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 a 2.0-L container, what is the partial pressure of IBr after equilibrium is reached?

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) . P_{Br2 (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 = P_IBr = 2.3×10^?2 - x = 0.023 - 0.01573 = 0.00727 atm .

Hence partial pressure of IBr at equilibrium =  0.00727 atm