Chemistry - Thermochemistry Question with Solution | TestHub
Some thermochemical details are given as:
3X(g) → X3(g), ΔH400 = -120 kJ
X(l) → X(g), Δvap H400 = +30 kJ/mol
X3(l) → X3(g), Δvap H500 = +60 kJ/mol
The standard boiling points of X(l) and X3(l) are 400 K and 500 K, respectively.
Molar heat capacities at constant pressure (in J/K-mol): X(l) = 50; X(g) = 30; X3(l) = 70; X3(g) = 40
Match the columns on the basis of these details:
Column I | Column II |
|---|---|
P) X(l) → X(g), Δvap H500 | i) -93 kJ/mol |
Q) X3(l) → X3(g), Δvap H400 | ii) +28 kJ/mol |
R) 3X(l) → X3(l), ΔH400 | iii) +63 kJ/mol |
S) 3X(g) → X3(g), ΔH500 | iv) -125 kJ/mol |
(Column I = Process; Column II = ΔH value)
Options:
Answer:
Solution:
Kirchhoff: ΔH(T2) = ΔH(T1) + ΔCP (T2 - T1). P: ΔvapH500 = 30 + (30 - 50)(100)/1000 = +28 kJ/mol, P-ii. Q: ΔvapH400 = 60 + (40 - 70)(400 - 500)/1000 = +63 kJ/mol, Q-iii. R: at 400 K, 3X(l) → 3X(g) costs 3 × 30 = +90, 3X(g) → X3(g) releases -120, condensing X3(g) at 400 K releases -63; total 90 - 120 - 63 = -93 kJ/mol, R-i. S: ΔH500 = -120 + (40 - 3 × 30)(100)/1000 = -125 kJ/mol, S-iv. Ans: P-ii; Q-iii; R-i; S-iv.
