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Chemistry Part-I

Equilibrium

At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO_{

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

Bromine monochloride,

B r Cl

, decomposes into broe and chlorine and reaches equilibrium as

2 B r C l_{(g)} ⇌ B r_{2 (g)} + C l_{2 (g)}

; for which

K_{c} = 32

500 K

B r Cl

1127 K

and

1 a t m

pressure, a gaseous mixture of

CO

and

C O_{2}

in equilibrium with solid carbon has

90.55%

CO

by mass.

C_{(s)} + C O_{2 (g)} ⇌ 2 C O_{(g)}

Calculate

K_{c}

for the reaction at the above temperature.

Calculate:

Δ G^{0}

b) the equilibrium constant for the formation of

N O_{2}

from NO and

O_{2}

at 298 K.

NO (g) + \frac{1}{2} O_{2} (g) ⇌ N O_{2} (g)

where

Δ_{f} G^{0} (N O_{2}) = 52.0

kJ/mol

Δ_{f} G^{0} (NO) = 87.0

kJ/mol

Δ_{f} G^{0} (O_{2}) = 0

kJ/mol.

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Question

Easy

Solving time: 4 mins

At $1127 K$ and $1 a t m$ pressure, a gaseous mixture of $CO$ and $C O_{2}$ in equilibrium with solid carbon has $90.55%$ of $CO$ by mass.

$C_{(s)} + C O_{2 (g)} ⇌ 2 C O_{(g)}$

Calculate $K_{c}$ for the reaction at the above temperature.

Name: Video solution 1: [Solved] At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2 in equilibrium with solid carbon has 90.55% of CO by mass.C(s)+CO2(g)⇌2CO(g)Calculate Kc for the reaction at the above temperature. - Chemistry Part-I
Uploaded: 2023-05-13T07:22:31.000Z
Duration: 8 min
Description: Video solution 1: [Solved] At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2 in equilibrium with solid carbon has 90.55% of CO by mass.C(s)+CO2(g)⇌2CO(g)Calculate Kc for the reaction at the above temperature. - Chemistry Part-I

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Updated on: May 13, 2023

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Chemistry Part-I (NCERT)

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1127 K

and

1 a t m

pressure, a gaseous mixture of

CO

and

C O_{2}

in equilibrium with solid carbon has

90.55%

CO

by mass.

C_{(s)} + C O_{2 (g)} ⇌ 2 C O_{(g)}

Calculate

K_{c}

for the reaction at the above temperature.

13 mins ago

Discuss this question LIVE

13 mins ago

Text solutionVerified

Let the mixture has 100g as total mass.

The masses of CO and

C O_{2}

are

90.55

g and

100 - 90.55 = 9.45

g respectively.

The number of moles of CO are

\frac{90.55}{28} = 3.234

The number of moles of

C O_{2}

are

\frac{9.45}{44} = 0.215

The mole fraction of CO is

\frac{3.234}{3.234 + 0.215} = 0.938

The mole fraction of

C O_{2}

1 - 0.938 = 0.062

The partial pressure of CO is the product of the mole fraction of CO and the total pressure.

It is

0.938 \times 1 = 0.938

atm.

The partial pressure of carbon dioxide is

0.062 \times 1 = 0.042

atm.

The expression for the equilibrium constant is:

K_{p} = \frac{P _{CO}^{2}}{P _{C O_{2}}} = \frac{( 0.938 ) ^{2}}{0.062} = 14.19

Δ n_{g} = 2 - 1 = 1

K_{c} = K_{p} (RT)^{- Δ n} = 14.19 \times (0.0821 \times 1127)^{- 1} = 0.153

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Practice more questions from Chemistry Part-I (NCERT)

Bromine monochloride,

B r Cl

, decomposes into broe and chlorine and reaches equilibrium as

2 B r C l_{(g)} ⇌ B r_{2 (g)} + C l_{2 (g)}

; for which

K_{c} = 32

500 K

B r Cl

1127 K

and

1 a t m

pressure, a gaseous mixture of

CO

and

C O_{2}

in equilibrium with solid carbon has

90.55%

CO

by mass.

C_{(s)} + C O_{2 (g)} ⇌ 2 C O_{(g)}

Calculate

K_{c}

for the reaction at the above temperature.

Calculate:

Δ G^{0}

b) the equilibrium constant for the formation of

N O_{2}

from NO and

O_{2}

at 298 K.

NO (g) + \frac{1}{2} O_{2} (g) ⇌ N O_{2} (g)

where

Δ_{f} G^{0} (N O_{2}) = 52.0

kJ/mol

Δ_{f} G^{0} (NO) = 87.0

kJ/mol

Δ_{f} G^{0} (O_{2}) = 0

kJ/mol.

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Question Text	At $1127 K$ and $1 a t m$ pressure, a gaseous mixture of $CO$ and $C O_{2}$ in equilibrium with solid carbon has $90.55%$ of $CO$ by mass. $C_{(s)} + C O_{2 (g)} ⇌ 2 C O_{(g)}$ Calculate $K_{c}$ for the reaction at the above temperature.
Updated On	May 13, 2023
Topic	Equilibrium
Subject	Chemistry
Class	Class 11
Answer Type	Text solution:1 Video solution: 2
Upvotes	225
Avg. Video Duration	7 min

At $1127 K$ and $1 a t m$ pressure, a gaseous mixture of $CO$ and $C O_{2}$ in equilibrium with solid carbon has $90.55%$ of $CO$ by mass.

$C_{(s)} + C O_{2 (g)} ⇌ 2 C O_{(g)}$

Calculate $K_{c}$ for the reaction at the above temperature.

Text solutionVerified

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At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2​ in equilibrium with solid carbon has 90.55% of CO by mass.C(s)​+CO2(g)​⇌2CO(g)​Calculate Kc​ for the reaction at the above temperature.

Text solutionVerified

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At $1127 K$ and $1 a t m$ pressure, a gaseous mixture of $CO$ and $C O_{2}$ in equilibrium with solid carbon has $90.55%$ of $CO$ by mass.

$C_{(s)} + C O_{2 (g)} ⇌ 2 C O_{(g)}$

Calculate $K_{c}$ for the reaction at the above temperature.