Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
Close
MY CART (5)

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

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

⇐ BackProceed to Checkout ⇒
There are no items in this cart.
Continue Shopping

the equilibrium constant kp for the following reaction at 191 degree celcius is 1.24. what is kc ? b(s) + 3/2f 2(g) = BF 3

the equilibrium constant kp for the following  reaction at 191 degree celcius is 1.24. what is kc ? 
               b(s) + 3/2f2(g) = BF3

Grade:12

1 Answers

Arun
25763 Points
3 years ago
Dear Rahul
 
Use formula
 
Kp = Kc * (RT)^(delta n)
 
Now I hope you can do it. 
 
In case of any difficulty, Please feel free to ask.

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

Other Related Questions on Physical Chemistry

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Register ICAT

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More