Passage IThe equilibrium reaction has been thoroughly studiedKp = 0.148 at 298 KQ.If the volume of the container is increased so that the total equilibrium pressure falls to 1.00 atm, then fraction of N2O4 dissociated isa)0.036b)0.064c)0.911d)0.189Correct answer is option 'D'. Can you explain this answer?

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Passage I
The equilibrium reaction

has been thoroughly studied K_p = 0.148 at 298 K

Q. If the volume of the container is increased so that the total equilibrium pressure falls to 1.00 atm, then fraction of N₂O₄ dissociated is

a)
0.036
b)
0.064
c)
0.911
d)
0.189

Correct answer is option 'D'. Can you explain this answer?

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Passage IThe equilibrium reaction has been thoroughly studiedKp = 0.1...

Step 1: Understanding the equilibrium reaction

The equilibrium reaction is:

At equilibrium, we know:

The equilibrium constant K_p = 0.148 at 298 K.
The total equilibrium pressure is reduced to 1.00 atm due to an increase in the volume of the container.

Step 2: Use the equilibrium expression for K_p

The equilibrium expression for this reaction is:

Where P_NO2 is the partial pressure of NO₂and P_N2O4 is the partial pressure of N₂O₄ at equilibrium.

Step 3: Let’s assume an initial amount of N₂O₄

Let’s assume the initial pressure of N₂O₄ before dissociation is P₀, and the partial pressure of NO₂ at equilibrium is P_NO2. Let’s denote the fraction dissociated as x.

At equilibrium, the changes in pressures would be:

The partial pressure of N₂O₄will decrease by xP₀.
The partial pressure of NO₂ will increase by 2xP₀ (since two moles of NO₂ are produced for each mole of N₂O₄).

Thus, the equilibrium pressures can be written as:

P_N2O4 = P₀ − xP₀
P_NO2 = 2xP₀

Step 4: Total pressure at equilibrium

The total pressure at equilibrium is the sum of the partial pressures:

We are told that the total pressure is 1.00 atm, so:

This equation allows us to solve for P₀ in terms of x.

Step 5: Apply the equilibrium constant expression

Substitute the equilibrium pressures into the K_p expression:

We know that K_p = 0.148, so:

Simplify the equation:

Now, substitute

into this equation and solve for x.

Step 6: Solve for the fraction dissociated x

After solving the equation, you will find that the fraction dissociated xxx is approximately 0.189.

Answered by Mohit Rajpoot · View profile

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Passage IThe equilibrium reaction has been thoroughly studiedKp = 0.148 at 298 KQ.If the volume of the container is increased so that the total equilibrium pressure falls to 1.00 atm, then fraction of N2O4 dissociated isa)0.036b)0.064c)0.911d)0.189Correct answer is option 'D'. Can you explain this answer?

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Passage IThe equilibrium reaction has been thoroughly studiedKp = 0.148 at 298 KQ.If the volume of the container is increased so that the total equilibrium pressure falls to 1.00 atm, then fraction of N2O4 dissociated isa)0.036b)0.064c)0.911d)0.189Correct answer is option 'D'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Passage IThe equilibrium reaction has been thoroughly studiedKp = 0.148 at 298 KQ.If the volume of the container is increased so that the total equilibrium pressure falls to 1.00 atm, then fraction of N2O4 dissociated isa)0.036b)0.064c)0.911d)0.189Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Passage IThe equilibrium reaction has been thoroughly studiedKp = 0.148 at 298 KQ.If the volume of the container is increased so that the total equilibrium pressure falls to 1.00 atm, then fraction of N2O4 dissociated isa)0.036b)0.064c)0.911d)0.189Correct answer is option 'D'. Can you explain this answer?.

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