Question

The entropy values (in $ J{{K}^{-1}}mo{{l}^{-1}} $ ) of $ {{H}_{2}}(g)=130.6,\text{ }C{{l}_{2}}(g)=223.0 $ and $ HCl(g)= $ $186.7$ at $298\, K$ and 1 atm pressure. Then entropy change for the reaction. $ {{H}_{2}}(g)+C{{l}_{2}}(g)\xrightarrow[{}]{{}}2HCl(g) $ is:

• A. 540.3
• B. 727.3
• C. $ -166.9 $
• D. 19.8

Answer 1

Answer: (D)

$ \Delta S={{S}_{p}}-{{S}_{R}} $
$ {{H}_{2}}(g)+C{{l}_{2}}(g)\xrightarrow[{}]{{}}2HCl(g) $
Given $ {{S}^{o}}H(g)=130.6J{{K}^{-1}}/mol $
$ {{S}^{o}}C{{l}_{2}}(g)=223.0\text{ }J{{K}^{-1}}/mol $
$ {{S}^{o}}HCl(g)=186.7\text{ }J{{K}^{-1}}/mol $
$ \Delta S=2{{S}^{o}}HCl-({{S}^{o}}{{H}_{2}}+{{S}^{o}}C{{l}_{2}}) $
$ =2\times 186.7-(130.6+223.0) $
$ =373.4-353.6=19.8\text{ }J{{K}^{-1}}mo{{l}^{-1}} $