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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.

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.

Grade:11

1 Answers

Chetan Mandayam Nayakar
312 Points
10 years ago

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

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