Chemistry - Thermochemistry Question with Solution | TestHub
Match the column
Column I | Column II |
|---|---|
P) C(s) + O2(g) → CO2(g) | i) ΔH = ΔU |
Q) PCl5(g) → PCl3(g) + Cl2(g) | ii) ΔH > ΔU and |ΔH| > |ΔU| |
R) 2C(s) + O2(g) → 2CO(g) | iii) ΔH > ΔU and |ΔH| < |ΔU| |
S) N2(g) + 3H2(g) → 2NH3(g) | iv) ΔH < ΔU and |ΔH| > |ΔU| |
(Column I = Reaction; Column II = Relation between ΔH and ΔU)
Options:
Answer:
Solution:
The relationship between enthalpy change (ΔH) and internal energy change (ΔU) is given by the equation:
ΔH = ΔU + Δng RT
where RT ≈ 2.5 kJ at 298 K.
For reaction P: Δng = 0, which implies ΔH = ΔU. This corresponds to P-i.
For reaction Q: Δng = +1, which implies ΔH > ΔU. The reaction is endothermic with ΔH ≈ +88 kJ, so ΔU ≈ +85.5 kJ. Therefore, |ΔH| > |ΔU|. This corresponds to Q-ii.
For reaction R: Δng = +1, which implies ΔH > ΔU. Here, ΔH = -221 kJ, and ΔU ≈ -223.5 kJ (which is more negative). Thus, |ΔH| < |ΔU|. This corresponds to R-iii.
For reaction S: Δng = -2, which implies ΔH < ΔU. With ΔH = -92 kJ, ΔU ≈ -87 kJ (which is less negative). Thus, |ΔH| > |ΔU|. This corresponds to S-iv.
The final answer is P-i; Q-ii; R-iii; S-iv.
