Chemistry - Thermochemistry Question with Solution | TestHub
Some thermochemical details are given as:
3A(g) → A3(g), ΔH300 = -100 kJ
A(l) → A(g), Δvap H300 = +25 kJ/mol
A3(l) → A3(g), Δvap H400 = +50 kJ/mol
The standard boiling points of A(l) and A3(l) are 300 K and 400 K, respectively.
Molar heat capacities at constant pressure (in J/K-mol): A(l) = 40; A(g) = 20; A3(l) = 50; A3(g) = 30
Match the columns on the basis of these details:
Column I | Column II |
|---|---|
P) A(l) → A(g), Δvap H400 | i) -103 kJ/mol |
Q) A3(l) → A3(g), Δvap H300 | ii) +23 kJ/mol |
R) 3A(l) → A3(l), ΔH300 | iii) +52 kJ/mol |
S) 3A(g) → A3(g), ΔH400 | iv) -77 kJ/mol |
(Column I = Process; Column II = ΔH value)
Options:
Answer:
Solution:
Kirchhoff's Law states: Δ H(T2) = Δ H(T1) + Δ CP (T2 - T1).
For process P:
ΔvapH400 = 25 + (20 - 40)(100)/1000 = +23 kJ/mol. This corresponds to P-ii.
For process Q:
ΔvapH300 = 50 + (30 - 50)(300 - 400)/1000 = +52 kJ/mol. This corresponds to Q-iii.
For process R:
At 300 K, the vaporization of 3A(l) to 3A(g) costs 3 × 25 = +75 kJ/mol.
The conversion of 3A(g) to A₃(g) releases -100 kJ/mol.
The condensation of A₃(g) at 300 K releases -52 kJ/mol.
The total enthalpy change is 75 - 100 - 52 = -77 kJ/mol. This corresponds to R-iv.
For process S:
Δ H400 = -100 + (30 - 3 × 20)(100)/1000 = -103 kJ/mol. This corresponds to S-i.
Therefore, the correct answer is: P-ii; Q-iii; R-iv; S-i.
