Chemistry - Thermodynamics - II Question with Solution | TestHub
Match Column I with Column II
Column I | Column II |
|---|---|
P) For the process: A(l) → A(s), Δ H and Δ V may be | i) −ve, +ve |
Q) A2(s) + B2(g) → C2(s) + D2(s), Δ H and Δ G may be | ii) +ve, −ve |
R) A2(g) → B(g) + C(g), Ea(forward) = 50 kJ/mol, Ea(backward) = 40 kJ/mol, Δ H and Δ G at very high temperature are | iii) +ve, +ve |
S) A(g) → B(g), Δ H and Δ G at very low temperature are | iv) −ve, −ve |
Options:
Answer:
Solution:
(P) Freezing is always exothermic (Δ H < 0); Δ V is positive for water-type substances and negative for most others — (−ve, +ve) or (−ve, −ve): i,iv. (Q) One mole of gas is consumed, so Δ S < 0; with Δ G = Δ H - TΔ S, the pairs (−,+), (+,+) and (−,−) are all possible depending on Δ H and T, but (+,−) is impossible — i,iii,iv. (R) Δ H = Ea(f) - Ea(b) = +10 kJ/mol; more gas moles give Δ S > 0, so at very high T, Δ G = Δ H - TΔ S < 0 — (+ve, −ve): ii. (S) Equal gas moles give Δ S ≈ 0, so at very low T, Δ G ≈ Δ H: both positive or both negative — iii,iv.
