CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

I2+II3. This reaction is set up in aqueous medium. We start with 1 mole of I2 and 0.5 mol of I in 1 L flask. After equilibrium, the excess of AgNO3 gave 0.25 moles of yellow ppt. Then the equilibrium constant is:

A
1.33
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
2.66
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2.00
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
3.00
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 1.33
I+AgNO3AgI+NO3
0.25 moles of yellow precipitate corresponds to 0.25 moles of iodide ions at equilibrium.
0.50.25=0.25 moles of iodide ions have reacted with 0.25 moles of iodine to form 0.25 moles of iodate ions.
10.25=0.75 moles of iodine remains at equilibrium.

Since total volume is 1 L, the number of moles is equal to molar concentration.
K=[I3][I2][I]=0.250.75×0.25=1.33

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