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Calculate the free energy change at 298 ...

Calculate the free energy change at 298 K for the reaction , `Br_(2)(l)+Cl_(2)(g)to2BrCl(g)`. For the reaction `DeltaH^(@)=29.3kJ` & the entropies of `Br_(2)(l),Cl_(2)(g)&BrCl(g)` at the 298 K are 152.3, 223.0, 239.7 J`mol^(-1)K^(-1)` respectively.

Text Solution

Verified by Experts

For the reaction,
`DeltaS = S("products") - S("reactants")`
` = 2 xx S(BrCI) - [S(Br_(2)) + S(CI_(2))]`
` 2 xx 239.7 - [152.3 + 223.0]`
= `479.4 - 375.3 `
`= 104.1 J`
Now, `DeltaG = DeltaH - T DeltaS`
`DeltaH = 29.3 kJ = 29300 J, T = 298 K`
`therefore DeltaG = 29300 - (298 xx 104.1)`
= 29300 - 31021.8
` = -1721.8 J`
Since `DeltaG` has negative value, the reaction is feasible at 298 K.
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