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Chemistry Part-I

Equilibrium

One mole of { H }_{ 2 }O and one mole of CO are taken in 10

Chemistry Part-I

Class 11

NCERT

Chapter 1: Some Basic Concepts of Chemistry

44 questions

Chapter 2: Structure of Atom

85 questions

Chapter 3: Classification of Elements and Periodicity in Properties

50 questions

Chapter 4: Chemical Bonding and Molecular Structure

44 questions

Chapter 5: States of Matter

28 questions

Chapter 6: Thermodynamics and Thermochemistry

36 questions

Chapter 7: Equilibrium

101 questions

The equilibrium constant expression for a gas reaction is:

K_{c} = \frac{[ N H _{3} ] ^{4} [ O _{2} ] ^{5}}{[ NO ] ^{4} [ H _{2} O ] ^{6}}

Write a balanced chemical equation corresponding to this expression.

One mole of

H_{2} O

and one mole of

CO

are taken in 10 L vessel and heated to 725 K. At equilibrium, 40% of water (by mass) reacts with

CO

according to the equation.

H_{2} O (g) + CO (g) ⇌ H_{2} (g) + CO_{2} (g)

Calculate the equilibrium constant for the reaction.

700 K

, the equilibrium constant for the reaction;

H_{2 (g)} + I_{2 (g)} ⇌ 2 H I_{(g)}

54.8

0.5 m o l l i t r e^{- 1}

H I_{(g)}

is present at equilibrium at

700 K

, what are the concentrations of

H_{2 (g)}

and

I_{2 (g)}

assug that we initially started with

H I_{(g)}

and allowed it to reach equilibrium at

700 K

View all question

Question

Medium

Solving time: 5 mins

One mole of $H_{2} O$ and one mole of $CO$ are taken in 10 L vessel and heated to 725 K. At equilibrium, 40% of water (by mass) reacts with $CO$ according to the equation.
$H_{2} O (g) + CO (g) ⇌ H_{2} (g) + CO_{2} (g)$
Calculate the equilibrium constant for the reaction.

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Chemistry Part-I (NCERT)

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Amelia Discussed

One mole of

H_{2} O

and one mole of

CO

are taken in 10 L vessel and heated to 725 K. At equilibrium, 40% of water (by mass) reacts with

CO

according to the equation.

H_{2} O (g) + CO (g) ⇌ H_{2} (g) + CO_{2} (g)

Calculate the equilibrium constant for the reaction.

14 mins ago

Discuss this question LIVE

14 mins ago

Text solutionVerified

The number of moles of water that have reacted are

\frac{1 \times 40}{100} = 0.4 m o l

Moles of water unreacted are

1.0 - 0.4 = 0.6

mol.

0.4 moles of water reacts with 0.4 mole of CO to form 0.4 moles of hydrogen and 0.4 moles of carbon dioxide.

The equilibrium molar concentrations are:

[H_{2}] = \frac{0.4}{10} = 0.04 M

[C O_{2}] = \frac{0.4}{10} = 0.04 M

[H_{2} O] = \frac{0.6}{10} = 0.06 M

[CO] = \frac{0.6}{10} = 0.06 M

The equilibrium constant is

K = \frac{[ H _{2} ] [ C O _{2} ]}{[ H _{2} O ] [ CO ]} = \frac{0.04 \times 0.04}{0.06 \times 0.06} = 0.444

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Practice more questions from Chemistry Part-I (NCERT)

The equilibrium constant expression for a gas reaction is:

K_{c} = \frac{[ N H _{3} ] ^{4} [ O _{2} ] ^{5}}{[ NO ] ^{4} [ H _{2} O ] ^{6}}

Write a balanced chemical equation corresponding to this expression.

One mole of

H_{2} O

and one mole of

CO

are taken in 10 L vessel and heated to 725 K. At equilibrium, 40% of water (by mass) reacts with

CO

according to the equation.

H_{2} O (g) + CO (g) ⇌ H_{2} (g) + CO_{2} (g)

Calculate the equilibrium constant for the reaction.

700 K

, the equilibrium constant for the reaction;

H_{2 (g)} + I_{2 (g)} ⇌ 2 H I_{(g)}

54.8

0.5 m o l l i t r e^{- 1}

H I_{(g)}

is present at equilibrium at

700 K

, what are the concentrations of

H_{2 (g)}

and

I_{2 (g)}

assug that we initially started with

H I_{(g)}

and allowed it to reach equilibrium at

700 K

View all

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Question Text	One mole of $H_{2} O$ and one mole of $CO$ are taken in 10 L vessel and heated to 725 K. At equilibrium, 40% of water (by mass) reacts with $CO$ according to the equation. $H_{2} O (g) + CO (g) ⇌ H_{2} (g) + CO_{2} (g)$ Calculate the equilibrium constant for the reaction.
Topic	Equilibrium
Subject	Chemistry
Class	Class 11
Answer Type	Text solution:1
Upvotes	41

One mole of $H_{2} O$ and one mole of $CO$ are taken in 10 L vessel and heated to 725 K. At equilibrium, 40% of water (by mass) reacts with $CO$ according to the equation.
$H_{2} O (g) + CO (g) ⇌ H_{2} (g) + CO_{2} (g)$
Calculate the equilibrium constant for the reaction.

Text solutionVerified

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One mole of H2​O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium, 40% of water (by mass) reacts with CO according to the equation.H2​O(g)+CO(g)⇌H2​(g)+CO2​(g) Calculate the equilibrium constant for the reaction.

Text solutionVerified

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One mole of $H_{2} O$ and one mole of $CO$ are taken in 10 L vessel and heated to 725 K. At equilibrium, 40% of water (by mass) reacts with $CO$ according to the equation.
$H_{2} O (g) + CO (g) ⇌ H_{2} (g) + CO_{2} (g)$
Calculate the equilibrium constant for the reaction.