1.

Calculate enthalpy change for the change 8S(g)toS_(8)(g), given that H_(2)S_(2)(g)to2H(g)+2S(g),DeltaH=239.0k"cal mol"^(-1) H_(2)S(g)to2H(g)+S(g),DeltaH=175.0k"cal mol"^(-1)

Answer»

`+512.0` k cal
`-512.0` k cal
`508.0` KCAL
`-508.0` kcal

Solution :`DeltaH_(S-S)+2DeltaH_(H-S)=239""2DeltaH_(H-S)=175`
Hence `DeltaH_(S-S)=239-175=64kcal "MOL"^(-1)`
Then `DeltaH` for `8S(G)toS_(8)` is `8xx(-64)=-512` kcal


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