For the given reaction:
`H_(2)(g)+Cl_(2)(g) to2H^(+)(aq)+2Cl^(-)(aq)`
`DeltaG^(@)=-262.4kJ`
The value of Gibbs free energy of formation `(DeltaG_(r)^(@))` for the ion `Cl^(-)(aq)` is:

Options

A. `- 131.2 kJ mol^(-1 )`
B. ` + 131.2 kJ mol^(-1 )`
C. `- 262.4 kJ mol^(-1)`
D. `+ 262.4 kJ mol^(-1)`

Answer

A

Explanation

`Delta_(r)G^(@)= Delta_(f)G^(@) ("products")-Delta_(f)G^(@) ("Reactants")`
`=[2Delta_(f)G^(@)(H^(+))+2Delta_(f)G^(@)(Cl^(-))]`
` -[Delta_(f)G^(@)(H_(2))+Delta_(f)G^(@)(Cl_(2))]`
`=[0+2Delta_(f)G^(@)(Cl_(-))]-[0+0]`
`or -262.4=2Delta_(f)G^(@))(Cl^(-1))`
or `Delta_fG^@(Cl^(-))= -131-.2k J mol^(-1)`