For a certain thermochemical reaction M → N at T = 400K, Δ H ^ o =...

Given,

 $T = 400K, {\Delta H}^{o} = 77 . 2kJ{mol}^{- 1}, {\Delta S}^{o} = 122{JK}^{- 1}$

${\Delta G}^{o} = {\Delta H}^{o} - {T \Delta S}^{o}$

$= 77 . 2 \times {10}^{3} - 400 \times 122 = 28400J$

$\Delta G ° = - 2 . 303RTlogK$

Where K is the equilibrium constant.

By putting the values in above equation,

$⇒ 28400 = - 2 . 303 \times 8 . 314 \times 400 \times logK$

$⇒ logK = - 3 . 708 = - 37 . 08 \times {10}^{- 1}$