The reaction for which $K_{c} = 2.3 \times 10^3 \ mol \ L^{-1}$ is $-$

  • A
    $2 SO_{2(g)} + O_{2(g)} \rightleftharpoons 2 SO_{3(g)}$
  • B
    $N_{2(g)} + 3 H_{2(g)} \rightleftharpoons 2 NH_{3(g)}$
  • C
    $N_{2(g)} + O_{2(g)} \rightleftharpoons 2 NO_{(g)}$
  • D
    $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$

Explore More

Similar Questions

The equilibrium $NH_4HS_{(s)} \rightleftharpoons NH_{3(g)} + H_2S_{(g)}$ is set up at $127 \,^oC$ in a closed vessel. The total pressure at equilibrium was $20 \,atm$. The $K_C$ for the reaction is: (in $,M^2$)

For the formation of ammonia from its constituent elements ($1 \ mol$ of $N_2$ and $3 \ mol$ of $H_2$) in a closed vessel of volume $V \ L$,the value of $K_C$ is [units of $K_C = mol^{-2} \ L^2$].

Consider the equilibria $(i)$ and $(ii)$ with equilibrium constants $K_1$ and $K_2$,respectively.
$SO_{2(g)} + 1/2 O_{2(g)} \rightleftharpoons SO_{3(g)} ..... (i)$
$2 SO_{3(g)} \rightleftharpoons 2 SO_{2(g)} + O_{2(g)} ..... (ii)$
$K_1$ and $K_2$ are related as

In a closed container of $1000\, cm^3$,$2\, mol$ of $PCl_5$,$2\, mol$ of $PCl_3$,and $3\, mol$ of $Cl_2$ are found to be at equilibrium at $27\, ^oC$. Then $K_P$ for the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ at $27\, ^oC$ is $.....$ $atm$.

For the following gaseous reaction $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$,the equilibrium constant relationship is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo