
What is the effect of halving the pressure by doubling the volume on the following system at \[500{}^\circ C\]?
${{H}_{{{2}_{(g)}}}}+{{I}_{{{2}_{(g)}}}}2H{{I}_{(g)}}$
(A) Shift towards the reactant side
(B) Shift towards the product side
(C) Liquefaction of $HI$
(D) No effect
Answer
164.4k+ views
Hint: In order to solve this question we should be familiar with the Le-Chatelier’s Principles as to how they explain different effects of volume, pressure, temperature, etc. on the shift of equilibrium.
Complete Step by Step Solution:
Chemical equilibrium refers to the situation in a chemical reaction where both the reactants and products are present in concentrations that have no further tendency to change over time, preventing any discernible change in the system's properties. But, equilibrium can be disturbed by changing different factors like pressure, volume and temperature. There's a set of principles or rules called Le-Chatelier's principle which deals with the shift of equilibrium with change in different factors.
According to Le-chatelier’s principle, there is no effect on equilibrium if the volume has changed(no matter if the volume has decreased or the volume has increased). But on the contrary, there is an effect on changing the pressure. If we increase the pressure then the equilibrium shifts towards that side which has less number of moles of gaseous molecules.
Now, according to the given question, the equilibrium chemical reaction given to us is:
${{H}_{{{2}_{(g)}}}}+{{I}_{{{2}_{(g)}}}}2H{{I}_{(g)}}$ and given in the question is that we are decreasing the pressure and increasing the volume. Since volume has no effect on shift of equilibrium we’ll look for the effect of pressure. And we have seen as the pressure increases the equilibrium shift towards that side which has less number of moles.
Since, the pressure is decreasing it’ll move towards that side which has lesser number of moles but if we look at the given chemical reaction we can see that the number of gaseous moles is same on both the sides. Hence, there will be no effect of decreasing the pressure to half or doubling the volume.
Therefore, the correct option is D. No effect
Note: The question is completely dependent on the factors that changes the direction of equilibrium. Le-Chatelier’s principle are a set of rules which tell us the different effect of changing the different factors. The thing to remember in case of changing pressure is that the number of moles of only gaseous molecules should be taken into consideration and solid and liquid molecules have no role to play in this.
Complete Step by Step Solution:
Chemical equilibrium refers to the situation in a chemical reaction where both the reactants and products are present in concentrations that have no further tendency to change over time, preventing any discernible change in the system's properties. But, equilibrium can be disturbed by changing different factors like pressure, volume and temperature. There's a set of principles or rules called Le-Chatelier's principle which deals with the shift of equilibrium with change in different factors.
According to Le-chatelier’s principle, there is no effect on equilibrium if the volume has changed(no matter if the volume has decreased or the volume has increased). But on the contrary, there is an effect on changing the pressure. If we increase the pressure then the equilibrium shifts towards that side which has less number of moles of gaseous molecules.
Now, according to the given question, the equilibrium chemical reaction given to us is:
${{H}_{{{2}_{(g)}}}}+{{I}_{{{2}_{(g)}}}}2H{{I}_{(g)}}$ and given in the question is that we are decreasing the pressure and increasing the volume. Since volume has no effect on shift of equilibrium we’ll look for the effect of pressure. And we have seen as the pressure increases the equilibrium shift towards that side which has less number of moles.
Since, the pressure is decreasing it’ll move towards that side which has lesser number of moles but if we look at the given chemical reaction we can see that the number of gaseous moles is same on both the sides. Hence, there will be no effect of decreasing the pressure to half or doubling the volume.
Therefore, the correct option is D. No effect
Note: The question is completely dependent on the factors that changes the direction of equilibrium. Le-Chatelier’s principle are a set of rules which tell us the different effect of changing the different factors. The thing to remember in case of changing pressure is that the number of moles of only gaseous molecules should be taken into consideration and solid and liquid molecules have no role to play in this.
Recently Updated Pages
Environmental Chemistry Chapter for JEE Main Chemistry

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

Difference Between Natural and Whole Numbers: JEE Main 2024

Vector Algebra Chapter For JEE Main Maths

JEE Main Physics: Comprehensive Guide Work, Energy, and Power Concepts

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Types of Solutions

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts of Chemistry

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE
