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Applications of equilibrium constant

The equilibrium constant of the reaction A_{2}({~g})+B_{2}({~g

Question

Hard

Solving time: 5 mins

The equilibrium constant of the reaction
$A_{2} (g) + B_{2} (g) ⇌ 2 A B (g)$
at $10 0^{\circ} C$ is 50 . If a one litre flask, containing one mole of $A_{2}$ is connected to a two litre flask, containing two mole of $B_{2}$ , how many mole of $A B$ will be formed at $37 3^{\circ} C$ ?

Name: Video solution 1: The equilibrium constant of the reactionA2( g)+B2( g)⇌2AB( g)at 100∘C is 50 . If a one litre flask, containing one mole of A2 is connected to a two litre flask, containing two mole of B2, how many mole of AB will be formed at 373∘C ?
Uploaded: 2023-09-01T04:29:16.000Z
Duration: 14 min
Description: Video solution 1: The equilibrium constant of the reactionA2( g)+B2( g)⇌2AB( g)at 100∘C is 50 . If a one litre flask, containing one mole of A2 is connected to a two litre flask, containing two mole of B2, how many mole of AB will be formed at 373∘C ?

0.8 68

1.8 68

2.886

3.886

Views: 6,145 students

Updated on: Sep 1, 2023

Found 7 tutors discussing this question

Sebastian Discussed

The equilibrium constant of the reaction

A_{2} (g) + B_{2} (g) ⇌ 2 A B (g)

10 0^{\circ} C

is 50 . If a one litre flask, containing one mole of

A_{2}

is connected to a two litre flask, containing two mole of

B_{2}

, how many mole of

A B

will be formed at

37 3^{\circ} C

12 mins ago

Discuss this question LIVE

12 mins ago

Text solutionVerified

∴ [A_{2}] = \frac{1 - x}{3}, [B_{2}] = \frac{2 - x}{3}, [A B] = \frac{2 x}{3} ∴ K = \frac{( 2 x /3 ) ^{2}}{[( 1 - x ) /3 ] [( 2 - x ) /3 ]} = 50

On solving we get,

23 x^{2} - 75 x + 50 = 0; x = 2.326

0.934

The value

2.326

is inadmissable because initial concentration of reactants is 2 moles and so

x = 0.943

∴

Moles of

A B

formed

= 2 \times 0.934 = 1.868

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Question Text	The equilibrium constant of the reaction $A_{2} (g) + B_{2} (g) ⇌ 2 A B (g)$ at $10 0^{\circ} C$ is 50 . If a one litre flask, containing one mole of $A_{2}$ is connected to a two litre flask, containing two mole of $B_{2}$ , how many mole of $A B$ will be formed at $37 3^{\circ} C$ ?
Updated On	Sep 1, 2023
Topic	Equilibrium
Subject	Chemistry
Class	Class 11
Answer Type	Text solution:1 Video solution: 2
Upvotes	252
Avg. Video Duration	9 min

The equilibrium constant of the reaction
$A_{2} (g) + B_{2} (g) ⇌ 2 A B (g)$
at $10 0^{\circ} C$ is 50 . If a one litre flask, containing one mole of $A_{2}$ is connected to a two litre flask, containing two mole of $B_{2}$ , how many mole of $A B$ will be formed at $37 3^{\circ} C$ ?

Text solutionVerified

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The equilibrium constant of the reactionA2​( g)+B2​( g)⇌2AB( g)at 100∘C is 50 . If a one litre flask, containing one mole of A2​ is connected to a two litre flask, containing two mole of B2​, how many mole of AB will be formed at 373∘C ?

Text solutionVerified

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The equilibrium constant of the reaction
$A_{2} (g) + B_{2} (g) ⇌ 2 A B (g)$
at $10 0^{\circ} C$ is 50 . If a one litre flask, containing one mole of $A_{2}$ is connected to a two litre flask, containing two mole of $B_{2}$ , how many mole of $A B$ will be formed at $37 3^{\circ} C$ ?