Question

The entropy values in $J{K}^{-1}{mol}^{-1}$ of $H_2\left(g\right)=130.6,{Cl}_2\left(g\right)=233$ and $HCl\left(g\right)=186.7$ at $298K$ and $1$ atm pressure. Then entropy change for the reaction

$H_2\left(g\right)+{Cl}_2\left(g\right)\to 2HCl\left(g\right)$ is:

• A. $+540.3$
• B. $+727.3$
• C. $-166.9$
• D. $+9.8$

Answer 1

The correct option is D $+9.8$

Entropy change for reaction = Entropy of product - Entropy of reactant

Entropy of product $=2\times$ entropy of HCl $=2\times 186.7J{K}^{-1}mo{l}^{-1}$

$=373.4J{K}^{-1}mo{l}^{-1}$

Entropy of reactant = entropy of H₂ + entropy of Cl₂ = (130.6 + 233) $J{K}^{-1}mo{l}^{-1}$

$=363.6J{K}^{-1}mo{l}^{-1}$

So, entropy change for reaction = (373.4 - 363.6) $J{K}^{-1}mo{l}^{-1}$

= 9.8 $J{K}^{-1}mo{l}^{-1}$
Option D is correct.