Chemistry - Thermochemistry Question with Solution | TestHub
Match the columns
Column I | Column II |
|---|---|
P) 2HBr(g) → H2(g) + Br2(g) | i) ΔH = -ve, ΔS = +ve |
Q) 2H2O2(l) → 2H2O(l) + O2(g) | ii) ΔH = +ve, ΔS = +ve |
R) CaCO3(s) → CaO(s) + CO2(g) | iii) ΔH = -ve, ΔS = -ve |
S) N2(g) + 3H2(g) → 2NH3(g) | iv) ΔH = +ve, ΔS ≈ 0 |
(Column I = Reaction; Column II = Signs of ΔH and ΔS)
Options:
Answer:
Solution:
P: The decomposition of HBr is the reverse of its exothermic formation, so ΔH > 0. With Δng = 0, the entropy change is nearly zero, corresponding to P-iv.
Q: The decomposition of hydrogen peroxide is strongly exothermic and produces a gas from liquids, so ΔH < 0 and ΔS > 0, corresponding to Q-i.
R: The thermal decomposition of solid CaCO₃ is endothermic and releases CO₂ gas, so ΔH > 0 and ΔS > 0, corresponding to R-ii.
S: Ammonia synthesis is exothermic, with 4 moles of gas becoming 2 moles, so ΔH < 0 and ΔS < 0, corresponding to S-iii.
Answer: P-iv; Q-i; R-ii; S-iii.
