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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???

6 years ago


Answers : (1)


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

6 years ago

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