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  • The variation of the equilibrium constant with temperature is given below <img alt="\begin{array}{|l|l|} \hline \text { Temp } & \text { eq. constant } \\ \hline T_{1}=25^{\circ} \mathrm{C} & K_{1}=10 \\ \hline T_{2}=100^{\circ} \mathrm{C} &amp

The variation of the equilibrium constant with temperature is given below \begin{array}{|l|l|} \hline \text { Temp } & \text { eq. constant } \\ \hline T_{1}=25^{\circ} \mathrm{C} & K_{1}=10 \\ \hline T_{2}=100^{\circ} \mathrm{C} & K_{2}=100 \\ \hline \end{array} The value of \Delta H_{o},\Delta G_{o} at T1 and \Delta G_{o} at T2 in kJ/mol respectively are close to:  
Option: 1 28.4, -7.14, -5.71
 
Option: 2 0.64, -7.14, -5.71
 
Option: 3 0.64, -5.71, -14.09  
 
Option: 4 28.4, -5.71. -14.29
 
 

Answers (1)

best_answer

\\\mathrm{(\Delta G^{0})_1 = -R T_{1} \ln k} \\\\ =-8.314 \times 298 \times \ln (10) \\\\ =-5.7 \mathrm{kJ}/\mathrm{mol}

\\\mathrm{\left(\Delta G^{o}\right)_{2} =-R T_{2} \ln K} \\\\ =(-8 \cdot 314) \times 373 \times \ln (100) \\\\ =-14.28 \mathrm{kJ} / \mathrm{mol}

\text { Now, from van't Hoff's Equation - }

\mathrm{\ln \left(\frac{k_{2}}{k_{1}}\right)=\frac{\Delta H^{0}}{R}\left[\frac{1}{T_{1}}-\frac{1}{T_{2}}\right]}

\mathrm{\Rightarrow \ln 10=\frac{\Delta H^{0}}{R}\left[\frac{1}{298}-\frac{1}{373}\right]}

\mathrm{\Rightarrow \quad \Delta H^{0}=\frac{(\ln 10) \times 298 \times 373 \times R}{75 \times 1000}}

=28.4 \mathrm{kJ} / \mathrm{mol}

Therefore, option (4) is correct.

Posted by

Kuldeep Maurya

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