Use the following data to calculate - S0 for the reaction- 2NH3g- N2g - 3H2g Compound SoJ-mol-K H2g N2g NH3g-98-74-114-99-304-99

The correct option is D 198.8 J/K 2NH_3 ⟶ N_2+3H_2 ⇒Δ S^o= n Δ S_product Δ S_reactant ⇒Δ S^o=3 ×Δ S_H_2+1 ×Δ S_N_2 2 ×Δ S_NH_3 ⇒Δ S^o= 3 × ...

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Entropy

Use the follo...

Question

Use the following data to calculate $Δ S^{0}$ for the reaction:
$2 N H_{3} (g) ⟶ N_{2} (g) + 3 H_{2} (g)$
Compound $S^{o}$ (J/mol-K)
$H_{2} (g)$
$N_{2} (g)$
$N H_{3} (g)$ -98.74
-114.99
-304.99

-198.8 J/K

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-91.26 J/K

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-106 J/K

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198.8 J/K

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Solution

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Similar questions

X \to Y + Z

X, Y

Z

120 J K^{- 1} m o l^{- 1}, 213.8 J K^{- 1} m o l^{- 1} a n d 197.9 J K^{- 1} m o l^{- 1}

Δ_{r} H

C_{p . m} (C H_{4}, g) = 38 J / K m o l

C_{p . m} (C O, g) = 31 J / K m o l

C_{p . m} (C H_{3} C H O . g) = 52 J K / m o l

Δ H

358 K

F e_{2} O_{3} (s) + 3 H_{2} (g) \to 2 F e (s) + 3 H_{2} O (l)

Δ H_{298} = - 33.29 k J m o l^{- 1}

C_{P} f o r F e_{2} O_{3} (s), F e (s), H_{2} O (l)

H_{2} (g)

103.8, 25.1, 75.3, a n d 28.8 J / K m o l

H_{2} (g) + C l_{2} (g) ⟶ 2 H C l (g)

298 K

S_{H_{2}}^{⊖} = 131 J K^{- 1} m o l^{- 1}

S_{C l_{2}}^{⊖} = 223 J K^{- 1} m o l^{- 1}

S_{H C l}^{⊖} = 187 J K^{- 1} m o l^{- 1}

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

k

2 N H_{3} (g) ⇌ N_{2} (g) + 3 H_{2} (g)

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Compound	$S^{o}$ (J/mol-K)
$H_{2} (g)$ $N_{2} (g)$ $N H_{3} (g)$	-98.74 -114.99 -304.99