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 |
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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, $\Delta S<0$.
