Question

For a. reaction $A + B \longrightarrow C + D , K _{ eq }$ represents equilibrium constant, The graph of $\ln K_{\text {eq }}$ versus $\frac{1}{T}$ is a straight line with a slope equal to $300$ . Also at $T =300 \,K$, the reaction is at equilibrium with
$[ A ]=[ B ]=[ C ]=1\, m$ and $[ D ] 10\, M .$ Calculate the value of $\left(1000 \times \Delta S ^{\circ}\right)$ in cal/Kelvin.
$[ R =2 \,cal / mol \,K ]$

Answer 1

$\ln K _{ eq }=\frac{-\Delta H ^{\circ}}{ RT }+\frac{\Delta S ^{\circ}}{ R }$
$\therefore $ Slope of $\ln K _{\text {eq }}$ vs $\frac{1}{ T }=\frac{-\Delta H ^{\circ}}{ R }$
$\Rightarrow \Delta H ^{\circ}=-300 \,R \Rightarrow-600\, cal$
at $300\, K\, K _{ eq }=10$
$\therefore \Delta G ^{\circ}=- RT \ln 10 $
$\Rightarrow-2.303 \times 300 \times 2$
$\therefore \Delta S ^{\circ}=\frac{\Delta H ^{\circ}-\Delta G ^{\circ}}{ T }=\frac{-600+600 \times 2.3 C 3}{300}$
$\Rightarrow 2.606 \,cal / K$
$\therefore 1000 \Delta S ^{\circ}=2606 \,cal / K$