Class 11Chemistry6-1.Equilibrium (Chemical Equilibrium)Mix Examples- 6-1.Equilibrium (Chemical Equilibrium)
DifficultAt $717 \ K$,$3.2 \ mol$ of $HI$ is heated in a closed tube. $20\%$ of $HI$ decomposes at equilibrium according to the reaction $2HI_{(g)} \rightleftharpoons H_{2_{(g)}} + I_{2_{(g)}}$. Find $K_c$ and the moles of $HI$,$H_2$,and $I_2$ at equilibrium.
(N/A) The reaction is $2HI_{(g)} \rightleftharpoons H_{2_{(g)}} + I_{2_{(g)}}$.
Initial moles: $HI = 3.2 \ mol$,$H_2 = 0 \ mol$,$I_2 = 0 \ mol$.
At equilibrium,$20\%$ of $HI$ decomposes,so the amount reacted is $3.2 \times 0.20 = 0.64 \ mol$.
Remaining $HI = 3.2 - 0.64 = 2.56 \ mol$.
According to the stoichiometry,$2 \ mol$ of $HI$ produces $1 \ mol$ of $H_2$ and $1 \ mol$ of $I_2$.
Thus,$H_2 = 0.64 / 2 = 0.32 \ mol$ and $I_2 = 0.64 / 2 = 0.32 \ mol$.
$K_c = [H_2][I_2] / [HI]^2 = (0.32 / V) \times (0.32 / V) / (2.56 / V)^2 = (0.32 \times 0.32) / (2.56)^2 = 0.1024 / 6.5536 = 0.015625$.
Initial moles: $HI = 3.2 \ mol$,$H_2 = 0 \ mol$,$I_2 = 0 \ mol$.
At equilibrium,$20\%$ of $HI$ decomposes,so the amount reacted is $3.2 \times 0.20 = 0.64 \ mol$.
Remaining $HI = 3.2 - 0.64 = 2.56 \ mol$.
According to the stoichiometry,$2 \ mol$ of $HI$ produces $1 \ mol$ of $H_2$ and $1 \ mol$ of $I_2$.
Thus,$H_2 = 0.64 / 2 = 0.32 \ mol$ and $I_2 = 0.64 / 2 = 0.32 \ mol$.
$K_c = [H_2][I_2] / [HI]^2 = (0.32 / V) \times (0.32 / V) / (2.56 / V)^2 = (0.32 \times 0.32) / (2.56)^2 = 0.1024 / 6.5536 = 0.015625$.
Explore More
Similar Questions
Which of the following equations is incorrect?
Medium
View SolutionOne mole of $N_2O_{4(g)}$ at $300 \ K$ is kept in a closed container under $1 \ atm$ pressure. It is heated to $600 \ K$ when $20 \%$ of $N_2O_{4(g)}$ decomposes to $NO_{2(g)}$. The resultant pressure at $600 \ K$ is $..... \ atm$.
Medium
View SolutionFor the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$,the value of $K_c$ is $0.04$ at $250 \ ^oC$ in a $3 \ L$ vessel. If the concentration of $Cl_2$ at equilibrium is $0.15 \ M$,then the initial moles of $PCl_5$ will be ...........
Difficult
View SolutionFor the equilibrium:
$CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$; $K_{p} = 1.64 \ atm$ at $1000 \ K$.
$50 \ g$ of $CaCO_{3}$ in a $10 \ L$ closed vessel is heated to $1000 \ K$. The percentage of $CaCO_{3}$ that remains unreacted at equilibrium is:
(Given $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$)
$CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$; $K_{p} = 1.64 \ atm$ at $1000 \ K$.
$50 \ g$ of $CaCO_{3}$ in a $10 \ L$ closed vessel is heated to $1000 \ K$. The percentage of $CaCO_{3}$ that remains unreacted at equilibrium is:
(Given $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$)
MediumKCET 2014
View Solution$X_{2(g)} + Y_{2(g)} \rightleftharpoons 2Z_{(g)}$
$X_{2(g)}$ and $Y_{2(g)}$ are added to a $1 \ L$ flask and it is found that the system attains the above equilibrium at $T \ K$ with the number of moles of $X_{2(g)}$,$Y_{2(g)}$ and $Z_{(g)}$ being $3$,$3$ and $9 \ mol$ respectively (equilibrium moles). Under these conditions of equilibrium,$10 \ mol$ of $Z_{(g)}$ is added to the flask and the temperature is maintained at $T \ K$. Then the number of moles of $Z_{(g)}$ in the flask when the new equilibrium is established is . . . . . . . (Nearest integer).
$X_{2(g)}$ and $Y_{2(g)}$ are added to a $1 \ L$ flask and it is found that the system attains the above equilibrium at $T \ K$ with the number of moles of $X_{2(g)}$,$Y_{2(g)}$ and $Z_{(g)}$ being $3$,$3$ and $9 \ mol$ respectively (equilibrium moles). Under these conditions of equilibrium,$10 \ mol$ of $Z_{(g)}$ is added to the flask and the temperature is maintained at $T \ K$. Then the number of moles of $Z_{(g)}$ in the flask when the new equilibrium is established is . . . . . . . (Nearest integer).
DifficultJEE MAIN 2026
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