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Enthalpy

The reactions...

Question

The reactions in which the yield of the products cannot be increased by the application of high pressure:

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)

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N H_{4} H S (s) ⇌ N H_{3} (g) + H_{2} S (g)

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N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

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

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Solution

The correct options are
B $N H_{4} H S (s) ⇌ N H_{3} (g) + H_{2} S (g)$
C $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$
In the reactions in which the yield of the products cannot be
increased by the application of high pressure, the number of moles of gaseous products has to be greater than no. of moles of gaseous reactants. In other words, $Δ n > 0$

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Similar questions

2 S O_{2_{(} g)} + O_{2_{(} g)} ⇋ 2 S O_{3_{(} g)} + Q k J

\frac{K_{p}}{K_{c}}

300 K

N_{2} (g) + O_{2} (g) ⇌ 2 N O (g)

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

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

[A t 300 K, R T = 24.62 d m^{3} a t m m o l^{- 1}]

C (s) + O_{2} (g) ⟶ C O_{2} (g)

Δ H = Δ U + R T

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

Δ H = Δ U

N H_{4} H S (s) ⟶ N H_{3} (g) + H_{2} S (g)

Δ H = Δ U - 2 R T

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

Δ H = Δ U + 2 R T

2 S O_{2} (g) + O_{2} (g) ⟶ 2 S O_{3} (g)

Δ H = Δ U - R T

C (s) + O_{2} (g) ⟶ C O_{2} (g)

Δ H = Δ U + R T

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

Δ H = Δ U

N H_{4} H S (s) ⟶ N H_{3} (g) + H_{2} S (g)

Δ H = Δ U - 2 R T

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

Δ H = Δ U + 2 R T

2 S O_{2} (g) + O_{2} (g) ⟶ 2 S O_{3} (g)

Δ H = Δ U - R T

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