PCl5g- PCl3g - Cl2gFor this reaction relate - H and - U-

The correct option is B Δ H > Δ U As we know, Δ H = Δ U+ Δ n_g RT here, nbsp; PCl_5(g)→ PCl_3(g) + Cl_2(g) Δ n_g = 1 so, Δ H ...

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Introduction to CFT

PCl5g→ PCl3g ...

Question

$P C l_{5} (g) \to P C l_{3} (g) + C l_{2} (g)$
For this reaction relate $Δ H$ and $Δ U$ .

Δ H

Δ U

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Δ H

Δ U

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Δ H

Δ U

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None of these

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Δ H > Δ U

Δ H < Δ U

C (g r a p h i t e) + O_{2} (g) ⟶ C O_{2} (g)

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

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

2 H I (g) ⟶ H_{2} (g) + I_{2} (g)

Δ H

Δ U

H_{2} (g) + O_{2} (g) ⟶ H_{2} O_{2} (g)

Δ H

Δ U

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