Modern family: Addition

State level Olympiad (7)

13. A 2.00 L balloon at 20.0˚C and 745 mmHg floats to an
altitude where the temperature is 10.0˚C and the air
pressure is 700 mmHg. What is the new volume of the
balloon?

(A) 0.94 L (B) 1.06 L (C) 2.06 L (D) 2.20 L

Sol. We can use the combined gas law to solve this question, which is below.
(P1V1/T1) = (P2V2/T2)

We are given that V1 = 2.00, T1 = 20 + 273 = 293 K, and T2 = 10 + 273 = 283 K. Now, we can substitute to solve for V2.
(745)(2.00)/293 = (700)(V2)/283
Solving, we get V2 = 2.06 L , or C .

14. Which family of elements has solid, liquid and gaseous
members at 25˚C and 1 atm pressure?

(A) alkali metals (Li – Cs) (B) pnictogens (N – Bi)
(C) chalcogens (O – Te) (D) halogens (F – I)

Sol . Only two elements are liquid in their standard state, which are bromine and mercury. Mercury’s family (Group 12) is not an option; it has no gases, anyways. That leaves us with the halogens. Fluorine and chlorine are gases
and iodine is a solid. Thus, the answer is halogens , or D . The trends of the boiling points and melting points of
halogens can be further explained by the intermolecular forces.

15. A gas diffuses one-third as fast as O2 at 100 ˚C. This gas
could be

(A) He (M = 4) (B) C2H5F (M = 48)
(C) C7H12 (M = 96) (D) C5F12 (M = 288)

Sol. We can use Graham’s Law of Diffusion to get the following relation.
R1/R2 = √(M2/M1)
We are given that R2 = (1/3)R1, and we know the molar mass of O2 is M1 = 32 g/mol. Substituting, we get
1/(1/3) = √(M2/32)
3 = √(M2/32)
M2 = 9 × 32 = 288
Thus, the correct answer is C5F12 , or D .

16. Moist air is less dense than dry air at the same
temperature and barometric pressure. Which is the best
explanation for this observation?

(A) H2O is a polar molecule but N2 and O2 are not.
(B) H2O has a higher boiling point than N2 or O2.
(C) H2O has a lower molar mass than N2 or O2.
(D) H2O has a higher heat capacity than N2 or O2.

Sol. Let us compare the molar masses of H2O, N2, O2. The molecular mass of H2O is 18 g/mol, N2 is 28 g/mol, and
O2 is 32 g/mol. The equation for molar mass M in terms of density d is
M = dRT/P
This can be derived from P V = nRT. We see that the molar mass of H2O is less than the other compounds, and thus will have a smaller density. Thus, the correct explanation is C .

17. Under certain conditions CO2 melts rather than sublimes. To
which transition in the phase diagram does this change
correspond?

(A) A → B (B) A → C (C) B → C (D) C → B

Sol. Melting refers to the phase transition from solid to liquid, and sublimation to the transition from solid to
gas. We additionally know that A represents the solid state, B represents the liquid state, and C represents the gas
state based on the pressure and temperature of each area. Thus, A → B corresponds to melting, and the answer is A .

18. The critical temperature of water is the

(A) temperature at which solid, liquid and gaseous water
coexist.
(B) temperature at which water vapor condenses.
(C) maximum temperature at which liquid water can
exist.
(D) minimum temperature at which water vapor can
exist.

Sol. The definition of the supercritical fluid in a phase diagram is when the liquid and gas phases are indistin- guishable, which is the area above critical temperature and critical pressure. So the critical temperature is also
the maximum temperature at which liquid water can exist , as water will be either gas or supercritical fluid under a
higher temperature. This description is most accurately depicted in answer choice C . Sketching a phase diagram to
label each phase comes handy to solve problems like this.

19. Which process is exothermic?

(A) condensation (B) fusion
(C) sublimation (D) vaporization

Sol. Heat is needed for phase transitions where the molecules are excited. Endothermic reactions or processes,
such as fusion (melting), vaporization (boiling), and sublimation, require heat. Exothermic reactions or processes, such as freezing, condensation , and deposition, release heat. Thus, the correct is A .

20. Use the thermodynamic information:
½ N2(g) + ½O2(g) → NO(g) ∆H˚ = 90.4 kJ/mol
½ N2(g) + O2(g) → NO2(g) ∆H˚ = 33.8 kJ/mol
2NO2(g) → N2O4(g) ∆H˚ = –58.0 kJ/mol
to calculate ∆H˚ in kJ/mol for the reaction:
2NO(g) + O2(g) → N2O4(g)

(A) –171.2 (B) –114.6 (C) 114.6 (D) 171.2

Sol. Reshuffle the reactions to get the desired final reaction. We see that the last reaction must remain the same to keep the 1 N2O4. However, let us double the first two reactions to eliminate the fractional coefficients.
N2 + O2 → 2 NO
N2 + 2 O2 → 2 NO2
2 NO2 → N2O4

However, seeing the final reaction, there is no N2 present, so we need to reverse the first reaction to eliminate N2.
2 NO → N2 + O2
N2 + 2 O2 → 2 NO2
2 NO2 → N2O4
Hess’ Law allows us to calculate ∆H◦ for the final reaction:
∆H◦ = 2(−1)(90.4) + 2(33.8) − 58.0
= −171.2 kJ/mol
This is A .

21. Determine the enthalpy change for the reaction of 5.00 g
of Fe2O3 with aluminum metal according to the equation
Fe2O3(s) + 2Al(s) → Al2O3(s) + 2Fe(l) Substance ∆Hf˚ kJ/mol
Fe2O3(s) –825.5
Al2O3(s) –1675.7
Fe(l) 12.4

(A) –25.8 kJ (B) –26.2 kJ
(C) –52.4 kJ (D) –77.9 kJ

Sol. The heat of a reaction can be calculated with:
∆H◦ _ rxn = ∆H◦ _ f (prod) − ∆H◦
_ f (reac)

To calculate the sum of enthalpy of formation for products, we have the following.
∆H◦ _ f (prod) = −1675.7 + 2(12.4) kJ/mol

To calculate the sum of enthalpy of formation for reactants, we have the following. Note that the standard enthalpy
of formation of Al, as with all standard elements, is 0.
∆H◦ _ f (reac) = −825.5 + 0 = −825.5 kJ/mol

∆H◦ _ rxn = −1675.7 + 2(12.4) − (−825.5) = −825.4 kJ/mol

∆H◦ _ rxn is ’per mole’, so we need to find the number of moles of reactant.

5.00 g Fe2O3 × (1 mol Fe2CO3 /
159.68 g Fe2CO3) × (−825 kJ/
1 mol Fe2O3) = −25.8 kJ
Thus, the answer is A .

22. Which reaction has the most positive entropy change
under standard conditions?

(A) H2O(g) + CO(g) → H2(g) + CO2(g)
(B) CaCO3(s) → CaO(s) + CO2(g)
(C) NH3(g) → NH3(aq)
(D) C8H18(l) → C8H18(s)

Sol. By just looking for the phases of the reactants and products, we can determine how the entropy changes, as the trend of molar entropy in general is S_m(gas) >> S_m(liquid) > S_m(solid). For Option A, there is no significant
change in entropy because there is no phase change or change in number of moles. For Options C and D, there is a
decrease in entropy, with a transition from gas to aqueous in Option C, and liquid to solid in Option D. The correct
answer is the reaction below, or B , because there is a transition from solid to gas.
CaCO3(s) → CaO(s) + CO2(g)

23. What are the signs of ∆H and ∆S for a reaction that is
spontaneous only at low temperatures?

(A) ∆H is positive, ∆S is positive
(B) ∆H is positive, ∆S is negative
(C) ∆H is negative, ∆S is negative
(D) ∆H is negative, ∆S is positive

Sol. In order for a reaction to be spontaneous, ∆G < 0. We know ∆G = (∆H) − (T ∆S). At low temperatures, the ∆H
term dominates; thus ∆H must be negative . At high temperatures, the T ∆S term dominates; T is in Kelvin and must always be positive in order to make ∆G > 0, so ∆S must be negative . Thus, the correct answer is confirmed
to be C .