Medium
The equilibrium constant of the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ is $64$. If the volume of the container is reduced to one-fourth of its original volume,the value of the equilibrium constant will be
- A$16$
- B$32$
- C$64$
- D$128$
Explore More
Similar Questions
$2 \ HI_{(g)} \rightleftharpoons H_{2(g)} + I_{2(g)}$
The equilibrium constant of the above reaction is $6.4$ at $300 \ K$. If $0.25 \ mol$ each of $H_{2}$ and $I_{2}$ are added to the system,the equilibrium constant will be
The equilibrium constant of the above reaction is $6.4$ at $300 \ K$. If $0.25 \ mol$ each of $H_{2}$ and $I_{2}$ are added to the system,the equilibrium constant will be
MediumKCET 2009
View SolutionFor the reaction ${H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}}$,the equilibrium constant $K_p$ changes with the change in:
Easy
View SolutionAmong the following,the correct statement about the chemical equilibrium is
MediumKVPY 2020
View SolutionThe equilibrium of the reaction $x \rightleftharpoons y$ is attained when......
Medium
View SolutionFor the equilibrium $A \rightleftharpoons B$,the variation of the rate of the forward $(a)$ and reverse $(b)$ reaction with time is given by:
DifficultJEE MAIN 2020
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo