wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

K1 and K2 are the respective equilibrium constant for the reactions:
(i) XeF6(g)+H2O(g)XeOF4(g)+2HF(g);K1
(ii) XeO4(g)+XeF6(g)XeOF4(g)+XeO1F2;K2
The equilibrium constant for the reaction
XeO4(g)+2HF(g)XeO3F2(g)+H2O(g) will be:

A
K1K22
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
K1K2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
K1K2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
K2K1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D K2K1
XeF6(g)+H2O(g)XeOF4(g)+2HF(g)
K1=[XeOF4][HF]2[XeF6][H2O](1)
XeO4(g)+XeF6(g)XeOF4(g)+XeO3F2(g)
K2=[XeO3F2][XeOF4][XeO4][XeF6](2)
Now for XeO4(g)+2HF(g)XeO3F2(g)+H2O(g)
K3=[XeO3F1][H2O][XeO4][HF]2(3)
Now, K2K1=[XeO3F2][XeOF4][XeO4][XeF6]×[XeF6][H2O][XeOF4][HF]2
K2K1=K3

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Equilibrium Constants
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon