Chemistry - Thermochemistry Question with Solution | TestHub
Match the column
Column I | Column II |
|---|---|
P) H2(g) + Cl2(g) → 2HCl(g) | i) ΔH = ΔU |
Q) 2C(s) + O2(g) → 2CO(g) | ii) ΔH > ΔU |
R) PCl5(g) → PCl3(g) + Cl2(g) | iii) ΔH < ΔU |
S) C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l) | iv) |ΔH| > |ΔU| |
| v) |ΔH| < |ΔU| |
(Column I = Reaction; Column II = Relation between ΔH and ΔU)
Options:
Answer:
Solution:
The relationship between enthalpy change (Δ H) and internal energy change (Δ U) is given by the equation:
Δ H = Δ U + Δ ng RT
where RT ≈ 2.5 kJ at 298 K.
For reaction P: Δ ng = 0, so Δ H = Δ U. This corresponds to P-i.
For reaction Q: Δ ng = +1, so Δ H > Δ U. Here, Δ H = -221 kJ and Δ U = Δ H - RT ≈ -223.5 kJ. Therefore, |Δ H| < |Δ U|. This corresponds to Q-ii,v.
For reaction R: Δ ng = +1, so Δ H > Δ U. The reaction is endothermic with Δ H ≈ +88 kJ. Δ U = Δ H - RT ≈ +85.5 kJ. Therefore, |Δ H| > |Δ U|. This corresponds to R-ii,iv.
For reaction S: Δ ng = 2 - 4 = -2, so Δ H < Δ U. Combustion is strongly exothermic with Δ H ≈ -1411 kJ. Δ U = Δ H - Δ ng RT ≈ -1406 kJ, which is less negative. Therefore, |Δ H| > |Δ U|. This corresponds to S-iii,iv.
The final answer is: P-i; Q-ii,v; R-ii,iv; S-iii,iv.
