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JEE Advanced 2011Paper 2
Exam
JEE Advanced
Year
2011
Paper
Paper 2
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Advanced 2011 · Paper 2 · Chemistry · Thermodynamics (C)

Match the transformation in Column I with appropriate options in...

Match the transformation in Column I with appropriate options in Column II. Match the transformation in Column I with appropriate options in Column II.

Options

  1. A(A) p,q,r,s, (B) r,s, (C) t, (D) p,q
  2. B(A) p,s, (B) p,s, (C) s, (D) p,q,s
  3. C(A) p,s, (B) q,r, (C) s, (D) p,q
  4. D(A) p,r,s, (B) r,s, (C) t, (D) p,q,t

Correct answer

D. (A) p,r,s, (B) r,s, (C) t, (D) p,q,t

Step-by-step solution

(A) CO2( s)⟶CO2( g)\mathrm{CO}_2(\mathrm{~s}) \longrightarrow \mathrm{CO}_2(\mathrm{~g}) It is just a phase transition (sublimation) as no chemical change has occurred. Sublimation is always endothermic. Product is gas, more disordered, hence Δs\Delta s is positive. (B) CaCO3( s)⟶CaO(s)+CO2( g)\mathrm{CaCO}_3(\mathrm{~s}) \longrightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~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, ΔS>0\Delta S>0. (C) 2H⟶H2( g)2 \mathrm{H} \longrightarrow \mathrm{H}_2(\mathrm{~g}) A new H−H\mathrm{H}-\mathrm{H} covalent bond is being formed, hence, ΔH<0\Delta H < 0. Also, product is less disordered than reactant, ΔS<0\Delta S < 0. (D) Allotropes are considered as different phase, hence P(white, solid) →P(red, solid) \mathrm{P}_{\text{(white, solid) }} \rightarrow \mathrm{P}_{\text{(red, solid) }} is a phase transition as well as allotropic change. Also, red phosphorus is more ordered than white phosphorus, ΔS<0\Delta S < 0.
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