MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: 11
        Two moles HI were heated in a sealed tube at 440C till the equilibrium was reached.HIwas found to be 22% decomposed.The equilibrium constant for dissociation is?
2 years ago

Answers : (1)

Arun
23045 Points
							
Dear Student
 
2HI(g) ⇐⇒ H2(g) + I2(g) 

K = [H2][ I2] / [HI]^2 
22% of 2 moles of HI decompose = 0.22(2) = 0.44 
0.44 moles HI x 1 mole H2 / 2 moles HI = 0.22 
moles of I2 = moles of H2 
moles of HI remaining = 2.0 – 0.44 = 1.56 
The 
K = 0.22(0.22) / (1.56)^2 
K = 0.0199
 
Regards
Arun (askIITians forum expert)
2 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 141 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details