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Thermodynamics question

2021 · 1 Sep · Shift 2 · Q16
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Thermodynamics question

2021 · 1 Sep · Shift 2 · Q16

JEE MainChemistryThermodynamicsNumerical+4 / −1
For the reaction, 2NO2NO_2NO2​(g) ⇌\rightleftharpoons⇌ N2O4N_2O_4N2​O4​(g), when Δ\DeltaΔ S = −-− 176.0 JK −-− 1 and Δ\DeltaΔ H = −-− 57.8 kJ mol −-− 1, the magnitude of Δ\DeltaΔ G at 298 K for the reaction is ‾\underline{\hspace{2cm}}​ kJ mol −-− 1. (Nearest integer)
Numerical answer
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Correct answer: 5

  1. Use the Gibbs free energy relation:
ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta SΔG=ΔH−TΔS
  1. Given:
ΔH=−57.8 kJ mol−1\Delta H = -57.8\ \text{kJ mol}^{-1}ΔH=−57.8 kJ mol−1 ΔS=−176.0 J K−1mol−1=−0.176 kJ K−1mol−1\Delta S = -176.0\ \text{J K}^{-1}\text{mol}^{-1} = -0.176\ \text{kJ K}^{-1}\text{mol}^{-1}ΔS=−176.0 J K−1mol−1=−0.176 kJ K−1mol−1 T=298 KT = 298\ \text{K}T=298 K
  1. Compute TΔST\Delta STΔS:
TΔS=298×(−0.176)=−52.448 kJ mol−1T\Delta S = 298 \times (-0.176) = -52.448\ \text{kJ mol}^{-1}TΔS=298×(−0.176)=−52.448 kJ mol−1
  1. Now calculate ΔG\Delta GΔG:
ΔG=−57.8−(−52.448)\Delta G = -57.8 - (-52.448)ΔG=−57.8−(−52.448) ΔG=−57.8+52.448=−5.352 kJ mol−1\Delta G = -57.8 + 52.448 = -5.352\ \text{kJ mol}^{-1}ΔG=−57.8+52.448=−5.352 kJ mol−1
  1. The question asks for the magnitude of ΔG\Delta GΔG:
∣ΔG∣=5.352 kJ mol−1|\Delta G| = 5.352\ \text{kJ mol}^{-1}∣ΔG∣=5.352 kJ mol−1
  1. Nearest integer:
555
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