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XY2 dissociates as, XY2(g)= XY(g) + Y(g) (reversible reacyion)..Initial pressure of XY2 is 600 mm Hg.THe total pressure at equlibrium is 800 mm hg.Assuming volume of system to reamin constant ,the value of Kp is.?

XY2 dissociates as, XY2(g)= XY(g) + Y(g) (reversible reacyion)..Initial pressure of XY2 is 600 mm Hg.THe total pressure at equlibrium is 800 mm hg.Assuming volume of system  to reamin constant ,the value of Kp is.?

Grade:12

4 Answers

Naveen Kumar
askIITians Faculty 60 Points
9 years ago
XY2(g).............= …...........XY(g) + Y(g)
600mm.............................0...........0......(initially)
600-p..................................p..........p....(at equilibrium)
[Here I have used directely the partial pressure because as the volume is constant, so number of moles of reqactant or product∝ partial pressure]
Now applying Dalton’s law of partial ppressure at equilibrium,( total pressure=sum of partial pressure of all the gas components in the mixture)
800=(600-p) + p+ p=600+p
p=200mm of Hg
writing Kp=p*p/(600-p);
put value of p and change the unit according to :105 Pa=760mm of Hg
Ramreddy IIT Patna
askIITians Faculty 49 Points
9 years ago
Kp= (pXY) x (pY)/ (pXY2)
XY2 ---------> XY + Y
600 mm 0 0
600 -x x x
Total pressure at equillibrium = 600+x = 800
x= 200
Kp= (200 x 200) / 600
..............................................................................................................
Sahil Saha
12 Points
6 years ago
XY2=XY+Y
[XY2] at equilibrium=600 atm
Total Pressure= 800 atm
Kp=[XY][Y]/[XY2]
[XY]=[Y]=x (let)
Total pressure at equilibrium = 600+x=800
x=200
Kp=200x200/600
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the solution to your question.
 
Kp= (pXY) x (pY)/ (pXY2)
XY2 ---------> XY + Y
600 mm       0      0
600 – x        x      x
Total pressure at equillibrium = 600 + x = 800
x = 200
Kp = (200 x 200) / 600 = 66.67 mm = 8772 Pa
 
Thanks and regards,
Kushagra

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