Calculate a) `DeltaG^(ɵ)` and b) the equilibrium constant for the formation of `NO_(2)` from `NO` and `NO_(2)` at `298 K`
`NO(g) + 1/2 O_(2) hArr NO_(2)(g)`
Where
`Delta_(f)G^(ɵ) (NO_(2)) = 52.0 kJ//"mol"`
`Delta_(f)G^(ɵ) (NO) = 87.0 kJ//"mol"`
`Delta_(f)G^(ɵ) (O_(2)) = 0 kJ//"mol"`

Explanation

(a) ` " " Delta_r G^(@) = sum Delta_f G^(@) ` ( product ) ` - sum Delta _f G^(@) ` (reactions)
or ` " " Delta _r G^(@) = =Delta _f G^(@) [NO_2] = {Delta_fG^(@) [NO]+(1)/(2) Delta _fG^(@) [O_2]} `
or ` " " Delta _r G^(@) = 52.0 - (87.0 + 0 ) = - 35 .0 " kJ mol"^(-1) `
(b)` " " Delta _r G^(@) = -2 .303 RT log_( 10 ) K`
or ` " " log_(10 ) K= - (Delta _rG^(@))/(2.303RT) `
` " " = ( (-350 .0))/(2.303xx 8.314 xx 10^(-3) xx 298)= 6.1341`
` " " ( because R = 8.314 xx 10 ^(-3) " kJ " K^(-1) " mol"^(-1) ) `
or `K = " antilog"_(10) 6.1341 = 1.362 xx 10 ^(-6)`