JEE Advance - Chemistry (2011 - Paper 2 Offline - No. 7)
Column I
(A) CO2(s) $$\to$$ CO2(g)
(B) CaCO3(s) $$\to$$ CaO(s) + CO2(g)
(C) 2H $$\to$$ H2(g)
(D) P(white, solid) $$\to$$ P(red, solid)
Column II
(p) phase transition
(q) allotropic change
(r) $$\Delta H$$ is positive
(s) $$\Delta S$$ is positive
(t) $$\Delta S$$ is negative
Explanation
Analysis of each transformation:
(A) CO2(s) $\to$ CO2(g)
This transformation represents the sublimation of solid carbon dioxide (dry ice) into gaseous carbon dioxide. This process is:
Phase transition (p): The substance changes from solid to gas phase.
$$\Delta H$$ is positive (r): Sublimation is an endothermic process, meaning it absorbs heat from the surroundings.
$$\Delta S$$ is positive (s): The entropy increases as the substance moves from a more ordered solid state to a less ordered gaseous state.
(B) CaCO3(s) $\to$ CaO(s) + CO2(g)
This transformation is the thermal decomposition of calcium carbonate to form calcium oxide and carbon dioxide. This process involves:
$$\Delta H$$ is positive (r): Decomposition reactions are typically endothermic, requiring energy to break chemical bonds.
$$\Delta S$$ is positive (s): The process results in an increase in entropy, primarily because a gas (CO2) is produced from a solid compound.
(C) 2H $\to$ H2(g)
This is the formation of hydrogen gas (H2) from atomic hydrogen (H). This transformation:
$$\Delta S$$ is negative (t): Two individual hydrogen atoms combine into a diatomic molecule, decreasing the system's entropy since fewer particles are free to move independently.
(D) P(white, solid) $\to$ P(red, solid)
Crystalline solid white phosphorous is converted to amorphous red phosphorous. This process involves a phase change. Since, white and red phosphorous are allotropic form of phosphorous, i.e., both are composed of phosphorous atoms but the connectivity between atom differ, it is therefore regarded as allotropic change.
The white form of phosphorous has lot of angular strain as compared to stable red phosphorous. Hence, randomness of system due to angular strain decreases as red phosphorus is formed. The entropy change is negative $(\Delta S<0)$
Option Evaluation:
Upon reviewing each transformation and comparing the mappings:
Option A agrees completely with the above analysis with proper mapping for each statement.
Thus, Option A is correct:
A $\to$ p,r,s; B $\to$ r,s; C $\to$ t; D $\to$ p,q,t
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