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JEE Advanced Chemistry Thermodynamics (C) 2021 JEE Advanced 2021 (Paper 1)

JEE Advanced Chemistry Question (2021) — Solution

Question

For the reaction, X (s) Y (s)+ Z (g), the plot of p_ Z p^ versus 10^ 4 T is given below (in solid line), where p_ Z is the pressure (in bar) of the gas Z at temperature T and p^ =1 bar. (Given, d ( K) d ( 1 T ) =- H^ R , where the equilibrium constant, K= p_ z p^ and the gas constant, R =8.314 ~J ~K ^ -1 ~mol ^ -1 ) The value of standard enthalpy,  ∆ H o  (in  kJ   mol - 1 ) for the given reaction is ___.

Options

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

Answer

A. A

Step-by-step solution

Slop  g  the Given graph d   ln p z p ϕ d   10 4 T = − 7 + 3 12 − 10 = − 2 d   ln p z p ϕ d 1 T − 2 × 10 4 = − ΔH o R ΔH o = 2 × 10 4 × R = 2 × 8 .314 × 10   kJ / mol = 166 . 28

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