If K1 and K2 are the respective equilibrium constants for the two reactions, XeF6(g)+H2O(g)⇌XeOF4(g)+2HF(g)XeO4(g)+XeF6(g)⇌XeOF4(g)+XeO3F2(g)The equilibrium constant for the reaction, XeO4(g)+2HF(g)⇌XeO3F2(g)+H2O(g) is

K1=XeOF4[HF]2XeF6H2O......(i)K2=XeOF4XeO3F2XeO4XeF6....(ii)Bv Eq. (ii)/(i) we have K2K1=XeO3F2H2OXeO4[HF]2=Kc

If K1 and K2 are the respective equilibrium constants for the two reactions, XeF6(g)+H2O(g)⇌XeOF4(g)+2HF(g)XeO4(g)+XeF6(g)⇌XeOF4(g)+XeO3F2(g)The equilibrium constant for the reaction, XeO4(g)+2HF(g)⇌XeO3F2(g)+H2O(g) is

K1=XeOF4[HF]2XeF6H2O......(i)K2=XeOF4XeO3F2XeO4XeF6....(ii)Bv Eq. (ii)/(i) we have K2K1=XeO3F2H2OXeO4[HF]2=Kc

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If K₁ and K₂ are the respective equilibrium constants for the two reactions,
$\begin{array}{l} {XeF}_{6} (g) + H_{2} O (g) ⇌ {XeOF}_{4} (g) + 2 HF (g) \\ {XeO}_{4} (g) + {XeF}_{6} (g) ⇌ {XeOF}_{4} (g) + {XeO}_{3} F_{2} (g) \end{array}$
The equilibrium constant for the reaction, ${XeO}_{4} (g) + 2 HF (g) ⇌ {XeO}_{3} F_{2} (g) + H_{2} O (g)$ is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$K_{1} / K_{2}$

$K_{1} K_{2}$

$K_{1} / K_{2}^{2}$

$K_{2} / K_{1}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} K_{1} = \frac{[{XeOF}_{4}] [HF]^{2}}{[{XeF}_{6}] [H_{2} O]} . . . . . . (i) \\ K_{2} = \frac{[{XeOF}_{4}] [{XeO}_{3} F_{2}]}{[{XeO}_{4}] [{XeF}_{6}]} . . . . (ii) \end{array}$
Bv Eq. (ii)/(i) we have $\frac{K_{2}}{K_{1}} = \frac{[{XeO}_{3} F_{2}] [H_{2} O]}{[{XeO}_{4}] [HF]^{2}} = K_{c}$