Jesse Salas Anh Nguyen Nizkorodov Chem 1B

Jesse Salas
Anh Nguyen
Nizkorodov Chem 1B
Midterm I Review
1. Rank the following by increasing strength of dispersion forces: AlCl3, AlBr3,
AlF3, AlH3, AI3
AlH3 < AlF3 < AlCl3 < AlBr < AlI3 (increasing mass)
2. Rank the following by increasing strength (generally speaking):
Dipole‐dipole, Dispersion forces, dipole‐induced diple, ionic, ion‐induced diple,
ion dipole, Hydrogen bonds,
Dispersion < Dipole‐induced dipole < Ion‐induced dipole < Dipole‐dipole
Hydrogen bonds < Ion dipole < Ionic
3. a.) Name the types of IMFs existing between the following molecules: HI, C2H6,
NaOH, HCl, H2O
HI : Dipole‐Dipole, VDW
NaOH: Ionic, Dipole‐dipole, Hydrogen bond, VDW
C2H6: VDW
HCl: Dipole‐Dipole, VDW
H2O: H‐bond, dipole‐dipole, VDW, Dipole‐Dipole
b.) Rank the same molecules above by increasing boiling points.
C2H6 < HCl < HI < H2O < NaOH
4. Rank the following substances according to decreasing boiling point:
CH3(CH2)3CH3, CaBr2, CH3OH, Ne (also put what IMFs hold them together)
CaBr2 > CH3OH > CH3(CH2)3CH3 > Ne
Ionic, Dipole, VDW > H‐bond, dipole, VDW > VDW > VDW
5. Krypton crystallizes with a facecentered cubic unit cell of edge 559 pm. What is the
density of solid Xenon? Jesse Salas
Anh Nguyen
Nizkorodov Chem 1B
Midterm I Review
Convert pm to cm:
559 pm x (1 cm/1010 pm) = 559 x 10¯10 cm = 5.59 x 10¯8 cm
Calculate the volume of the unit cell:
(5.59 x 10¯8 cm)3 = 1.7468 x 10¯22 cm3
Calculate the average mass of one atom of Kr:
139.29 g mol¯1 divided by 6.022 x 1023 atoms mol¯1 = 2.313 x 10¯22 g
Calculate the mass of the 4 krypton atoms in the facecentered cubic unit cell:
2.313 x 10¯22 g times 4 = 9.252 x 10¯22 g
Calculate the density (value from step 4 divided by value from step 2):
9.252 x 10¯22 g / 1.7468 x 10¯22 cm3 = 5.30 g/cm3
6. Palladium crystallizes in a facecentered cubic unit cell. Its density is 12.023 g/cm3.
Calculate the atomic radius of palladium.
Calculate the average mass of one atom of Pd:
106.42 g mol¯1 ÷ 6.022 x 1023 atoms mol¯1 = 1.767187 x 10¯22 g/atom
Calculate the mass of the 4 palladium atoms in the facecentered cubic unit cell:
1.767187 x 10¯22 g/atom times 4 atoms/unit cell = 7.068748 x 10¯22 g/unit cell
Use density to get the volume of the unit cell:
7.068748 x 10¯22 g ÷ 12.023 g/cm3 = 5.8793545 x 10¯23 cm3
Determine the edge length of the unit cell:
[cube root of] 5.8793545 x 10¯23 cm3 = 3.88845 x 10¯8 cm
Determine the atomic radius:
Remember that a facecentered unit cell has an atom in the middle of each face of the
cube. The square represents one face of a facecentered cube:
Jesse Salas
Anh Nguyen
Nizkorodov Chem 1B
Midterm I Review
Using the Pythagorean Theorem, we find:
d2 + d2 = (4r)2
2d2 = 16r2
r2 = d2 ÷ 8
r = d ÷ 2(√2)
r = 1.3748 x 10¯8 cm
7. The volume of an ideal gas is increased from 9.1 L to 16.4 L at a constant pressure of 1.5
atm. Calculate the work (in J) associated with this process. What sign does it have and
why? What kind of reaction is taking place? Is work being done on the system or by the
system?
W=PΔV
ΔV=16.4 L – 9.1 L = 7.3L
W= 1.5atm(3.8L)= 10.95 L*atm
1L*atm = 101.325 J
W= 10.95 L*atm (101.325 J/L*atm)= 1109 J negative sign because energy is being
released; exothermic
Work is being done by the system to the surroundings.
8. Ethane burns in air according to the equation 2 C2H6 + 7 O2 6H2O + 4 CO2,
Jesse Salas
Anh Nguyen
Nizkorodov Chem 1B
Midterm I Review
132 g of ethane are used up in the combustion, what mass of water is
produced?
9. In one process, 111 g of Al are reacted with 584 g of Fe2O3 in the reaction Al +
Fe2O3 Al2O3 + Fe, Calculate the maximum mass of Al2O3 that can be formed. If
167g of Al2O3 is actually formed what is the reaction yield?
10. Calculate the final volume of a balloon when it is heated with 1.5*105 J of
energy at an initial volume is 3.5*103L while the total internal energy of the
system is 1.2*105 J . Assume balloon expands at constant pressure of 1.5atm.
U = q+w
U = 9.00*107 J
Q = 1.5 *108 J
W = ‐PdV = ‐1.5(Vf ‐ 3.5*103L) * 101.395J/L*atm
1.2*105 J = 1.5 *105 J + ‐1.5(Vf ‐ 3.5*103L) * 101.395J/L*atm
Vf = 3.7
11. Consider H2 + Cl2 ‐> 2HCl ΔHrxn =‐184.6kJ/mol
If 5mol of H2 reacts with 5mol of Cl2, calculate the work done at STP, and ΔE for
the reaction.
W = ‐PΔV = ‐1Δ0. 0 because 5mol+5mol ‐> 10mol, n stays the same!
E = q + w. q = 10mol HCl * ‐184.6kJ/mol = 1846kJ
E = 1846kJ + 0kJ = 1846kJ
12. Determine amount of energy released when 3.45*103 g of NO2 are produced
from the following equation: 2NO(g) + O2 ‐> 2NO2(g) ΔHrxn = ‐114.6kJ/mol
Jesse Salas
Anh Nguyen
Nizkorodov Chem 1B
Midterm I Review
Given: 3.45*103 g of NO2
Required: kJ released from NO2
Plan: g of NO2 to mol of NO2 to kJ of NO2
3.45*103 g of NO2 * 1mol NO2 * ‐114.6kJ = 3594.27 kJ = 3.59*103 kJ
55g NO2 2mol NO2
13. Consider H2(g) ‐> H (g)+ H(g). ΔHrxn =436.4 kJ/mol
Calculate enthalpy of formation of H(g)
ΔHrxn = ΣProducts – Σreactants
436.4 kJ/mol = (0) – (x + x)
x = ‐ 218.2 kJ/mol
‐‐‐‐‐ ΔHf of H2,O2,N2 = 0
14. An object of mass 25g at 150ºC is placed in a calorimeter containing 90ml of
water at room temperature (25ºC). Calculate the final temperature of the
mixture if the specific heat capacity of the metal is 23.5 J/ g*K
Given: 25g object at 150ºC, and 90ml of water at 25ºC.
System 1 = object, q=mcΔT = 25(23.5J/g*K) (150+273 –Tf)
System 2 = water, q = mcΔT = 90ml = 90g(4.816J/g*K)(Tf – 25+273)
Conservation of energy = System1 = System2, therefore
25g(23.5J/g*K)(423K – Tf) = 90g(4.816J/g*K)(Tf – 298)
248512.5J – 587.5J/K Tf = 433.44J/K Tf – 129165.12J
377677.62J = 1020.94 J/K Tf
369.93 K = Tf = 96.93ºC
15. Two 150g icecube at 0ºC is placed in a 500mL water bath of 100ºC. Will there
be any remaining ice in the solution? If yes, calculate the amount that is left. If
no, calculate the final temperature that the mixture will be at.
Hfus of ice = 6.010kJ/mol. Specific heat of water = 4.816 J/g*C
Jesse Salas
Anh Nguyen
Nizkorodov Chem 1B
Midterm I Review
Energy it takes to melt the ice:
2*150g ice * 1mol * 6.010kJ/mol = 100.166kJ
18gice
Energy it takes to cool down water:
Q=mcat = 500(4.816J/g*C)(100) = 240800J = No ice will remain!
System1 = two 150g Ice @ 0ºC changing state to total 300g water at 0ºC
System 2 = 300g Water at 0ºC going up to final temperature
System 3 = 200ml Water at 70ºC going down to final temperature
System1 = 2*150g ice * 1mol * 6.010kJ/mol = 100.166kJ
18gice
System 2 = q=mcat = 300g * 4.816J/g*C (Tf – 0) = 1444.8J/K Tf
System 3 = q=mcat = 500g * 4.816J/g*C (100 – Tf) = 240900J –2408J /K Tf
U1 + U2 (energy system of moving ice up) = U3 (energy system of moving water
down)
100.166kJ + 1444.8J/K Tf = 240900J –2408J /K Tf
Tf = 36.5ºC
16. Find the enthalpy of reaction for N2(g) + 2O2(g) ‐> 2NO2(g), given that
N2(g) + 3H2(g) → 2NH3(g) ΔH = ‐115 kJ 2NH3(g) + 4H2O(l) → 2NO2(g) + 7H2(g) ΔH = ‐142.5 kJ H2O(l) → H2(g) + 1/2O 2(g) ΔH = ‐43.7 kJ answer = ‐83 kJ
a + 3(c) + b – 7(c) = ‐82.7kJ

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