Class 11Chemistry6-1.Equilibrium (Chemical Equilibrium)Mix Examples- 6-1.Equilibrium (Chemical Equilibrium)
MediumBromine monochloride,$BrCl$ decomposes into bromine and chlorine and reaches the equilibrium:
$2 BrCl_{(g)} \longleftrightarrow Br_{2(g)} + Cl_{2(g)}$
for which $K_c = 32$ at $500 \ K$.
If initially pure $BrCl$ is present at a concentration of $3.3 \times 10^{-3} \ mol \ L^{-1}$,what is its molar concentration in the mixture at equilibrium?
$2 BrCl_{(g)} \longleftrightarrow Br_{2(g)} + Cl_{2(g)}$
for which $K_c = 32$ at $500 \ K$.
If initially pure $BrCl$ is present at a concentration of $3.3 \times 10^{-3} \ mol \ L^{-1}$,what is its molar concentration in the mixture at equilibrium?
(D) Let the amount of $Br_2$ and $Cl_2$ formed at equilibrium be $x \ mol \ L^{-1}$.
The reaction is: $2 BrCl_{(g)} \longleftrightarrow Br_{2(g)} + Cl_{2(g)}$
Initial concentration: $3.3 \times 10^{-3} \ M$,$0$,$0$
Equilibrium concentration: $(3.3 \times 10^{-3} - 2x)$,$x$,$x$
$K_c = \frac{[Br_2][Cl_2]}{[BrCl]^2} = 32$
$\frac{x^2}{(3.3 \times 10^{-3} - 2x)^2} = 32$
Taking the square root on both sides: $\frac{x}{3.3 \times 10^{-3} - 2x} = \sqrt{32} \approx 5.657$
$x = 5.657 \times (3.3 \times 10^{-3} - 2x)$
$x = 18.668 \times 10^{-3} - 11.314x$
$12.314x = 18.668 \times 10^{-3}$
$x \approx 1.516 \times 10^{-3} \ mol \ L^{-1}$
At equilibrium,$[BrCl] = 3.3 \times 10^{-3} - 2x$
$[BrCl] = 3.3 \times 10^{-3} - 2(1.516 \times 10^{-3})$
$[BrCl] = 3.3 \times 10^{-3} - 3.032 \times 10^{-3} = 0.268 \times 10^{-3} \ mol \ L^{-1} \approx 2.7 \times 10^{-4} \ mol \ L^{-1}$
The reaction is: $2 BrCl_{(g)} \longleftrightarrow Br_{2(g)} + Cl_{2(g)}$
Initial concentration: $3.3 \times 10^{-3} \ M$,$0$,$0$
Equilibrium concentration: $(3.3 \times 10^{-3} - 2x)$,$x$,$x$
$K_c = \frac{[Br_2][Cl_2]}{[BrCl]^2} = 32$
$\frac{x^2}{(3.3 \times 10^{-3} - 2x)^2} = 32$
Taking the square root on both sides: $\frac{x}{3.3 \times 10^{-3} - 2x} = \sqrt{32} \approx 5.657$
$x = 5.657 \times (3.3 \times 10^{-3} - 2x)$
$x = 18.668 \times 10^{-3} - 11.314x$
$12.314x = 18.668 \times 10^{-3}$
$x \approx 1.516 \times 10^{-3} \ mol \ L^{-1}$
At equilibrium,$[BrCl] = 3.3 \times 10^{-3} - 2x$
$[BrCl] = 3.3 \times 10^{-3} - 2(1.516 \times 10^{-3})$
$[BrCl] = 3.3 \times 10^{-3} - 3.032 \times 10^{-3} = 0.268 \times 10^{-3} \ mol \ L^{-1} \approx 2.7 \times 10^{-4} \ mol \ L^{-1}$
Explore More
Similar Questions
Two solids dissociate as follows:
$A_{(s)} \rightleftharpoons B_{(g)} + C_{(g)}$; $K_{p_1} = x \, atm^2$
$D_{(s)} \rightleftharpoons C_{(g)} + E_{(g)}$; $K_{p_2} = y \, atm^2$
The total pressure when both the solids dissociate simultaneously is:
$A_{(s)} \rightleftharpoons B_{(g)} + C_{(g)}$; $K_{p_1} = x \, atm^2$
$D_{(s)} \rightleftharpoons C_{(g)} + E_{(g)}$; $K_{p_2} = y \, atm^2$
The total pressure when both the solids dissociate simultaneously is:
DifficultJEE MAIN 2019
View Solution$28 \, g$ of $N_2$ and $6 \, g$ of $H_2$ were kept at $400 \, ^oC$ in a $1 \, L$ vessel. The equilibrium mixture contained $27.54 \, g$ of $NH_3$. The approximate value of $K_c$ for the reaction $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ is (in $L^2 \, mol^{-2}$):
DifficultAIPMT 1990
View SolutionDissociation of a gas $A_2$ takes place according to the following chemical reaction. At equilibrium, the total pressure is $1 \ bar$ at $300 \ K$.
$A_{2(g)} \rightleftharpoons 2A_{(g)}$
The standard Gibbs energy of formation of the involved substances has been provided below:
Substance $\Delta G_f^{\circ} / kJ \ mol^{-1}$ $A_2$ $-100.00$ $A$ $-50.832$
The degree of dissociation of $A_{2(g)}$ is given by $(x \times 10^{-2})^{1/2}$ where $x =$ . . . . . . . (Nearest integer).
[Given: $R = 8.3 \ J \ mol^{-1} \ K^{-1}$, $\ln 2 = 0.693$]
$A_{2(g)} \rightleftharpoons 2A_{(g)}$
The standard Gibbs energy of formation of the involved substances has been provided below:
| Substance | $\Delta G_f^{\circ} / kJ \ mol^{-1}$ |
| $A_2$ | $-100.00$ |
| $A$ | $-50.832$ |
The degree of dissociation of $A_{2(g)}$ is given by $(x \times 10^{-2})^{1/2}$ where $x =$ . . . . . . . (Nearest integer).
[Given: $R = 8.3 \ J \ mol^{-1} \ K^{-1}$, $\ln 2 = 0.693$]
DifficultJEE MAIN 2026
View SolutionConsider the following gas phase reaction being carried out in a closed vessel at $25^\circ\text{C}$: $2A(g) \rightarrow 4B(g) + C(g)$. The table provides the total pressure of the system at different time intervals. Calculate the pressure of $C(g)$ at $30$ minutes time interval.
Time (min) Total pressure (mm Hg) $30$ $300$ $\infty$ $600$
| Time (min) | Total pressure (mm Hg) |
| $30$ | $300$ |
| $\infty$ | $600$ |
DifficultJEE MAIN 2026
View SolutionReaction between $N_{2}$ and $O_{2}$ takes place as follows:
$2 N_{2(g)} + O_{2(g)} \longleftrightarrow 2 N_{2}O_{(g)}$
If a mixture of $0.482 \ mol$ of $N_{2}$ and $0.933 \ mol$ of $O_{2}$ is placed in a $10 \ L$ reaction vessel and allowed to form $N_{2}O$ at a temperature for which $K_{c} = 2.0 \times 10^{-37}$,determine the composition of the equilibrium mixture.
$2 N_{2(g)} + O_{2(g)} \longleftrightarrow 2 N_{2}O_{(g)}$
If a mixture of $0.482 \ mol$ of $N_{2}$ and $0.933 \ mol$ of $O_{2}$ is placed in a $10 \ L$ reaction vessel and allowed to form $N_{2}O$ at a temperature for which $K_{c} = 2.0 \times 10^{-37}$,determine the composition of the equilibrium mixture.
Difficult
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