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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  ∆ S o  (in  kJ   mol - 1 ) for the given reaction, at 1000   K is ___.

Options

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

Answer

A. A

Step-by-step solution

According to graph when  T = 1000   K  and  10 4 T = 10  then  ln p z p ϕ = − 3 Now,  ΔG o = ΔH o − T   ΔS o − RT   lnK p = ΔH o − T   ΔS o R   ln   K p = ΔS o − ΔH o T ΔS o = RlnK p + ΔH o T lnK p = ln p z p ϕ = − 3 ΔS o = − 3 R + 166 .28 × 10 3 J 1000 166 .28 − 3 × 8 .314 ΔS o = 141 .34

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