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Chemistry Part-I

Equilibrium

Ethyl acetate is formed by the reaction between ethanol and ac

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

K_{p} = 0.04 a t m

at 899 K for the equilibrium shown below. What is the equilibrium concentration of

C_{2} H_{6}

when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?

C_{2} H_{6} (g) ⇌ C_{2} H_{4} (g) + H_{2} (g)

Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:

C H_{3} COO H (l) + C_{2} H_{5} O H (l) \to C H_{3} COO C_{2} H_{5} (l) + H_{2} O (l)

(i) Write the concentration ratio (reaction quotient), Qc , for this reaction (note: water is not in excess and is not a solvent in this reaction)
(ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
(iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?

A sample of pure

PCl_{5}

was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of

PCl_{5}

was found to be

0.5 \times 10^{- 1} m o l L^{- 1}

. If value of

K_{c}

8.3 \times 10^{- 3}

, what are the concentrations of

PCl_{3}

and

Cl_{2}

at equilibrium?

PCl_{5} (g) ⇌ PCl_{3} (g) + Cl_{2} (g)

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Question

Hard

Solving time: 5 mins

Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as: $C H_{3} COO H (l) + C_{2} H_{5} O H (l) \to C H_{3} COO C_{2} H_{5} (l) + H_{2} O (l)$
(i) Write the concentration ratio (reaction quotient), Qc , for this reaction (note: water is not in excess and is not a solvent in this reaction)
(ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
(iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?

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Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:

C H_{3} COO H (l) + C_{2} H_{5} O H (l) \to C H_{3} COO C_{2} H_{5} (l) + H_{2} O (l)

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Text solutionVerified

(i) The concentration ratio is

Q_{c} = \frac{[ C H _{3} COO C _{2} H _{5} ] [ H _{2} O ]}{[ C H _{3} COO H ] [ C _{2} H _{5} O H ]}

(ii) The initial concentrations are

[C H_{3} COO H] = 1.00

[C_{2} H_{5} O H] = 0.18

[C H_{3} CO C_{2} H_{4}] = 0

[H_{2} O] = 0

The equilibrium concentrations are:

[C H_{3} COO H] = \frac{1 - 0.171}{V} = \frac{0.829}{V}

[C_{2} H_{5} O H] = \frac{0.18 - 0.171}{V} = \frac{0.009}{V}

[C H_{3} CO C_{2} H_{4}] = \frac{0.171}{V}

[H_{2} O] = \frac{0.171}{V}

The equilibrium constant expression is

K_{c} = \frac{[ C H _{3} COO C _{2} H _{5} ] [ H _{2} O ]}{[ C H _{3} COO H ] [ C _{2} H _{5} O H ]} = \frac{\frac{0.171}{V} \times \frac{0.171}{V}}{\frac{0.829}{V} \times \frac{0.009}{V}} = 3.92

(iii) The initial concentrations are:

[C H_{3} COO H] = 1.00

[C_{2} H_{5} O H] = 0.500

[C H_{3} CO C_{2} H_{4}] = 0

[H_{2} O] = 0

The equilibrium concentrations are:

[C H_{3} COO H] = \frac{1 - 0.214}{V} = \frac{0.786}{V}

[C_{2} H_{5} O H] = \frac{0.500 - 0.214}{V} = \frac{0.286}{V}

[C H_{3} CO C_{2} H_{4}] = \frac{0.214}{V}

[H_{2} O] = \frac{0.214}{V}

The expression for the reaction quotient is:

Q_{c} = \frac{[ C H _{3} COO C _{2} H _{5} ] [ H _{2} O ]}{[ C H _{3} COO H ] [ C _{2} H _{5} O H ]} = \frac{\frac{0.214}{V} \times \frac{0.214}{V}}{\frac{0.786}{V} \times \frac{0.286}{V}} = 0.204

As the value of the reaction quotient is not equal to the value of the equilibrium constant, the equilibrium is not yet achieved.

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Practice more questions from Chemistry Part-I (NCERT)

K_{p} = 0.04 a t m

at 899 K for the equilibrium shown below. What is the equilibrium concentration of

C_{2} H_{6}

when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?

C_{2} H_{6} (g) ⇌ C_{2} H_{4} (g) + H_{2} (g)

Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:

C H_{3} COO H (l) + C_{2} H_{5} O H (l) \to C H_{3} COO C_{2} H_{5} (l) + H_{2} O (l)

A sample of pure

PCl_{5}

was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of

PCl_{5}

was found to be

0.5 \times 10^{- 1} m o l L^{- 1}

. If value of

K_{c}

8.3 \times 10^{- 3}

, what are the concentrations of

PCl_{3}

and

Cl_{2}

at equilibrium?

PCl_{5} (g) ⇌ PCl_{3} (g) + Cl_{2} (g)

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Question Text	Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as: $C H_{3} COO H (l) + C_{2} H_{5} O H (l) \to C H_{3} COO C_{2} H_{5} (l) + H_{2} O (l)$ (i) Write the concentration ratio (reaction quotient), Qc , for this reaction (note: water is not in excess and is not a solvent in this reaction) (ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant. (iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?
Topic	Equilibrium
Subject	Chemistry
Class	Class 11
Answer Type	Text solution:1
Upvotes	34

Text solutionVerified

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