Question

The variation of equilibrium constant with temperature is given below :

Temperature	Equilibrium constant
$T_1=25^{\circ }C$	$K_1=10$
$T_2=100^{\circ }C$	$K_2=100$

The values of $\Delta H^{\circ },\Delta G^{\circ }$ at $T_1$ and $\Delta G^{\circ }$ at $T_2$ (in $kJmo{l}^{-1}$ ) respectively, are close to
[Use $R=8.314J{K}^{-1}mo{l}^{-1}]$

• A. $0.64,-5.71$ and $-14.29$
• B. $28.4,-7.14$ and $-5.71$
• C. $28.4,-5.71$ and $-14.29$
• D. $0.64,-7.14$ and $-5.71$

Answer 1

Answer: (C)

$T_1=298K;K_1=10$

$T_2=373K;K_2=100$

For $\Delta H$ calculation

$\log \left(\frac{K_2}{K_1}\right)=\frac{\Delta H}{2.303R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$

$\log \left(\frac{100}{10}\right)=\frac{\Delta H}{2.303\times \left(\frac{8.314}{1000}\right)}\left(\frac{1}{298}-\frac{1}{373}\right)$

$\Delta H=28.4kJ/mole$

for $\left(\Delta G_1^0\right)$ and $\left(\Delta G_2^0\right)$ calculation.

$\Delta G_1^0=-2.303RT\log K$

$\Delta G_1^0=-2.303\times \frac{8.314}{1000}\times 298\times \log 10$

$=-5.71kJ/mole$

$\Delta G_2^0=-2.303\times \frac{8.314}{1000}\times 373\times \log 100$

$=-14.29kJ/mole$