Chemistry - Thermochemistry Question with Solution | TestHub
Match the columns
Column I | Column II |
|---|---|
P) Sb(s) | i) Δf H° = +ve, Δf S° = +ve |
Q) O3(g) | ii) Δf H° = 0, Δf S° = 0 |
R) I2(g) | iii) Δf H° = +ve, Δf S° = -ve |
S) CO(g) | iv) Δf H° = -ve, Δf S° = +ve |
(Column I = Substance; Column II = Signs of Δf H° and Δf S°)
Options:
Answer:
Solution:
P: Sb(s) is the standard state of antimony. Therefore, both Δf H° and Δf S° are zero (P-ii).
Q: The reaction 3/2O2(g) tarrow O3(g) is endothermic, with Δf H° = +143 kJ/mol. In this reaction, 1.5 mol of gas is converted to 1 mol, which means the entropy decreases (Q-iii).
R: The process I2(s) tarrow I2(g) represents sublimation. This is an endothermic process with a significant increase in entropy as the substance transitions from solid to gas (R-i).
S: For the reaction C(s) + 1/2O2(g) tarrow CO(g), Δf H° = -110.5 kJ/mol. The moles of gas increase from 0.5 to 1, indicating that Δf S° > 0 (S-iv).
Answer: P-ii; Q-iii; R-i; S-iv.
