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Law of Mass Action

The following...

Question

The following reactions carried in open vessel. The reaction for which $Δ H = Δ U$ will be:

P C l_{5} (g) \to P C l_{3} (g) + C l_{2} (g)

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2 C O (g) + O_{2} (g) \to 2 C O_{2} (g)

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N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)

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H_{2} (g) + I_{2} (g) \to 2 H I (g)

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Solution

The correct option is B $P C l_{5} (g) \to P C l_{3} (g) + C l_{2} (g)$
Open vessel reaction
$Δ H = Δ U$
when $Δ n = 0$
(a) ${P C l}_{5} (g) ⇌ {P C l}_{3} (g) + {C l}_{2} (g) Δ n = 1 + 1 - 1 = 1 \to (\times)$
(b) $2 C O (g) + O_{2} (g) ⇌ 2 {C O}_{2} (g) Δ n = 2 - 2 - 1 = - 1 \to (\times)$
(c) $N_{2} (g) + 3 H_{2} (g) ⇌ 2 {N H}_{3} (g) Δ n = 2 - 3 - 1 = - 2 \to (\times)$
(d) $H_{2} (g) + I_{2} (g) ⇌ 2 H I (g) Δ n = 2 - 1 - 1 = 0$
So option ( $D$ ) is correct.

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Similar questions

2 H I_{(g)} ⟶ H_{2 (g)} + I_{2 (g)}

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

2 N O_{(g)} ⟶ N_{2 (g)} + O_{2 (g)}

P C l_{5 (g)} ⟶ P C l_{3 (g)} + C l_{2 (g)}

P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)

2 H I (g) ⇌ H_{2} (g) + I_{2} (g)

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

N_{2} (g) + O_{2} (g) \to 2 N O (g)

Δ H = 181

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H_{2} (g) + I_{2} (g) \to 2 H I (g); Δ H = - 94

H_{2} (g) + F_{2} (g) \to 2 H F (g); Δ H = - 541

I . 2 H I_{(g)} ⇌ H_{2 (g)} + I_{2 (g)}

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I I I . 2 N O_{(g)} ⇌ N_{2 (g)} + O_{2 (g)}

I V . P C l_{5 (g)} ⇌ P C l_{3 (g)} + C l_{2 (g)}

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2 H I (g) ⇌ H_{2} (g) + I_{2} (g)

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

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