The variation of equilibrium constant with temperature is given...

${T}_{1} = 323K{T}_{2} = 373K$
${k}_{1} = 10{k}_{2} = 100$
$\log \left(\dfrac{{k}_{2}}{{k}_{1}}\right) = \dfrac{\DeltaH}{2 . 303R}\left[\dfrac{1}{{T}_{1}} - \dfrac{1}{{T}_{2}}\right]$

$\log \left(\dfrac{100}{10}\right) = \dfrac{\DeltaH}{2 . 303 \times 8 . 314}\left[\dfrac{1}{298} - \dfrac{1}{373}\right]$

$\log 10 = \dfrac{\DeltaH}{2 . 303 \times 8 . 314}\left[\dfrac{75}{298 \times 373}\right]$

$\DeltaH = \dfrac{2 . 303 \times 8 . 314 \times 298 \times 373}{75} = 28 . 4KJ$

At ${T}_{1} = 25 ° C = 298K, {K}_{1} = 10$

$\DeltaG = - 2 . 303{RT}_{1}{\text{logk}}_{1}$

$= - 2 . 303 \times 8 . 314 \times 298 \times \log ( 10 )$

$= - 2 . 303 \times 8 . 314 \times 298 \times 1$

$= - 5 . 7KJ$

At ${T}_{2} = 100 ° C = 373K{K}_{2} = 100$

$\DeltaG = - 2 . 303{RT}_{2}{logK}_{2}$

$= - 2 . 303 \times 8 . 314 \times 373 \times \log ( 10{)}^{2}$

$= - 2 . 303 \times 2 \times 8 . 314 \times 373 \times 1$

$= - 14283 . 7J$

$= - 14 . 29KJ$