Question

The equilibrium constant for the reaction given below is $2.0\times {10}^{-7}$ at 300 K. Calculate the standard entropy change if $△H^{\circ }=28.40kJmo{l}^{-1}$ for the reaction:
$PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$

• A. $-33.6Jmo{l}^{-1}{K}^{-1}$
• B. $33.6Jmo{l}^{-1}{K}^{-1}$
• C. $-43.6Jmo{l}^{-1}{K}^{-1}$
• D. $43.6Jmo{l}^{-1}{K}^{-1}$

Answer 1

The correct option is A $-33.6Jmo{l}^{-1}{K}^{-1}$

$K=2.0\times {10}^{-7}T=300K△H^{\circ }=28.4kJmo{l}^{-1}△G^{\circ }=-2.303RT\log k=-2.303\times 8.314\times {10}^{-3}\times 300\times \log (2\times {10}^{-7})=-5.744(\log 2-7)=-5.744(0.3010-7)=38.48kJ$
From $△G^{\circ }=△H^{\circ }-T△S^{\circ }\Rightarrow T△H^{\circ }-△G^{\circ }△S^{\circ }=\frac{△H^{\circ }-△G^{\circ }}{T}=\frac{28.4-38.48}{300}=0.0336kJmo{l}^{-1}$
or $△S=-33.6Jmo{l}^{-1}{K}^{-1}$