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
```        A 2:1 mixture by volume of SO2 and O2 at an initial pressure of 15atm is allowed to reach equilibirum at 250'C.
2SO2(g) + O2(g) <-> 2SO3(g)
The equilibirum pressure was found to be 11atm.
(i) Calculate the partial pressure of each gas at equilibrium.```
7 years ago

Chetan Mandayam Nayakar
312 Points
```										let initial partial pressures be p1,p2,p3.
p1/p2 = 2/1, thus p1=10atm,p2=5atm and p3=zero.
initial number of moles= 2n,n,0.
let x moles of O2 react. 3n/(3n-x)=15/11, n/x=5/4
p(SO2)=((2n-2x)/2n)*10 = 2atm
p(O2) =((n-x)/2n)*10 = 1atm
p(SO3) = 11-2-1 = 8atm
```
7 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Physical Chemistry

View all Questions »
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details