Quantrex Academy · Free JEE Advanced PYQ solutions
JEE Advanced Chemistry Thermodynamics (C) 2023 JEE Advanced 2023 (Paper 2)

JEE Advanced Chemistry Question (2023) — Solution

Question

The entropy versus temperature plot for phases α  and β at 1 bar pressure is given. S T and S 0  are entropies of the phases at temperatures T and 0   K , respectively. The transition temperature for α to β phase change is 600   K and C p , β - C p , α = 1   J   mol - 1   K - 1 . Assume C p , β - C p , α is independent of temperature in the range of 200 to 700   K . C p , α and C p , β are heat capacities of α and β phases, respectively. The value of enthalpy change, H β - H α (in Jmol - 1 ), at 300   K is

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

According to Kirchhoff's law  ΔH T 2 - ΔH T 1 = n × C p , β - C p , α ( T 2 - T 1 ) ΔH 600 - ΔH 300 = 1 × C p , β - C p , α ( 600 - 300 ) Now, at transition temperature,  ∆ G = 0 ΔH 600 = TΔS 600 = 600 × ( 6 - 5 ) 600 - ΔH 300 = 1 × 1 × 300 ΔH 300 = 600 - 300 = 300   J   mol - 1

Practice more on Quantrex App →

Related: Chemistry — Thermodynamics (C) · All PYQ Banks