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.

Calculate free energy change for the reaction, `H_(2)(g)+Cl_(2)(g)to2H

1.	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» Correct Answer - 190.9kJ `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`

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