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Le Chatelier's Principle for Del N Equals to 0

Backward reac...

Question

Backward reaction is favoured by increase in the pressure of which of the following equilibrium?

2 S O_{2} + O_{2} ⇋ 2 S O_{3}

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N_{2} + O_{2} ⇋ 2 N O

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N_{2} + 3 H_{2} ⇋ 2 N H_{3}

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P C l_{5} ⇋ P C l_{3} + C l_{2}

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Solution

The correct option is D $P C l_{5} ⇋ P C l_{3} + C l_{2}$
The increase of pressure will shift the equilibrium position in the direction of a decrease in the number of moles.
For the reaction $P C l_{5} ⇋ P C l_{3} + C l_{2}$ , the forward reaction occurs with an increase in the number of moles from 1 to 2. Also, the reverse reaction occurs with a decrease in the number of moles from 2 to 1.
When the pressure of the system is increased, the backward reaction will be favoured as the reverse reaction occurs with a decrease in the number of moles.

Option D is correct.

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Similar questions

P {C l}_{5} \to P {C l}_{3} + {C l}_{2}

2 {S O}_{2} + O_{2} \to 2 {S O}_{3}

N_{2} + 3 H_{2} \to 2 {N H}_{3}

2 K C l O_{3} \to 2 K C l + 3 O_{2}

In which of the following reaction, the value of $K_{p}$ will be equal to $K_{c}$

N_{2} + 3 H_{2} ⇌ 2 {N H}_{3}

K_{1}

N_{2} + O_{2} ⇌ 2 N O

K_{2}

H_{2} + \frac{1}{2} O_{2} ⇌ H_{2} O

K_{3}

2 {N H}_{3} + \frac{5}{2} O_{2} ⇌ 2 N O + 3 H_{2} O

K_{1}, K_{2}

K_{3}

N_{2} + 3 H_{2} \to 2 N H_{3} - K_{1}

N_{2} + O_{2} \to 2 N O - K_{2}

H_{2} + 1 / 2 O_{2} \to H_{2} O - K_{3}

3 H_{2} + 5 / 2 O_{2} \to 2 N O + 3 H_{2} O

K_{1}, K_{2}, K_{3}

N_{2} + 3 H_{2} ⇌ 2 N H_{3}; K_{1}

N_{2} + O_{2} ⇌ 2 N O; K_{2}

H_{2} + \frac{1}{2} O_{2} \to H_{2} O; K_{3}

(K)

2 N H_{3} + \frac{5}{2} O_{2} K ⇌ 2 N O + 3 H_{2} O

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