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H(2)(g)+Br(2)(l) to 2HB r (g),DeltaH=-72...

`H_(2)(g)+Br_(2)(l) to 2HB r (g),DeltaH=-72.8kJ("1 atm ",25^(@)C)` if molar entropies of `H_(2)(g),Br_(2)(l),HB r(g)` are 130.5, 152. 3 & 198.3`J*K^(-1)*mol^(-1)` respectively then at which temperature the reaction will be spontaneous?

Text Solution

Verified by Experts

Change in entropy for the given reaction,
`DeltaS=2S_(HB r(g))-[S_(H_(2)(g))+S_(Br_(2)(l))]`
`=[2xx198.3-(130.5+152.3)]J*K^(-1)=113.8J*K^(-1)`
As `DeltaH lt 0` and `DeltaS gt0` for the given reaction so the reaction will be spontaneously at any temperature.
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