How many moles of the iodine remain unreacted at equilibrium?

0.388

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.112

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.25

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.125

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 0.112
The equilibrium number of moles of hydrogen and iodine will be $(0.5 - x)$ each respectively.
The equilibrium number of moles of HI will be 2x.
The expression for the equilibrium constant will be $49 = (\frac{2 x}{0.50 - x})^{2}$ .
Hence, x=0.388.
At equilibrium, the number of moles of iodine unreacted is $0.50 - 0.388 = 0.112 m o l e s .$

Suggest Corrections

Similar questions

How many moles of iodine are liberated when 1 mole of potassium dichromate reacts with potassium iodide?

2

K M n O_{4}

K I

H S C N

I^{-}

10 ml

0.2 M K M n O_{4}

1

Join BYJU'S Learning Program