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 $kJ mol ^{-1}$ ) respectively, are close to
[Use $\left. R =8.314 JK ^{-1} mol ^{-1}\right]$

• 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)

$T _{1}=298 K ; K _{1}=10$

$T _{2}=373 K ; K _{2}=100$

For $\Delta H$ calculation

$\log \left(\frac{ K _{2}}{ K _{1}}\right)=\frac{\Delta H }{2.303 R }\left(\frac{1}{ T _{1}}-\frac{1}{ T _{2}}\right)$

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

$\Delta H =28.4 kJ / mole$

for $\left(\Delta G_{1}^{0}\right)$ and $\left(\Delta G_{2}^{0}\right)$ calculation.

$\Delta G _{1}^{0} =-2.303 RT \log K $

$ \Delta G _{1}^{0} =-2.303 \times \frac{8.314}{1000} \times 298 \times \log 10 $

$=-5.71 kJ / mole $

$\Delta G _{2}^{0} =-2.303 \times \frac{8.314}{1000} \times 373 \times \log 100 $

$=-14.29 kJ / mole $