Question

For the equilibrium 2IBr(g)?I2(g)+Br2(g) Kp=8.5×10?3 at 150 ?C. If 2.3×10?2 atm of IBr is placed in...

For the equilibrium 2IBr(g)?I2(g)+Br2(g) Kp=8.5×10?3 at 150 ?C. If 2.3×10?2 atm of IBr is placed in a 2.0-L container, what is the partial pressure of IBr after equilibrium is reached?

Homework Answers

Answer #1

Sol:-

For the equilibrium

..............2IBr(g) <------------------> I2 (g) ...........+.................Br2 (g)

I........... 2.3×10-2 atm........................0 atm............................0 atm

C ......... - x .......................................+ x/2 ............................ +x/2

E ......(2.3×10?2 - x) atm ................... x/2 ...............................x/2

Now expression of Kp is :

Kp = P I2(g) . PBr2 (g) / (P IBr(g) )2

8.5 X10-3 = x2 / 4 X (2.3×10?2 - x)

on cross multiplication we have

x2 + 0.034 x - 0.000782 = 0

on solving by quadratic formula , we have

x = 0.01573

therefore partial pressure of IBr at equilibrium = PIBr = 2.3×10?2 - x = 0.023 - 0.01573 = 0.00727 atm .

Hence partial pressure of IBr at equilibrium =  0.00727 atm

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