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Law of Mass Action

The change of...

Question

The change of pressure does not affect which of the following equilibrium?

A

C O + H_{2} ⇌ C + H_{2} O

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B

N_{2} + O_{2} ⇌ 2 N O

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C

2 N O_{2} ⇌ 2 N O + O_{2}

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D

2 S O_{2} + O_{2} ⇌ 2 S O_{3}

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E

2 C O + O_{2} ⇌ 2 C O_{2}

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Solution

The correct option is B $N_{2} + O_{2} ⇌ 2 N O$
when pressure changes according to le-chatlier
principle: -
1) If $Δ n g > 0$ , the reaction proceeds backward or forwards according to pressure increased
or decreased and vice-versa for $△ n g < 0$
but if $Δ$ ng=o there is no effect on the equilibrium.
So, option B) $N_{2} + O_{2} \Rightarrow 2 N O (Δ n g = 0)$ is correct.

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Similar questions

Q. Equilibrium constant for the reactions.

2 N O + O_{2} ⇌ 2 N O_{2}

K_{C_{1}}

;

N O_{2} + S O_{2} ⇌ S O_{3} + N O

K_{C_{2}}

2 S O_{3} ⇌ 2 s O_{2} + O_{2}

K_{C_{3}}

then correct reaction is:

Q. What is the relation between the equilibrium constants

K

K_{1}

K_{2}

for the following reactions?

N_{2} + 2 O_{2} ⇋ 2 N O_{2}, K

N_{2} + O_{2} ⇋ 2 N O, K_{1}

2 N O + O_{2} ⇋ 2 N O_{2}, K_{2}

Q. The following equilibrium are given:

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}

The equilibrium constant of the reaction

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

K_{1}, K_{2}

K_{3}

Q. Statement - I : The reactions

2 N O + O_{2} \to 2 N O_{2}

2 C O + O_{2} \to 2 C O_{2}

proceed at the same rate because they are similar.
Statement - II Above two reactions will have different activation energies.

For which of the following process, the value of $Δ$ H negative?

I) $N_{2} + O_{2} \to 2 N O$

N_{2} + 3 H_{2} F e + M o - --- \to 2 N H_{3}

2 S O_{2} + O_{2} V_{2} O_{5} - -- \to 2 S O_{3}

H_{2} + I_{2} p t - \to 2 H I

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