Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

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
`        VD of N2O4 which dissociates is 25.67 at 100 °c and pressure of 1 atm. calculate degree of dissociation and Kp for this.`
2 years ago

Vikas TU
8468 Points
```							Degree of dissociation,1 + a = Mthe./Mobs. => 92/(2*25.67)a = 1.791 – 1 => 0.791 Now the reaction is given as:\                      N2O4        ---------------->         2NO2        T=0         P        T=Teq.    P(1-a)                                    2PaKp = (2Pa)^2/P(1-a)     = 4P^2a^2/P(1-a)    = 4Pa^2/(1-a)put P= 1 atm and a = 0.791we get,Kp = 4*(0.791)^2/(1-0.791) => 11.9714 atm.
```
2 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Physical Chemistry

View all Questions »

### 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