Question

The equilibrium constant for the reaction given below is $2.0\times {10}^{-7}at300K$.
$PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$. The value of $|S^{\circ }|$ is $-xJmo{l}^{-1}{K}^{-1}$. Calculate the value of $x$ if $\Delta H^{\circ }=28.4kJmo{l}^{-1}$ (Give your answer upto one deimal only)

Answer 1

$K_c=2.0\times {10}^{-7}$
$T=300K$
$\Delta H^{\circ }=28.4kJmo{l}^{-1}$
$\Delta G^{\circ }=-2.303RTlog{K}_c$
$=-2.303\times 8.314\times {10}^{-3}\times 300\times log(2\times {10}^7$
$=-5.744(log2-7)$
$=-5.744(0.3010-7)$
$=38.48kJ$
From $\Delta G^{\circ }=\Delta H^{\circ }-T\Delta S^{\circ }$
$T\Delta S^{\circ }=\Delta H^{\circ }-\Delta G^{\circ }$
$\Delta S^{\circ }=\frac{\Delta H^{\circ }-T\Delta S^{\circ }}{T}$
$=\frac{28.4-38.48}{300}$
$=-0.0336kJmo{l}^{-1}{K}^{-1}$
$\Delta S^o=-33.6Jmo{l}^{-1}{K}^{-1}$