Medium
The decomposition of $N_2O_4$ to $NO_2$ is carried out at $280 \ K$ in chloroform. When equilibrium has been established,$0.2 \ mol$ of $N_2O_4$ and $2 \times 10^{-3} \ mol$ of $NO_2$ are present in $2 \ L$ solution. The equilibrium constant for the reaction $N_2O_4 \rightleftharpoons 2NO_2$ is:
- A$1 \times 10^{-2}$
- B$2 \times 10^{-3}$
- C$1 \times 10^{-5}$
- D$2 \times 10^{-5}$
Explore More
Similar Questions
For the given hypothetical reactions,the equilibrium constants are as follows:
$X \rightleftharpoons Y ; K_1=1.0$
$Y \rightleftharpoons Z ; K_2=2.0$
$Z \rightleftharpoons W ; K_3=4.0$
The equilibrium constant for the reaction $X \rightleftharpoons W$ is (in $.0$)
$X \rightleftharpoons Y ; K_1=1.0$
$Y \rightleftharpoons Z ; K_2=2.0$
$Z \rightleftharpoons W ; K_3=4.0$
The equilibrium constant for the reaction $X \rightleftharpoons W$ is (in $.0$)
If the volume of the container is $1 \ L$ and at equilibrium the amounts are $SO_3 = 48 \ g$,$SO_2 = 12.8 \ g$,and $O_2 = 9.6 \ g$,find the value of $K_c$ for the reaction $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$.
Difficult
View SolutionThe compounds $A$ and $B$ are mixed in equimolar proportion to form the products,$A + B \rightleftharpoons C + D$. At equilibrium,one-third of $A$ and $B$ are consumed. The equilibrium constant for the reaction is:
Medium
View SolutionThe equilibrium constant for the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ is $32$ at a given temperature. The equilibrium concentrations of $I_2$ and $HI$ are $0.5 \times 10^{-3} \ M$ and $8 \times 10^{-3} \ M$ respectively. The equilibrium concentration of $H_2$ is:
Medium
View SolutionConsider the following reactions in which all the reactants and products are in gaseous state:
$2PQ \rightleftharpoons P_2 + Q_2\,;\,K_1 = 2.5 \times 10^5$
$PQ + \frac{1}{2}R_2 \rightleftharpoons PQR\,;\,K_2 = 5 \times 10^{-3}$
The value of the equilibrium constant for the reaction:
$\frac{1}{2}P_2 + \frac{1}{2}Q_2 + \frac{1}{2}R_2 \rightleftharpoons PQR$ is
$2PQ \rightleftharpoons P_2 + Q_2\,;\,K_1 = 2.5 \times 10^5$
$PQ + \frac{1}{2}R_2 \rightleftharpoons PQR\,;\,K_2 = 5 \times 10^{-3}$
The value of the equilibrium constant for the reaction:
$\frac{1}{2}P_2 + \frac{1}{2}Q_2 + \frac{1}{2}R_2 \rightleftharpoons PQR$ 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