For the reaction $P C l_{5} (g) \to P C l_{3} (g) + C l_{2} (g)$

△ H = △ E

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△ H > △ E

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△ H < △ E

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None of these

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Solution

The correct option is B $△ H > △ E$
$Δ H = Δ E + P δ V = Δ E + n_{g} R T$
$P C l_{5} (g) \to P C l_{3} (g) + C l_{2} (g)$
$Δ n_{g} = n_{(p) g} - n_{(r) g} = 2 - 1 = 1$
$∴ Δ H = Δ E + 1 \times R T$
$∴ Δ H > Δ E$

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The relationship between $Δ$ H and $Δ$ E for the reaction $P C l_{3} (g) + C l_{2} (g) \to P C l_{5} (g)$ is given as;

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

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

P C l_{5} (g) \Leftrightarrow P C l_{3} (g) + C l_{2} (g), K_{c}

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

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