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A 2:1 mixture by volume of SO 2 and O 2 at an initial pressure of 15atm is allowed to reach equilibirum at 250°C. 2SO 2 (g) + O 2 (g) 2SO 3 (g) The equilibirum pressure was found to be 11atm. (i) Calculate the partial pressure of each gas at equilibrium. How to do this question???

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???

Grade:11

1 Answers

Srinivas Rao
11 Points
11 years ago

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

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