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Vant Hoff's Equation and Effect of Temperature on Equilibrium Constant

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Question

What happens when pressure is increased in the following equilibrium rection
(i) N₂O₄(g)=2NO₂(g)
(ii) H₂(g)+I₂(g)=2HI(g)

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Solution

According to lechatlier principle the pressure in a system increases, the equilibrium will shift to favor the side of the reaction that involves fewer moles of gas.
Thus,
Part -I
N₂O₄(g)â†’2NO₂(g)
Reaction move in backward direction.
Part -II
H₂(g)+I₂(g)â†’2HI(g)
No effect due to same no. of moles in reactant and product.

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Similar questions

Q. For the reaction

I_{2} (g) + H_{2} (g) ⇌ 2 H I (g)

,
what is the effect on the equilibrium position when the pressure of the reaction vessel is increased?

As pressure is increased, what will happen to the following reactions?

Match the column:

Column - I	Column - II
(P) $P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$	(1) Shift to forward
(Q) $2 N O_{2} (g) ⇌ N_{2} O_{4} (g)$	(2) Shift to backward
(R) $H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)$	(3) No effect
(S) $a A (g) + b B (g) ⇌ c C (g)$	(4) Cannot predict

Q. An increase in pressure in the reaction

2 H I (g) ⇌ H_{2} (g) + I_{2} (g)

2 H I (g) ⇋ H_{2} (g) + I_{2} (g)

An increase in pressure in the above reaction would :

Q. If the equilibrium constant of the reaction,

2 H I (g) ⇌ H_{2} (g) + I_{2} (g)

0.25

, the equilibrium constant of

H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)

, under similar conditions, will be:

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