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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

Views: 6,120 students

Updated on: May 13, 2023

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Chemistry Part-I (NCERT)

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Lucas Discussed

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.

6 mins ago

Discuss this question LIVE

6 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.