Thermodynamics and Thermochemistry - Result Question 55
55. Match the transformations in Column I with appropriate options in Column II.
(2011)
| Column I | Column II | ||
|---|---|---|---|
| A. | $\quad CO _2(s) \longrightarrow CO _2(g)$ | p. | Phase transition |
| B. | $\quad CaCO _3(s) \longrightarrow CaO(s)+CO _2(g)$ | q. | Allotropic change |
| C. | $\quad 2 H \bullet \longrightarrow H _2(g)$ | r. | $\Delta H$ is positive |
| D. | $\quad P _{\text {(white, solid) }} \rightarrow P _{\text {(red, solid) }}$ | s. | $\Delta S$ is positive |
| t. | $\Delta S$ is negative |
Show Answer
Answer:
Correct Answer: 55. $\mathrm{A} \rightarrow \mathrm{p}, \mathrm{r}, \mathrm{s} ; \mathrm{B} \rightarrow \mathrm{r}, \mathrm{s} ; \mathrm{C} \rightarrow \mathrm{t} ; \mathrm{D} \rightarrow \mathrm{p}, \mathrm{q}, \mathrm{t}$
Solution:
(A) $CO _2(s) \longrightarrow CO _2(g)$
It is just a phase transition (sublimation) as no chemical change has occurred. Sublimation is always endothermic. Product is gas, more disordered, hence $\Delta S$ is positive.
(B) $CaCO _3(s) \longrightarrow CaO(s)+CO _2(g)$
It is a chemical decomposition, not a phase change. Thermal decomposition occur at the expense of energy, hence endothermic. Product contain a gaseous species, hence, $\Delta S>0$.
(C) $2 H \longrightarrow H _2(g)$
A new $H-H$ covalent bond is being formed, hence, $\Delta H < 0$. Also, product is less disordered than reactant, $\Delta S < 0$.
(D) Allotropes are considered as different phase, hence $P _{\text {(white, solid) }} \longrightarrow P _{\text {(red, solid) }}$ is a phase transition as well as allotropic change.
Also, red phosphorus is more ordered than white phosphorus, $\Delta S < 0$.
BETA
