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The standard free energy change for a re...

The standard free energy change for a reaction is `-213.3 KJ mol^(-1) "at" 25^(@)C`. If the enthalpy change of the reaction is `-217.77 KJ "mole"^(-1)` . Calculate the magnitude of entropy change for the reaction in Joule `"mole"^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
-15
Given : `" "DeltaH =-271.77 KJ K^(-1) mol^(-1)`
`" "DeltaG =- 313.3 KJ mol^(-1)`
`" "T=273 + 25 =298 K`
Asked : The entropy change for the reaction=?
Formula: `DeltaG = DeltaH -TDeltaS`
Explanation: `DeltaG` = Gibb's free energy change
`" "DeltaH`= Enthalpy change
`" "`T=temperature in K
`" "DeltaS` = Entropy change
Substitution & Calculation:
`" "-213.3 xx 10^(3) =- 217.77 xx 10^(3) - 298 xx DeltaS`
`" "or" "DeltaS= (-(217.77 xx 10^(3) - 213.3 xx 10^(3)))/(298)`
`" "DeltaS=-15` J/mol K
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