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First Law of Thermodynamics

The standard ...

Question

The standard entropies of $N_{2} (g)$ , $H_{2} (g)$ and $N H_{3} (g)$ are $191.5$ , $130.5$ , $192.6$ $J K^{- 1}$ $m o l^{- 1}$ . The value of $Δ S^{o}$ of formation of ammonia is:

- 98.9

J . K^{- 1}

m o l^{- 1}

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0 J . K^{- 1} m o l^{- 1}

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+ 129.4

J . K^{- 1}

m o l^{- 1}

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- 29.4

J . K^{- 1}

m o l^{- 1}

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Solution

The correct option is A $- 98.9$ $J . K^{- 1}$ $m o l^{- 1}$
According to the problem :
$\frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) \to N H_{3} (g)$
$Δ S_{3} Δ S_{2} Δ S_{1}$

Therefore
$Δ S_{3} = \frac{1}{2} \times 191.5$
$= 95.755 k^{- 1} m o l^{- 1}$
For $Δ S_{2} = \frac{3}{2} \times 130.5$
$= 195.75 5 K^{- 1} m o l^{- 1}$
Hence,
$Δ S^{0} = Δ S_{1} - Δ S_{2} - Δ S_{3}$
$= 192.6 - 195.75 - 95.75$
$= - 98.9 J K^{- 1} m o l^{- 1}$

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Similar questions

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The standard entropies of N2(g), H2(g) and NH3(g) are 191.5, 130.5, 192.6 JK−1 mol−1. The value of ΔSo of formation of ammonia is:

The correct option is A −98.9 J.K−1 mol−1According to the problem : 12N2(g)+32H2(g)→NH3(g)ΔS3ΔS2 ΔS1Therefore ΔS3=12×191.5 =95.755k−1mol−1For ΔS2=32×130.5 =195.755K−1mol−1Hence, ΔS0=ΔS1−ΔS2−ΔS3 =192.6−195.75−95.75 =−98.9JK−1mol−1

The standard entropies of $N_{2} (g)$ , $H_{2} (g)$ and $N H_{3} (g)$ are $191.5$ , $130.5$ , $192.6$ $J K^{- 1}$ $m o l^{- 1}$ . The value of $Δ S^{o}$ of formation of ammonia is: