Chem 10113, Exam 2 Answer Key October 25, 2017

1. (10 points) SHOW ALL WORK. Given the following thermochemical data, determine the standard

heat of formation (∆H°f) of N2O4(g) in kJ/mole.

compound: H2O(l) N2H4(l) N2O(g)

∆H°f (kJ/mole): - 285.8 95.4 81.6

N2O4(l) + N2H4(g) → 2 N2O(g) + 2 H2O(l) ∆H° = - 484.3 kJ

∆H° = Σ ∆H°f (products) - Σ ∆H°f (reactants)

-484.3 kJ = {2 mole N2O (81.6 kJ/mole) + 2 mole H2O (-285.8 kJ/mole)}

- {1 mole N2H4 (95.4 kJ/mole) + 1 mole N2O4 [∆H°f (N2O4)]}

∆H°f (N2O4) = -19.5 kJ/mole

2. (8 points) Write a balanced chemical equation for the process that occurs when each of the

following substances are mixed with water. If the substance is a weak electrolyte, indicate that by

using the appropriate symbol(s) in your equation. (a) (CH2)2NH + H2O (CH2)2NH2

+ + OH-

(b) Li2O + H2O → 2 LiOH

(c) HSCN + H2O H3O+ + SCN-

3. (5 points) Write the oxidation number of phosphorus in each of the following.

P2F4 +2 K3P -3 H2PO2- +1 P4 0 P4O10 +5

4. (10 points) SHOW ALL WORK. Initially, a mixture of N2O(g) and O2(g) in a 0.300 L container at

500 °C has a total pressure of 20.0 atm. After some time, the total pressure increases to 22.4 atm

due to the formation of some NO(g). Determine the mass in grams of NO that forms.

2 N2O + O2 → 4 NO

If x atm of O2 react, 2x atm of N2O are consumed, and 4x atm of NO are formed.

After some NO is formed, the total pressure, 22.4 atm, is equal to the sum of the

partial pressures.

(20.0 - 2x - x) + 4x = 22.4 20.0 + x = 22.4

so, x = 2.4 atm and PNO = 4x = 9.60 atm

moles NO = n = PV/RT = (9.60 atm) (0.300 L) / (0.0821 L.atm) (773 K)

n = 0.0454 mole NO

(0.0454 mole NO) (30.0 g/mole) = 1.36 g NO

5. (10 points) SHOW ALL WORK. Magnesium metal reacts with hydrochloric acid as follows.

Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) ∆Hrxn = ??? When 0.125 g of Mg is combined with enough HCl to make 50.0 mL of solution in an insulated

container, all of the Mg reacts, raising the temperature of the solution from 22.3 °C to 33.8 °C.

Determine ∆Hrxn (in kJ) for the reaction as written above. (If necessary. assume the solution has

the same density and specific heat as pure water.)

Heat produced = (50.0 g) (4.184 J/g.°C) (33.8 °C - 22.3 °C) = 2406 J

Since heat is produced, the reaction is exothermic, i.e., ∆Hrxn is negative.

∆Hrxn = (- 2.406 kJ) / [(0.125 g) (1 mole / 24.3 g)] = - 468 kJ (per mole of Mg)

6. (4 points) Write the short-hand electron configuration for platinum (Pt).

[Xe]54 6s2 4f14 5d8

7. (5 points) Give the orbital diagram for the valence shell configuration of each of the following.

(a) Sb (b) Os3+

Sb

5p5s 5dOs3+

8. (2 points) In basic solution, I3-(aq) reacts with thiosulfate ion to produce iodide ion and sulfate ion.

Write the chemical formula for the reducing agent in this process.

S2O32- (thiosulfate)

9. (10 points) SHOW ALL WORK. The bromine molecule (Br2) can be photo-dissociated (broken

apart with light) into Br atoms in the gas phase using light with a wavelength of 620 nanometers. A

250.0-mL glass bulb contains 3.80 torr of gaseous Br2 at 27.0 °C. Determine the minimum amount

of energy (in joules) that must be absorbed by the bromine to dissociate 25.0 % of the molecules.

(h = 6.626 x 10-34 J.sec)

E = hc/λ = (6.626 x 10-34 J.sec) (3.00 x 108 m/sec) / (620 x 10-9 m)

E = 3.206 x 10-19 J (per photon)

n = PV/RT = (3.80/760)atm (0.250 L) / (0.0821 L.atm/mole.K) (300 K)

n = 5.075 x 10-5 mole Br2

Br2 molecules dissociated?

(0.25) (5.075 x 10-5 mole) (6.022 x 1023 molecules/mole) = 7.641 x 1018 molecules

= 7.641 x 1018 photons

energy required = (3.206 x 10-19 J/photon) (7.641 x 1018 photons) = 2.45 J

10. (12 points) In the space below each of the following reactions, clearly write the balanced, net ionic

equation. (Only the net ionic equation will be graded!) Use subscripts [(s), (aq), (g), etc.] to

indicate the phase of each compound or ion. If no reaction occurs, write No Rx.

(a) (NH4)2CO3(aq) + HBr(aq) → (gas formation)

CO32-

(aq) + 2 H+(aq) → H2O + CO2(g)

(b) HNO2(aq) + RbOH(aq) → (weak electrolyte formation)

HNO2(aq) + OH-(aq) → NO2

-(aq) + H2O

(c) Fe(NO3)3(aq) + (NH4)2SO4(aq) → No Reaction

(d) HNO3(aq) + BaS(s) → (gas formation)

2 H+(aq) + BaS(s) → Ba2+

(aq) + H2S(g)

11. (2 points) The niobium(IV) ion, Nb4+, has one unpaired electron. List all possible quantum numbers

for this unpaired electron. n = 4 l = 2 ml = -2, -1, 0, +1, +2

12. (3 points) Circle any of the following that are not soluble in water.

NH4IO3 AlPO4 Pb(C2H3O2)2 BaSO4 HN3 Fe2S3

13. (2 points) Circle any of the following that have the exact same valence shell electron configuration

as argon (Ar). Ca2+ O2- K Mg Xe V3+ P3- Al3+

14. (2 points) The anhydride of H3BO3 is B2O3.

15. (3 points) The values listed below are the atomic radii (in picometers) for the following atoms:

Mg, Al, Si, Na, N, K, C. Match each radius with the corresponding element.

227 K 186 Na 160 Mg 143 Al 118 Si 77 C 70 N

16. (2 points) Which element in question 15 above has the lowest third ionization energy? Al

17. (10 points) SHOW ALL WORK. In a certain gas-separation system, 15.0 mL of N2 gas effuses

through a membrane in 1.00 minute. In the same system, under the same conditions, 53.0 mL of an

unknown gas effuses in 5.00 minutes. Assuming that the unknown gas is a hydrocarbon, determine

its molecular formula.

effusion rate of unknown gas = (53.0 L) / (5 min) = 10.6 L/min

Graham's Law of Effusion:

ER (N2) / ER (unknown) = { (molar mass of unknown) / molar mass N2) }1/2

(15.0 L/min) / (10.6 L/min) = { x / (28.0 g/mole) }1/2

x = 56.1 g/mole ∴ unknown hydrocarbon must be C4H8