Question

Given that:
$S_{H_2}^{\circ }=131J{K}^{-1}mo{l}^{-1}{S}_{C{l}_2}^{\circ }=223J{K}^{-1}mo{l}^{-1}and{S}_{HCl}^{\circ }=187J{K}^{-1}mo{l}^{-1}$
The standard entropy change in formation of $1$ mole of $HCl\left(g\right)$ from $H_2\left(g\right)$ and $C{l}_2\left(g\right)$ will be:
(Given, reaction for formation of $HCl$ : $H_2+C{l}_2\to 2HCl$)

• A. $20J{K}^{-1}$
• B. $10J{K}^{-1}$
• C. $187J{K}^{-1}$
• D. $374J{K}^{-1}$

Answer 1

The correct option is B $10J{K}^{-1}$

Balanced chemical equation for formation of one mole of $HCl$ is :
$\frac{1}{2}{H}_2\left(g\right)+\frac{1}{2}C{l}_2\left(g\right)\to HCl\left(g\right)\Delta S^{\circ }=\Sigma S^{\circ }\text{(Product)}-\Sigma S^{\circ }\text{(reactants)}=S_{HCl}^{\circ }-[\frac{1}{2}{S}_{H_2}^{\circ }+\frac{1}{2}{S}_{C{l}_2}^{\circ }]=187-\frac{1}{2}[131+223]=10J{K}^{-1}$