One mole of $A_{(g)}$ is heated to $T(K)$ until the following equilibrium is obtained:
$A_{(g)} \rightleftharpoons B_{(g)}$
The equilibrium constant of this reaction is $10^{-1}$. After reaching the equilibrium,$0.5 \ mol$ of $A_{(g)}$ is added and heated. The equilibrium is again established. The value of $\frac{[A]}{[B]}$ is:

  • A
    $10^{-1}$
  • B
    $10$
  • C
    $10^{-2}$
  • D
    $100$

Explore More

Similar Questions

For the reaction, $2 SO_{2(g)} + O_{2(g)} \rightleftharpoons 2 SO_{3(g)}$ at $300 \ K$, the value of $\Delta G^{\circ}$ is $-690.9 R$. The equilibrium constant value for the reaction at that temperature is ($R$ is gas constant).

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

For the reaction $C_{(s)} + CO_{2_{(g)}} \rightleftharpoons 2CO_{(g)}$,the partial pressures of $CO$ and $CO_2$ at equilibrium are $2.0 \ atm$ and $4.0 \ atm$ respectively. The value of $K_p$ for the reaction is:

In which of the following equilibria,$K_p$ and $K_c$ are $NOT$ equal?

The reaction rate for the reaction $[PtCl_4]^{2-} + H_2O \rightleftharpoons [Pt(H_2O)Cl_3]^- + Cl^-$ was measured as a function of concentrations of different species. It was observed that $\frac{-d[[PtCl_4]^{2-}]}{dt} = 4.8 \times 10^{-5} [[PtCl_4]^{2-}] - 2.4 \times 10^{-3} [[Pt(H_2O)Cl_3]^-] [Cl^-]$,where square brackets are used to denote molar concentrations. The equilibrium constant $K_c = ...$. (Nearest integer)

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