Medium
What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of $ICl$ was $0.78 \, M$?
$2ICl_{(g)} \leftrightarrow I_{2(g)} + Cl_{2(g)}; \, K_c = 0.14$
$2ICl_{(g)} \leftrightarrow I_{2(g)} + Cl_{2(g)}; \, K_c = 0.14$
(N/A) The given reaction is:
$2ICl_{(g)} \leftrightarrow I_{2(g)} + Cl_{2(g)}$
Initial concentration: $[ICl] = 0.78 \, M$,$[I_2] = 0 \, M$,$[Cl_2] = 0 \, M$
At equilibrium: $[ICl] = (0.78 - 2x) \, M$,$[I_2] = x \, M$,$[Cl_2] = x \, M$
The equilibrium constant expression is:
$K_c = \frac{[I_2][Cl_2]}{[ICl]^2} = 0.14$
Substituting the values:
$\frac{x \cdot x}{(0.78 - 2x)^2} = 0.14$
Taking the square root on both sides:
$\frac{x}{0.78 - 2x} = \sqrt{0.14} \approx 0.374$
$x = 0.374(0.78 - 2x)$
$x = 0.2917 - 0.748x$
$1.748x = 0.2917$
$x \approx 0.167 \, M$
Therefore,at equilibrium:
$[I_2] = [Cl_2] = 0.167 \, M$
$[ICl] = 0.78 - 2(0.167) = 0.446 \, M$
$2ICl_{(g)} \leftrightarrow I_{2(g)} + Cl_{2(g)}$
Initial concentration: $[ICl] = 0.78 \, M$,$[I_2] = 0 \, M$,$[Cl_2] = 0 \, M$
At equilibrium: $[ICl] = (0.78 - 2x) \, M$,$[I_2] = x \, M$,$[Cl_2] = x \, M$
The equilibrium constant expression is:
$K_c = \frac{[I_2][Cl_2]}{[ICl]^2} = 0.14$
Substituting the values:
$\frac{x \cdot x}{(0.78 - 2x)^2} = 0.14$
Taking the square root on both sides:
$\frac{x}{0.78 - 2x} = \sqrt{0.14} \approx 0.374$
$x = 0.374(0.78 - 2x)$
$x = 0.2917 - 0.748x$
$1.748x = 0.2917$
$x \approx 0.167 \, M$
Therefore,at equilibrium:
$[I_2] = [Cl_2] = 0.167 \, M$
$[ICl] = 0.78 - 2(0.167) = 0.446 \, M$
Explore More
Similar Questions
The figure shows the change in concentration of species $A$ and $B$ as a function of time. The equilibrium constant $K_C$ for the reaction $2A_{(g)} \rightleftharpoons B_{(g)}$ is
Medium
View SolutionIf the equilibrium constant for the reaction,$2 SO_2 + O_2 \rightleftharpoons 2 SO_3$ is $64$ at $500 \ K$,then the equilibrium constant for the reaction $SO_3 \rightleftharpoons SO_2 + \frac{1}{2} O_2$ at the same temperature is
For the reaction,$A_{(g)} + 2B_{(g)} \rightleftharpoons 2C_{(g)}$ at $25 \, ^oC$,$2 \, moles$ of $A$,$1 \, mole$ of $B$ and $1 \, mole$ of $C$ are present in a $1 \, L$ vessel. If $K_c$ for the reaction is $2$,then the reaction will proceed in:
Medium
View SolutionFor the gaseous reactions $(I)$ and $(II)$,the equilibrium constants are $X$ and $Y$,respectively.
$I. \frac{1}{2} N_{2(g)} + O_{2(g)} \rightleftharpoons NO_{2(g)}$
$II. 2 NO_{2(g)} \rightleftharpoons N_2O_{4(g)}$
Using the above reactions,the equilibrium constant $Z$ for the reaction $(III)$ given below is:
$III. N_2O_{4(g)} \rightleftharpoons N_{2(g)} + 2 O_{2(g)}$
$I. \frac{1}{2} N_{2(g)} + O_{2(g)} \rightleftharpoons NO_{2(g)}$
$II. 2 NO_{2(g)} \rightleftharpoons N_2O_{4(g)}$
Using the above reactions,the equilibrium constant $Z$ for the reaction $(III)$ given below is:
$III. N_2O_{4(g)} \rightleftharpoons N_{2(g)} + 2 O_{2(g)}$
DifficultAP EAMCET 2018
View SolutionThe equilibrium constant $K_{c}$ at $298 \ K$ for the reaction $A + B \rightleftharpoons C + D$ is $100$. Starting with an equimolar solution with concentrations of $A$,$B$,$C$ and $D$ all equal to $1 \ M$,the equilibrium concentration of $D$ is $....... \times 10^{-2} \ M$. (Nearest integer)
Vedclass 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