If \[{{K}_{1}}\] and \[{{K}_{2}}\] are the respective equilibrium constants for the two reactions: [AIPMT 1998] |
\[Xe{{F}_{6}}(g)+{{H}_{2}}O(g)~XeO{{F}_{4}}(g)+2HF(g)\] |
\[Xe{{O}_{4}}(g)+Xe{{F}_{6}}(g)~XeO{{F}_{4}}(g)+Xe{{O}_{3}}{{F}_{2}}(g)\] |
the equilibrium constant of the reaction |
\[Xe{{O}_{4}}(g)+2HF(g)~Xe{{O}_{3}}{{F}_{2}}(g)+{{H}_{2}}O(g)\]will be : |
A) \[{{K}_{2}}/{{({{K}_{2}})}^{2}}\]
B) \[{{K}_{1}}\,.\,{{K}_{2}}\]
C) \[{{K}_{1}}/{{K}_{2}}\]
D) \[{{K}_{2}}/{{K}_{1}}\]
Correct Answer: D
Solution :
....(i) |
...(ii) |
For reaction, |
From Eqs. (i) and (ii) |
You need to login to perform this action.
You will be redirected in
3 sec