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Le Chatelier's Principle for Del N Equals to 0

For a hypothe...

Question

For a hypothetical reaction of the kind
$A B_{2} (g) + \frac{1}{2} B_{2} (g) ⇌ A B_{3} (g); Δ H = - x k J$ More $A B_{3}$ could be produced at equilibrium by :

A

using a catalyst

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B

removing some of

B_{2}

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C

increasing the temperature

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D

increasing the pressure

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Solution

The correct option is D increasing the pressure
Since the number of moles decreases. So reaction is favoured forward by increasing pressure,
$Δ n = 1 - (1 + \frac{1}{2}) = - \frac{1}{2}$

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Q. For a hypothetical reaction of the kind

A B_{2} (g) + \frac{1}{2} B_{2} (g) ⇌ A B_{3} (g); Δ H = - x k J

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Q. For hypothetical reversible reaction,

1 / 2 A_{2} (g) + 3 / 2 B_{2} (g) ⟶ A B_{3} (g); Δ H = - 20 K J

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1 / 2 A_{2} (g) + 3 / 2 B_{2} (g) ⟶ A B_{3} (g); Δ H = - 20 k J

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Q. For hypothetical reversible reaction:

\frac{1}{2} A_{2} (g) + \frac{3}{2} B_{2} (g) ⟶ A B_{3} (g); Δ H = - 20 k J

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