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For the reaction
$2 H_{2} (g) + O_{2} (g) ⟶ 2 H_{2} O (l)$ . The standard entropies of $H_{2} (g), O_{2} (g)$ and $H_{2} O (l)$ at $27$ $° C$ are $30 {calK}^{- 1} {mol}^{- 1}$ , $50 {calK}^{- 1} {mol}^{- 1}$ and $15 {calK}^{- 1} {mol}^{- 1}$ respectively.
Assuming $H_{2}$ and $O_{2}$ to behave as ideal gas (without vibrational degree of freedom) and $C_{P_{m} (H_{2} O (n))} = 15.5 {calK}^{- 1} {mol}^{- 1}$ . Determine magnitude of standard ontropy (in ${calK}^{- 1}$ ) change for the reaction at $177 ° C$ .
[Given : $R = 2 {calK}^{- 1} {mol}^{- 1}, \ln 1.5 = 0.4$ ]

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answer is 76.

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Detailed Solution

$Δ S_{300 K} = 2 \times 15 (2 \times 30 + 50)$

$= 30 - 110 = - 80 {calK}^{- 1}$

$C_{P_{m} O_{2} (g)} = \frac{R γ}{(γ - 1)} = \frac{2 \times 7}{5 (\frac{7}{5} - 1)} = 7 {calK}^{- 1} {mol}^{- 1}$

$Δ C_{P_{m}} = 31 - (2 \times 7 + 7) = 10 {calK}^{- 1} {mol}^{- 1}$

$Δ S_{450 K} = - 80 + 10 \ln \frac{450}{300} = - 80 + 4 = - 76$

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