1. Tardigrade
  2. Question
  3. Chemistry
  4. If K1 and K2 are the respective equilibrium constants for the two reactions, XeF6(g)+H2O(g) leftharpoons XeOF4(g)+2HF(g) XeO4(g)+XeF6(g) leftharpoons XeOF4(g)+ XeO3F2(g) The equilibrium constant of the reaction, XeO4(g)+2HF(g) leftharpoons XeO3F2(g)+H2O(g)

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. If $K_1$ and $K_2$ are the respective equilibrium constants for the two reactions,
$ XeF_6(g)+H_2O(g) \rightleftharpoons XeOF_4(g)+2HF(g)$
$ XeO_4(g)+XeF_6(g) \rightleftharpoons XeOF_4(g)+ XeO_3F_2(g)$
The equilibrium constant of the reaction,
$XeO_4(g)+2HF(g) \rightleftharpoons XeO_3F_2(g)+H_2O(g)$

IIT JEEIIT JEE 1998Aldehydes Ketones and Carboxylic Acids

Solution:

$XeF_6(g)+H_2O(g) \rightleftharpoons XeOF_4(g)+2HF(g)$
$K_1=\frac {[XeOF_4][HF]^2}{[XeF_6][H_2O]} $ ...(i)
$XeO_4(g)+XeF_6(g) \rightleftharpoons XeOF_4(g)+ XeO_3F_2(g)$
$K_2=\frac {[XeOF_4][XeO_3F_2]}{[XeO_4][XeF_6]}$ ...(ii)
For the reaction,
$ XeO_4(g)+2HF(g) \rightleftharpoons XeO_3F_2(g)+H_2O(g) $ $K=\frac {[XeO_3F_2][H_2O]}{[XeO_4][HF]^2}$ ...(iii)
$\therefore $ From Eqs. (i), (ii) and (iii)
$K=\frac {K_2}{K_1}$