Thermochemistry Quiz-13

Thermochemistry Quiz-13

Dear Readers,

Thermochemistry is the part of thermodynamics that studies the relationship between heat and chemical reactions. As we know thermodynamics is a very important topic from examination point of view. Hence as a part of it, thermochemistry also becomes important for exams. So if you want to be best in thermodynamics then you must study thermochemistry properly. So go ahead and prepare well. All the best !

Q1. The value of ΔH° for the reaction Cu⁺(g) + I⁻(g) → CuI(g) is −446 kJ/mol. If the ionisation energy of Cu(g) is 745 kJ/mol and the electron affinity of I(g) is 295 kJ/mol , then the value of ΔH° for the formation of one mole of CuI(g) from Cu(g) and I(g) is :

•  −446 kJ/mol
•  450 kJ/mol
•  594 kJ/mol
•  4 kJ/mol

Solution:-

Q2. H₂(g) + Cl(g) = 2HCl(g); ΔH(298 K) = 22.06 kcal. For this reaction, ΔU is equal to:

•  −22.06 + 2×10⁻³ × 298 × 2 kcal
•  −22.06 + 2 × 298 kcal
•  −22.06 − 2 × 298 × 4 kcal
•  −22.06 kcal

Solution:-
Δn = 0,   ∴ ΔH = ΔU

Q3. The enthalpy of fusion of water is 1.435 kcal/mol. The molar entropy change for the melting of ice at 0°C is:

•  5.260 cal/(mol K)
•  0.526 cal/(mol K)
•  10.52 cal/(mol K)
•  21.04 cal/(mol K)

Solution:-
ΔS = ΔH/T = 5.260 cal/(mol K)

Q4. The enthalpy changes of formation of the gaseous oxides of nitrogen (N₂O and NO) are positive because of:

•  The high bond energy of the nitrogen molecule
•  The high electron affinity of oxygen atoms
•  The high electron affinity of nitrogen atoms
•  The tendency of oxygen to form O²⁻

Solution:-
Due to high bond energy of N≡N, more heat is absorbed to break up N₂ molecule.

Q5. C_graphite + O₂(g) → CO₂(g);   ΔH = −94.05 kcal/mol
  C_diamond + O₂(g) → CO₂(g);   ΔH = −94.05 kcal/mol , therefore:

•  C_diamond → C_graphite ;   ΔH°_298K = +450 cal/mol
•  C_graphite → C_diamond ;   ΔH°_298K = −450 cal/mol
•  Diamond is harder than graphite
•  Graphite is the stabler allotrope

Solution:-
Graphite possesses lesser energy than diamond.

Q6. The energy absorbed by each molecule (A₂) of a substance is 4.4 × 10⁻¹⁹ J and bond energy per molecule is 4.0 × 10⁻¹⁹ J. The kinetic energy of the molecule per atom will be :

•  4.0 × 10⁻²⁰ J
•  2.0 × 10⁻²⁰ J
•  2.2 × 10⁻¹⁹ J
•  2.0 × 10⁻¹⁹ J

Solution:-

Q7. Which of the following is an intensive property?

•  Temperature
•  Viscosity
•  Surface tension
•  All of these

Solution:-
The properties of the system whose value is independent of the amount of substance present in the system are called intensive properties e.g., viscosity, surface tension, temperature, pressure etc.

Q8. For the reaction, C(graphite) + 1/2O₂(g) → CO(g) at 298 K and 1 atm, ΔH = −26.4 kcal. What is ΔE, if the molar volume of graphite is 0.0053 L? (R = 0.002 kcal mol⁻¹ K⁻¹)

•  −26.7 kcal
•  +26.7 kcal
•  −52.4 kcal
•  +52.4 kcal

Solution:-
C(graphite) + 1/2O₂(g) → CO(g)
  Δn_g = 1 − 0.5 = 0.5
  ΔH = ΔE + Δn_gRT
  ΔE = ΔH − Δn_gRT
  = −26.4 − 0.5×0.002×298
  = −26.7 kcal

Q9. The heat of formation is the change in enthalpy accompanying the formation of a substance from its elements at 298 K and 1 atm pressure. Since, the enthalpies of elements in their most stable state are taken to be zero, the heat of formation of compounds is :

•  Always negative
•  Always positive
•  Standard heat enthalpy of that compound
•  Zero

Solution:-
  C + O₂ → CO₂(g);   ΔH°f = ? , if reaction is made at 25°C and 1 atm.
  ΔH°f = H°_CO2 − H°_C − H°_O2 = H°_CO2 − 0 − 0
  ΔH°f = H°_CO2 (H°_C and H°_O2 are assumed arbitrarily zero)

Q10. The internal energy change when a system goes from state A to B is 40 kJ/mol. If the system goes from A to B by a reversible path and returns to state A by an irreversible path, what would be the net change in internal energy?

•  40 kJ
•  >40 kJ
•  <40 kJ
•  Zero

Solution:-
In a cyclic process, ΔE = 0.