Calculate free energy change for the reaction,
`H_(2)(g)+Cl_(2)(g)to2H-Cl(g)`
By using the bond energy and entropy data.
Bond energies of `H-H,Cl-Cl and H-Cl ` bonds are 435kJ `mol^(-1),240jK" "mol^(-1) and 430jk" "mol^(-1)` respectively standard entropies of `H_(2),Cl_(2) and HCl` are 130.59,222.95 and 186.68 `JK^(-1)" "mol^(-1)` respectively.

Answer

190.9kJ

Explanation

`DeltaG^(@)` can be calculated by using:
`DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)`
`DeltaH^(@)=sum(BE)_("reactants")-sum(BE)_("products")`
`=435+240-2xx430=-185kJ`
`DeltaS^(@)=sumS_("products")^(@)-sumS_("reactants")^(@)`
`=2xx186.68-130.59-222.95`
`=19.82JK^(-1)=19.82xx10^(-3)kJ" "K^(-1)`
`DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)`
`=-185-298xx19.82xx10^(-3)=-190.9kJ`