`Zn(s)+2AgCl(s) hArr ZnCl_(2) (0.555 M)+2Ag(s)` `E_(0^(@)C)=1.015" volt"((dE)/(dT))_(P)=-4.02xx10^(-4)` volt per degree. Find `DeltaG, DeltaS`
`Zn(s)+2AgCl(s) hArr ZnCl_(2) (0.555 M)+2Ag(s)`
`E_(0^(@)C)=1.015" volt"((dE)/(dT))_(P)=-4.02xx10^(-4)` volt per degree. Find `DeltaG, DeltaS`
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Correct Answer - `DeltaG=-195.895 kJ, DeltaS=-18.55` cal
We know that,
`DeltaH=nF [T((dE)/(dT))_(P)-E]`
`=2xx96500 [-273xx4.02xx10^(-4)-1.015]`
`=-217075.98" joule"=-217.075 kJ`
`DeltaG=-nFE=-2xx96500xx1.015=-195895 J`
`=-195.895 kJ`
`DeltaG=DeltaH-TDeltaS`
`-195.895=-217.075-273xxDeltaS`
`DeltaS=-0.07758 kJ =-77.58 J=-18.55` cal
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