Chemistry - Thermochemistry Question with Solution | TestHub
Match the columns
Column I | Column II |
|---|---|
P) 2HCl(g) → H2(g) + Cl2(g) | i) ΔH = +ve, ΔS = +ve |
Q) 2O3(g) → 3O2(g) | ii) ΔH = -ve, ΔS = +ve |
R) MgCO3(s) → MgO(s) + CO2(g) | iii) ΔH = -ve, ΔS = -ve |
S) 2NO2(g) → N2O4(g) | iv) ΔH = +ve, ΔS ≈ 0 |
(Column I = Reaction; Column II = Signs of ΔH and ΔS)
Options:
Answer:
Solution:
P: The decomposition of HCl is the reverse of its exothermic formation, so Δ H > 0. With Δ ng = 0, the entropy change is near zero (Δ S° ≈ -20 J/K-mol), which corresponds to P-iv.
Q: Ozone decomposition is exothermic, and 2 moles of gas become 3 moles, so Δ H < 0 and Δ S > 0, which corresponds to Q-ii.
R: The thermal decomposition of solid MgCO₃ is endothermic and releases a gas, so Δ H > 0 and Δ S > 0, which corresponds to R-i.
S: The dimerization of NO₂ is exothermic, with 2 moles of gas becoming 1, so Δ H < 0 and Δ S < 0, which corresponds to S-iii.
Answer: P-iv; Q-ii; R-i; S-iii.
