$K_{c}$ for the reaction $A_{2(g)} \rightleftharpoons B_{2(g)}$ is $39.0$. In a closed one litre flask,one mole of $A_{2(g)}$ was heated to $T \ K$. What are the concentrations of $A_{2(g)}$ and $B_{2(g)}$ (in $mol \ L^{-1}$) respectively at equilibrium?

  • A
    $0.025, 0.975$
  • B
    $0.975, 0.025$
  • C
    $0.05, 0.95$
  • D
    $0.02, 0.98$

Explore More

Similar Questions

For the reversible reaction,$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ at $500 \ ^oC$,the value of $K_P$ is $1.44 \times 10^{-5}$ when partial pressure is measured in atmospheres. The corresponding value of $K_c$ with concentration in $\text{mol L}^{-1}$ is:

For the reaction $2NOCl_{(g)} \rightleftharpoons 2NO_{(g)} + Cl_{2_{(g)}}$ at $1060 \ K$,the equilibrium constant $K_p$ is $0.033 \ atm$. Find the value of $K_c$. (Given $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$)

Calculate $K_{C}$ for the reversible process given below if $K_{P}=167$ and $T=800^{\circ}C$.
$CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$

For the reaction ${N_2}_{(g)} + {O_2}_{(g)} \rightleftharpoons 2NO_{(g)}$,the equilibrium constant ${K_C}$ is $4 \times 10^{-4}$ at temperature $T$. Calculate ${K_P}$.

For a reaction $aA_{(g)} \rightleftharpoons bB_{(g)}$ at equilibrium,the heat of reaction at constant volume is $1500 \text{ cal}$ more than that at constant pressure. If the temperature is $27^\circ\text{C}$,then:

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