Question

$S_{H_{2\left(g\right)}}^O=130.6$J $K^{-1}$ $mol6-1$; $S_{H_2{O}_{\left(l\right)}}^O=69.9$J $K^{-1}mo{l}^{-1}$ $S_{O_{2\left(g\right)}}^O=205$ J$K^{-1}$$mo{l}^{-1}$, then the absolute entropy change of $H_{2_{\left(g\right)}}+\frac{1}{2}{O}_{2\left(g\right)}\to H_2{O}_{\left(l\right)}$ is?

• A. $-163.2$J $mo{l}^{-1}{K}^{-1}$
• B. $+163.2$J $mo{l}^{-1}{K}^{-1}$
• C. $-3-3$J $mo{l}^{-1}{K}^{-1}$
• D. $+303$J $mo{l}^{-1}{K}^{-1}$

Answer 1

Answer: (A)

$-163.2$J $mo{l}^{-1}{K}^{-1}$

$H_2$ + $\frac{1}{2}{O}_2$ -------> $H_{2}O$

Entropy Change of reaction = Entropy of products - Entropy of reactants

$\Delta S^o=\Delta S_{H_{2}O}^o-\frac{1}{2}\times \Delta S_{O_2}^o-\Delta S_{H_2}^o$

$\Delta S^o=69.9J{K}^{-1}mo{l}^{-1}-\frac{1}{2}\times 205J{K}^{-1}mo{l}^{-1}-130.6J{K}^{-1}mo{l}^{-1}$

$\Delta S^o=-163.2Jmo{l}^{-1}{K}^{-1}$