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A reaction is at equilibrium at `100^(@)C` and the enthalpy change for the reaction is `"42.6 kJ mol"^(-1)`. What will be the value of `DeltaS` in `"J K"^(-1)"mol"^(-1)`?A. 120B. 426.2C. 373.1D. 114.2 |
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Answer» Correct Answer - D `DeltaG=DeltaH-TDeltaS` At equilibiurm, `DeltaS=(DeltaH)/(T)=("42600 J mol"^(-1))/("373 K")="114.2 J K"^(-1)"mol"^(-1)` |
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