For the reaction $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$ at $298 \ K$,the $K_c$ is $7 \times 10^{25}$. Calculate $K_c$ for the reaction $SO_{3(g)} \rightleftharpoons SO_{2(g)} + \frac{1}{2}O_{2(g)}$.

  • A
    $1.195 \times 10^{-13}$
  • B
    $1.428 \times 10^{-26}$
  • C
    $3.78 \times 10^{-13}$
  • D
    $7.0 \times 10^{-25}$

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:

In the gas phase reaction,$C_2H_4 + H_2 \rightleftharpoons C_2H_6$,the equilibrium constant can be expressed in units of

Consider the equilibrium,$H_2 + I_2 \rightleftharpoons 2 HI$. Calculate the equilibrium constant of the reverse reaction when the equilibrium concentrations of $H_2$,$I_2$,and $HI$ are $1.14 \times 10^{-2} \ mol \ L^{-1}$,$0.12 \times 10^{-2} \ mol \ L^{-1}$,and $2.52 \times 10^{-2} \ mol \ L^{-1}$,respectively.

For the reaction $CO_{(g)} + 1/2 O_{2(g)} \rightleftharpoons CO_{2(g)}$,the value of $K_p/K_c$ is .....

$A + B \rightleftharpoons C + D$. If the final equilibrium concentrations of $A$ and $B$ are equal,and the equilibrium concentration of $D$ is twice that of $A$,what is the equilibrium constant $(K_c)$ of the reaction?

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