IIT JAM Exam  >  IIT JAM Questions  >  For the reactionBr2 BF2--->2BrF3,the equilibr... Start Learning for Free
For the reactionBr2 BF2--->2BrF3,the equilibrium constant at 2000k and 1.0bar is 5.25.when the presure is increased by 8 fold, the equilibrium constant is ?
Most Upvoted Answer
For the reactionBr2 BF2--->2BrF3,the equilibrium constant at 2000k and...
Equilibrium Constant and Pressure

The equilibrium constant (K) of a chemical reaction is a measure of the ratio of the concentrations (or partial pressures) of the products to the concentrations (or partial pressures) of the reactants, with each concentration or partial pressure raised to the power of its stoichiometric coefficient in the balanced chemical equation.

Reaction:
Br2 + BF2 → 2BrF3

Given Information:
Equilibrium constant (K) at 2000K and 1.0 bar = 5.25

Effect of Pressure on Equilibrium Constant:

When the pressure is increased by 8-fold, it means that the new pressure is 8 times the initial pressure. Let's denote the initial pressure as P and the final pressure as 8P.

Pressure is directly proportional to the concentration of gases (according to the ideal gas law). Therefore, when the pressure is increased, the concentration of the reactants and products also increases.

Le Chatelier's Principle:
According to Le Chatelier's Principle, when a system at equilibrium is subjected to a change, it tends to counteract the change and reestablish equilibrium. In this case, the change is an increase in pressure.

Effect on Concentrations:
Since pressure and concentration are directly related, an increase in pressure will cause an increase in the concentration of the reactants and products.

Effect on Equilibrium Constant:
The equilibrium constant (K) is a ratio of the concentrations (or partial pressures) of the products to the concentrations (or partial pressures) of the reactants.

When the pressure is increased, the concentrations of both reactants and products increase. Therefore, the new equilibrium constant (K') will have a higher value than the initial equilibrium constant (K).

Calculating the New Equilibrium Constant:

To find the new equilibrium constant (K'), we need to consider the change in pressure. Since the pressure is increased by 8-fold, the new pressure is 8P.

The new equilibrium constant (K') can be determined using the following equation:

K' = K * (new pressure / initial pressure)

K' = 5.25 * (8P / P)

K' = 5.25 * 8

K' = 42

Therefore, when the pressure is increased by 8-fold, the new equilibrium constant is 42.

Summary:

- The equilibrium constant (K) is a measure of the ratio of the concentrations (or partial pressures) of the products to the concentrations (or partial pressures) of the reactants.
- When the pressure is increased, the concentrations of the reactants and products increase.
- According to Le Chatelier's Principle, the system at equilibrium will counteract the change and reestablish equilibrium.
- The new equilibrium constant (K') will have a higher value than the initial equilibrium constant (K) when the pressure is increased.
- The new equilibrium constant (K') can be calculated using the equation K' = K * (new pressure / initial pressure).
- In this case, when the pressure is increased by 8-fold, the new equilibrium constant is 42.
Explore Courses for IIT JAM exam
For the reactionBr2 BF2--->2BrF3,the equilibrium constant at 2000k and 1.0bar is 5.25.when the presure is increased by 8 fold, the equilibrium constant is ?
Question Description
For the reactionBr2 BF2--->2BrF3,the equilibrium constant at 2000k and 1.0bar is 5.25.when the presure is increased by 8 fold, the equilibrium constant is ? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about For the reactionBr2 BF2--->2BrF3,the equilibrium constant at 2000k and 1.0bar is 5.25.when the presure is increased by 8 fold, the equilibrium constant is ? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For the reactionBr2 BF2--->2BrF3,the equilibrium constant at 2000k and 1.0bar is 5.25.when the presure is increased by 8 fold, the equilibrium constant is ?.
Solutions for For the reactionBr2 BF2--->2BrF3,the equilibrium constant at 2000k and 1.0bar is 5.25.when the presure is increased by 8 fold, the equilibrium constant is ? in English & in Hindi are available as part of our courses for IIT JAM. Download more important topics, notes, lectures and mock test series for IIT JAM Exam by signing up for free.
Here you can find the meaning of For the reactionBr2 BF2--->2BrF3,the equilibrium constant at 2000k and 1.0bar is 5.25.when the presure is increased by 8 fold, the equilibrium constant is ? defined & explained in the simplest way possible. Besides giving the explanation of For the reactionBr2 BF2--->2BrF3,the equilibrium constant at 2000k and 1.0bar is 5.25.when the presure is increased by 8 fold, the equilibrium constant is ?, a detailed solution for For the reactionBr2 BF2--->2BrF3,the equilibrium constant at 2000k and 1.0bar is 5.25.when the presure is increased by 8 fold, the equilibrium constant is ? has been provided alongside types of For the reactionBr2 BF2--->2BrF3,the equilibrium constant at 2000k and 1.0bar is 5.25.when the presure is increased by 8 fold, the equilibrium constant is ? theory, EduRev gives you an ample number of questions to practice For the reactionBr2 BF2--->2BrF3,the equilibrium constant at 2000k and 1.0bar is 5.25.when the presure is increased by 8 fold, the equilibrium constant is ? tests, examples and also practice IIT JAM tests.
Explore Courses for IIT JAM exam
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev