Question

Combustion of Hydrogen in a fuel cell at $300 \, K$ is represented as $2 H_{2} \left(g\right) + O_{2} \left(g\right) \rightarrow 2 H_{2} O \left(g\right).$ If $\Delta H$ and $\Delta G$ are $-241.60 \, kJ \, mol^{- 1}$ and $-228.40 \, kJ \, mol^{- 1}$ respectively for $H_{2}O$ then the value of $\Delta S$ for the above process is:

• A. $44\,J \, K^{- 1}$
• B. $-88\,J \, K^{- 1}$
• C. $+88\,J \, K^{- 1}$
• D. $-44\,J \, K^{- 1}$

Answer 1

Answer: (B)

$\Delta S=\frac{\Delta H \, - \, \Delta G}{T}=\frac{- 241.6 - \left(- 228.4\right)}{300}$
For $1$ mole of $H_{2}O=-44J \, K^{- 1}$
Hence, for $2$ moles $=-88J \, K^{- 1}$