Chemistry - Thermodynamics - II Question with Solution | TestHub

$T_1=323\mathrm{K} {\mathrm{T}}_2=373\mathrm{K}$
${\mathrm{k}}_1=10 {\mathrm{k}}_2=100$
$\log \left(\frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}\right)=\frac{\Delta \mathrm{H}}{2.303\mathrm{R}}\left[\frac{1}{{\mathrm{T}}_1}-\frac{1}{{\mathrm{T}}_2}\right]$

$\log \left(\frac{100}{10}\right)=\frac{\Delta \mathrm{H}}{2.303\times 8.314}\left[\frac{1}{298}-\frac{1}{373}\right]$

$\log 10=\frac{\Delta \mathrm{H}}{2.303\times 8.314}\left[\frac{75}{298\times 373}\right]$

$\Delta H=\frac{2.303\times 8.314\times 298\times 373}{75}=28.4\mathrm{KJ}$

At $T_1=25°\mathrm{C}=298\mathrm{K},{\mathrm{K}}_1=10$

$\Delta \mathrm{G}=-2.303{\mathrm{RT}}_1{\text{logk }}_1$

$=-2.303\times 8.314\times 298\times \log \left(10\right)$

$=-2.303\times 8.314\times 298\times 1$

$=-5.7\mathrm{KJ}$

At $T_2=100°\mathrm{C}=373\mathrm{K} {\mathrm{K}}_2=100$

$\Delta \mathrm{G}=-2.303{\mathrm{RT}}_2{\mathrm{logK}}_2$

$=-2.303\times 8.314\times 373\times \log (10{)}^2$

$=-2.303\times 2\times 8.314\times 373\times 1$

$=-14283.7\mathrm{J}$

$=-14.29\mathrm{KJ}$