate CO + CO2
MM He
MM CO + CO 2
⎛ rate He
=⎜
⎜ rate CO + CO
⎝
2
2
⎞
⎛ 14.55 mL ⋅ min −1 ⎞
4.00 g ⋅ mol −1 = 35.7 g ⋅ mol −1 ⎟ × MM He = ⎜
−1 ⎟
⎟
⋅
4.87
mL
min
⎝
⎠
⎠
2
Now, to determine the percent of CO versus CO2, represent the percent of one gas versus the other as 1 variable (x and 1 – x): MM CO2 ( x ) + MM CO (1 − x ) = 35.7 g ⋅ mol −1
x=
36.0 − 28.01
= 0.568
44.01 − 28.01
%CO2 = 48.1% %CO = 51.9%
Chang.6.31 Because one metal is at a higher temperature than the other, the heat flow will (spontaneously) go from a hotter body to a colder body, so the only correct answer is (d). How much heat will flow and (assuming the transfer is perfect from copper to aluminum), what is the final temperature of both metals? q Al = − qCu
(
)
(
)
m Al s Al T f − TiAl = mCu sCu TiCu − T f Tf =
m Al s Al TiAl + mCu sCu TiCu
m Al s Al + mCu sCu
= 51 o C
Chang.6.37 To solve this, you must understand calorimetry and the process (heat exchange) in the reaction. The heat absorbed or evolved by the calorimeter is equal in magnitude but opposite in sign to the heat absorbed or evolved by the reaction. Also, recognize that the calorimeter does not have a uniform composition (being composed of a number of substances) and so has a heat capacity and not a specific heat – therefore, you will only need to know the change in temperature and the heat capacity to determine the heat evolved or absorbed by the calorimeter. qcal = −qrxn
qcal = Ccal Δtcal = ( 3024 J ⋅ o C−1 )(1.126 o C) = 3405 J
Now, use this information to determine the heat evolved (since it is endothermic with regards to the calorimeter, so it is exothermic with regard to the reaction) for the reaction in the requested units: qrxn = −3405 J
Heat evolved per g (with heat in kJ):
−3.405 kJ
= −24.8 kJ ⋅ g −1
0.1375 g
Heat evolved per g (with heat in kJ):
−24.8 kJ ⋅ g −1 × 24.31 g ⋅ mol−1 = −603 kJ ⋅ mol−1
OW.9.a (Chapter 6) What is the volume at 25 ºC and at 1 atm of pressure of methane needed to boil (to completion) 1.50 kg of ice initially at −15.0 ºC? Assume all are complete combustion reactions (in excess oxygen) at 50% efficiency under constant pressure and all fuels behave as ideal gases. ΔHvap(water) = 40.656 kJ⋅mol−1 (plus all other constants the previous problem). To solve this problem, you will be best served by again considering a visual image of what is happening: the solid water is heated (which takes energy), the solid water is melting (which takes energy), the liquid water is heated (which takes energy), and the liquid water is boiled (which takes energy) – al l of this energy is delivered at 50% efficiency from the combustion of a fuel (for this item, methane) – you will need to consider the reaction of this, therefore, as well. The heat needed is in four steps: qice = mice sice Δtice
fusion
liq
liq liq liq
vap
H 2 O ( solid, − 15o C to 0o C ) ⎯⎯⎯⎯⎯→
H 2 O ( solid to liquid ) ⎯⎯⎯→
H 2 O ( liquid, 0o C to 100o C ) ⎯⎯⎯⎯⎯
→ H 2 O ( liquid to gas ) ⎯⎯⎯
→ H 2 O ( gas ) ΔH
q = m s Δt
ΔH
qice = mice sice Δtice = (1500 g ) ( 2.06 J ⋅ g −1 ⋅ o C −1 )(15 o C ) = 46.4 kJ
⎛
⎞
1500 g
ΔH ice →liquid = ( 6.00 kJ ⋅ mol−1 ) ⎜
= 500 kJ
−1 ⎟
⎝ 18.02 g ⋅ mol ⎠
qliq = mliq sliq Δtliq = (1500 g ) ( 4.18 J ⋅ g −1 ⋅ o C−1 )(100 o C ) = 627 kJ
⎛
⎞
1500 g
= 3380 kJ
ΔH liquid → gas = ( 40.656 kJ ⋅ mol−1 ) ⎜
−1 ⎟
⎝ 18.02 g ⋅ mol ⎠
Total = 4550 kJ → Therefore will need − 9100 kJ of energy from the reaction
Now, consider the reaction which will supply this energy from the difference in heat content (enthalpy) of the reactants versus the products: CH 4 ( g ) + 2O2 ( g ) → 2H 2 O ( g ) + CO 2 ( g )
ΔH rxn = −802.3 kJ which is for 1 mole of methane. Since the water needs 9100 kJ of heat, the reaction will need to combust 11.3 mol of methane (maintain the sign in order to have the correct sign on the number of moles). Now use the ideal gas law to determine the volume of methane: VCH 4 =
nCH 4 RT
P
=
(11.3 mol ) ( 0.08206 L ⋅ atm ⋅ mol−1 ⋅ K −1 ) ( 298 K )
1 atm
= 276 L How much would this cost if methane was selling for $0.10 / ft3? If natural gas is assumed to be 80% methane and 20% ethane, how would you solve this problem if natural gas was the fuel? Chang.7.79 To solve this problem, you need to remember the rules for determining quantum numbers. The principal quantum number, n, begins at 1, 2, 3, … The angular momentum quantum number, l, has the value(s) of l = 0, 1, … (n – 1). The magnetic quantum number, ml, has the values of ml = –l, …, 0, … +l. The last quantum number, the spin quantum number, ms, has values of –½ or ½. (a) Not acceptable. n and l are acceptable but ml is not. The only acceptable value of ml would be 0. The value of ms is acceptable. If the set is altered to (1, 0, 0, ½), it would be acceptable and refer to a 1s electron. (b) Acceptable. Referring to a 3s electron. (c) Not acceptable. n is acceptable, but l is not. Acceptable values of l for this value of n are 0 or 1. Both ml and ms values are acceptable. Two possible correct sets of quantum number for n = 2 are (2, 0, 0, ½) and (2, 1, 1, ½), among others. The first set of quantum numbers refers to a 2s electron while the second set refers to a 2p electron. (d) Acceptable. Referring to a 4f electron. (e) Not acceptable. All quantum numbers are correct except for the spin quantum number which must be ½ or –½. This refers to a 3d electron. M.11 Which set(s) of quantum numbers is/are not possible for a ground state electron in antimony? Choose at least one answer. A. (1, 1, 0, 1/2) B. (3, 2, ‐2, ‐1/2) C. (5, 1, 0, 1/2) D. (6, 3, 0, 1/2) E. (2, 0, 1, 1/2) F. (4, 1, ‐1, ‐1/2) To answer this, first consider the ground state electron configuration for antimony (Sb, Z = 51): 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p3 Now, consider each set of quantum numbers, remembering the rules for quantum numbers. A.
B.
C.
D.
E.
F.
(1, 1, 0, ½) – this is not allowed because n ≠ ℓ
(3, 2, −2, −½) – this is allowed; would it be an electron in Sb? This is referring to a 3d electron, so yes. One correct response is B. (5, 1, 0, ½) – this is allowed; would it be an electron in Sb? This is referring to a 5p electron, so yes. One correct response is C. (6, 3, 0, ½) – this is allowed; would it be an electron in Sb? This is referring to a 6f electron, so no. (2, 0, 1, ½) – this is not allowed because mℓ = −ℓ…0…+ℓ. (4, 1, −1, −½) – this is allowed; would it be an electron in Sb? This is referring to a 4p electron, so yes. One correct response is F. S.7 “What is the maximum number of electrons that can have n = 4, ml = –1 in an atom?” When n = 4, the possible values of l and ml are given in the table. n 4 4 4 4 l 0 1 2 3 ml 0 –1, 0, 1 –2, –1, 0, 1, 2 –3, –2, –1, 0, 1, 2, 3 Each value of ml has two possible values of ms. Another way to recognize the number of electrons, each value of ml denotes an atomic orbital (which can vary in orientation for the same value of l) and each atomic orbital hold a maximum of 2 electrons. Regardless, there are 3 atomic orbitals with a value of ml = –1 when n = 4 for a total of 6 electrons. Would the answer change if this was a tin atom in the ground state? OW.1 (part g through k only) (Chapter 7) What is the ground state electronic configuration for the species, what is the orbital diagram for the highest energy (partially filled in most cases) sublevel, and is the species paramagnetic or diamagnetic? g. Zn2+ h. Pb i. Pb2+ j. S k. S2‐ To solve these, you will need to know the order of the sublevels or subshells in a non‐hydrogen atom. You will also need to remember the Aufbau Principle (fill in lowest to highest energy subshells), Pauli’s Exclusion Principle (no electrons in one atom can have the same (complete) set of quantum numbers) and Hund’s Rule (electrons will fill degenerate orbitals to yield the greatest number of parallel spins). Lastly, you will need to know the definition of paramagnetic (attracted by a magnet – contain unpaired spins) and diamagnetic (slightly repelled by a magnet – do not contain net unpaired spins). (g) A zinc atom has 30 electrons, so a Zn2+ ion has 28 electrons. The ground state electron configuration is [Ar] 4s0 3d10 (since it is a transition metal – ns electrons will be lost when forming an ion before (n–1)d electrons). The orbital diagram for 3d10 will have all 5 orbitals with all electrons paired. This ion is diamagnetic. (h) A lead atom has 82 electrons. The ground state electron configuration is [Xe] 6s2 4f14 5d10 6p2. The orbital diagram for 6p2 will have both electrons in different orbitals with the same spin. This atom is paramagnetic. (i) A Pb2+ ion has 80 electrons. The ground state electron configuration is [Xe] 6s2 4f14 5d10. The orbital diagram for 5d10 will have all 5 orbitals with all electrons paired. This ion is diamagnetic. (j) A sulfur atom has 16 electrons. The ground state electron configuration is [Ne] 3s2 3p4. The orbital diagram for 3p4 will have two electrons paired in the same orbital and the remaining two electrons in different orbitals with the same spin. This atom is paramagnetic. (k) A S2– ion has 18 electrons. The ground state electron configuration is [Ne] 3s2 3p6. The orbital diagram for 3p6 will have all 3 orbitals with all electrons paired. This ion is diamagnetic. S.8 “Which of the following species are isoelectronic? Kr Cl– Fe3+ S2+ Fe2+ Rb+ Se2– Fe Br– Sr2+ Fe+” To solve this problem, you must know the definition of isoelectronic (same number of electrons and same ground state electron configuration), the number of electrons in atoms or ions and the ground state electron configurations. Filling the number of electrons: Kr Cl– Fe3+ S2+ Fe2+ Rb+ Se2– Fe Br– Sr2+ Fe+ 36 18 23 16 24 36 36 26 36 36 25 Now recognize that there is only one common number of electrons, 36. Considering the species with 36 electrons, they all have the ground state electron configuration of: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 Therefore, Kr, Rb+, Se2–, Br– and Sr2+ are isoelectronic. What zirconium ion is isoelectronic with these species? Chang.8.73 To answer this, you must consider what you know about the properties of the elements. The first is easiest to answer. Noting that the periodic table has the nonmetals concentrated in the upper left corner, metallic character decreases across a period from left to right and increase from top to bottom in a group. Effective nuclear charge increases across a period from left to right – similarly atomic size decreases across period in the same manner. As the number of electrons in the core increase in the same group, the size of the atom also increases, so size increases from top to bottom in a group. As atoms get smaller, the outer electrons are closer to the positively charged nucleus – it takes more energy to remove this electron. Therefore ionization energy follows the opposite trends of atomic size or increases across a period from left to right and decrease from top to bottom in a group. Finally, as metallic character increases, acidity of oxides decreases (since metal oxides tend to be basic). Therefore, acidity of oxides will increase across a period from left to right and decrease from top to bottom in a group. Of the elements, carbon, nitrogen, phosphorus and sulfur which has the largest atomic radius, the most metallic character, the smallest first ionization energy and the least acidic oxide? Do you see how these are trends and not absolute rules? M.51 Using the equation in your textbook (9.2, p. 291) as an approximation, which ionic compound will have the greatest lattice energy? A. LiF B. LiCl C. NaF D. NaCl To answer this, you must consider two factors: ionic size and ionic charge. Because the charges on the ions is the same throughout (all +1/−1), the only factor to consider here is the size of the ions. Because this equation has the distance between the ions (r) in the denominator, as the distance between the ions gets smaller, the lattice energy increases. Therefore, the smallest ions will have the greatest lattice energy. Which ions are the smallest? Li+ vs Na+ Li+ is smallest (furthest up the group) Cl− vs F− F− is smallest (furthest up the group) Therefore, LiF will have the greatest lattice energy. Chang.9.26 (Slightly different question – same substance) “What additional data is needed to determine the lattice energy of calcium chloride?” In the problem you are given the heat of sublimation of calcium and the standard molar enthalpy of formation of calcium chloride. You begin with free elements in the standard states, Ca(s) and Cl2(g) You will need ions in the gas state. You are given the enthalpy of sublimation of calcium Æ Ca(g) You will need the first and second ionization energies of Ca(g) Æ Ca2+(g) You will need the bond enthalpy of chlorine Æ 2Cl(g) You will need the electron affinity of chlorine Æ 2Cl–(g) Now, you have the ions in the gas state and the formation of CaCl2(s) from these ions corresponds to the lattice energy of calcium chloride. Why don’t you need the ionization energy of chlorine since you are forming the chloride ion? M.55 Which molecule has the longest carbon‐carbon bond length? A. actylene, HCCH B. ethylene, C2H4 C. ethane, C2H6 D. allene, H2CCCH2
To answer this, you will need to generate the Lewis dot structures and consider the CC bond(s) in the molecules. Once you do that you will determine: A.
B.
C.
D.
HCCH contains a CC triple bond C2H4 contains a CC double bond C2H6 contains a CC single bond H2CCCH2 contains two CC double bonds Now, you need to consider that the greater the number of bonds, the shorter the bond (between the same atoms – can compare C−O to C=O but not C=C to C−H). Therefore between all of these carbon‐carbon bonds, the smallest number of bonds occurs between the carbon atoms in ethane with a single bond. Therefore, ethane (C) has the longest carbon‐carbon bond. In Chapter 10, we will find a better way to describe the “number of bonds” between atoms – we will call this bond order.