Thermodynamics and Thermochemistry 1 Question 52
55. Match the transformations in Column I with appropriate options in Column II.
(2011)
| Column I | Column II | |
|---|---|---|
| A. $\quad \mathrm{CO}{2}(s) \longrightarrow \mathrm{CO}{2}(g)$ | p. | Phase transition |
| B. $\quad \mathrm{CaCO}{3}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}{2}(g)$ | q. | Allotropic change |
| C. $\quad 2 \mathrm{H} \bullet \longrightarrow \mathrm{H}_{2}(g)$ | r. | $\Delta H$ is positive |
| D. $\quad \mathrm{P}{\text {(white, solid) }} \rightarrow \mathrm{P}{\text {(red, solid) }}$ | s. | $\Delta S$ is positive |
| t. | $\Delta S$ is negative |
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Solution:
- (A) $\mathrm{CO}{2}(s) \longrightarrow \mathrm{CO}{2}(g)$
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) $\mathrm{CaCO}{3}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}{2}(g)$
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) $2 \mathrm{H} \longrightarrow \mathrm{H}_{2}(\mathrm{~g})$
A new $\mathrm{H}-\mathrm{H}$ 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 $\mathrm{P}{\text {(white, solid) }} \longrightarrow \mathrm{P}{\text {(red, solid) }}$ is a phase transition as well as allotropic change.
Also, red phosphorus is more ordered than white phosphorus, $\Delta S<0$.
