1. Tardigrade
  2. Question
  3. Chemistry
  4. A certain reaction is at equilibrium at 82° C and the enthalpy change for this reaction is 21.3 kJ. The value of Δ S(JK-1mol-1) for the reaction is

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Q. A certain reaction is at equilibrium at $ 82{}^\circ C $ and the enthalpy change for this reaction is 21.3 kJ. The value of $ \Delta S(J{{K}^{-1}}mo{{l}^{-1}}) $ for the reaction is

MGIMS WardhaMGIMS Wardha 2014

Solution:

Gibbs free energy. $ \Delta G=\Delta H-T\Delta S $ But, at equilibrium, $ \Delta G=0 $ $ \therefore $ $ \Delta H=T\Delta S $ or $ \Delta S=\frac{\Delta H}{T}=\frac{21.3\times {{10}^{3}}}{355}=60\,J{{K}^{-1}}mo{{l}^{-1}} $