wiz-icon
MyQuestionIcon
MyQuestionIcon
13
You visited us 13 times! Enjoying our articles? Unlock Full Access!
Question

Four moles of hydrogen iodide were taken in a 10 litre flask kept at 800K. When equilibrium was attained, the mixture was found to contain 0.42 mole of iodine. Calculate the equilibrium constant for dissociation of HI.

Open in App
Solution

2HIH2+I2
Four moles of hydrogen iodide were present initially. 0.42 mole of iodine were formed from 2×0.42=0.84 moles of HI. 0.42 moles of hydrogen were also formed. 40.84=3.16 moles of HI were present at equilibrium.

The equilibrium constant Kc=[HI]2[H2][I2]
Kc=[ 3.16 mol 10 L ]2[ 0.42 mol 10 L ][0.42 mol 10 L ]
Kc=[0.316]2[0.042][0.042]
Kc=56.6

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