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

Total Price: Rs.

There are no items in this cart.
Continue Shopping
Mohamed Faisal Grade: 11

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.


How to do this question???

7 years ago

Answers : (1)

Srinivas Rao
11 Points

SO2 and O2 are in 2:1 ratio by volume and total pressure is 15 atm. So, Pressure of SO2 is 15 x (2/3) = 10 atm; Pressure of O2 is 15 x (1/3) = 5 atm

2 X moles of SO2 combines with X moles of O2 to form 2X moles of SO3. So, the total decrease in number of moles is X.

From the given data, X is (15-11) = 4

At equilibrium, partial pressure of SO2 is 10-2X= 10 – 8 = 2 atm

At equilibrium, partial pressure of O2 is 5-X= 5 – 4 = 1 atm

At equilibrium, partial pressure of SO3 is 2X= 8 atm

7 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete JEE Main/Advanced Course and Test Series
  • OFFERED PRICE: Rs. 15,900
  • View Details

Ask Experts

Have any Question? Ask Experts

Post Question

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