Medium
The equilibrium concentrations of $HI$,$I_2$,and $H_2$ are $0.7 \ M$,$0.1 \ M$,and $0.1 \ M$ respectively. The equilibrium constant for the reaction $I_2 + H_2 \rightleftharpoons 2HI$ is:
- A$36$
- B$49$
- C$0.49$
- D$0.36$
Explore More
Similar Questions
For the reaction $2HI \rightleftharpoons H_2 + I_2$,the equilibrium constant is $K_1 = 0.25$. What is the value of $K_2$ for the reaction $H_2 + I_2 \rightleftharpoons 2HI$?
Medium
View SolutionExplain why pure liquids and solids can be ignored while writing the equilibrium constant expression?
Medium
View SolutionThe equilibrium constant expression for a gas reaction is,
$K_{C} = \frac{[NH_{3}]^{4}[O_{2}]^{5}}{[NO]^{4}[H_{2}O]^{6}}$
Write the balanced chemical equation corresponding to this expression.
$K_{C} = \frac{[NH_{3}]^{4}[O_{2}]^{5}}{[NO]^{4}[H_{2}O]^{6}}$
Write the balanced chemical equation corresponding to this expression.
Easy
View SolutionIf the equilibrium constant for the formation of $NH_3$ is $K_c$,then the dissociation constant of $NH_3$ at the same temperature is:
Medium
View SolutionWhen $2 \ mol$ of $HI$ is heated in a closed vessel at $440 \ ^\circ C$,$22\%$ of $HI$ dissociates until equilibrium is reached. The equilibrium constant $K_c$ for the reaction is ..........
Medium
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