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
        when 3 mol of a and 1 mole of b are mixed in a 1 litre vessel the following reaction takes place a+b→2c    the equilibrium mixture contains 0.5 moles of c. what is the value of equilibrium constant for the reaction
10 months ago

Answers : (1)

Arun
14829 Points
							
Note: we should always put the concentratin with their unit/(3-x)*(1-x)Put here the value of x and calculate K2/[A]*[B]=(2x)2please follow the answers and try to understand it properly, if still having some doubts then don’t hesitate in posting on the forum.in the given question, volume=1L, so initial concentration of A=3mol/1L=3MSimilarly for B=1MNow,A(g) + B(g)= 2C(g)3mol/L.....1mol/L..0.......(initially)3-x..........1-x …......2x...at equilibriumBut given that, mole of C=0.5 at equilibrium, Hence its equilibrium concentration=0.5/1L=0.5M=2x(from our equation)So x=0.25Mwriting for the equilibrium constantK=[C]  RegardsArun (askIITians forum expert)
10 months ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Get Extra Rs. 1,590 off

COUPON CODE: SELF10


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


Get Extra Rs. 276 off

COUPON CODE: SELF10

Ask Experts

Have any Question? Ask Experts

Post Question

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