Question

For the equilibrium
2IBr(g)?I2(g)+Br2(g) Kp=8.5×10?3 at 150 ?C.

A) If 2.1×10^?2 atm of IBr is placed in a 2.0-L container, what is the partial pressure of IBr after equilibrium is reached?

Express your answer to two significant figures and include the appropriate units.

B) If 2.1×10^?2 atm of IBr is placed in a 2.0-L container, what is the partial pressure of I2 after equilibrium is reached?

Express your answer to two significant figures and include the appropriate units.

C) If 2.1×10^?2 atm of IBr is placed in a 2.0-L container, what is the partial pressure of Br2 after equilibrium is reached?

Express your answer to two significant figures and include the appropriate units.

Answer 1

2IBr(g)          I₂(g)+Br₂(g)   , Here Kp = 8.5 x 10^-3 at 150 ^oC.

                            2IBr(g)           I₂(g)         +        Br₂(g)

Initial                (2.1x10^-2)                        0                          0

Equilibrium       (2.1x10^-2)-2x                 x                           x       

Now,

Kp = [P(I₂) x P(Br₂)] / (P_IBr)²

8.5 x 10^-3 = [P(I₂) x P(Br₂)] / (P_IBr)²

8.5 x 10^-3 = x² / (2.1 x 10^-2 - 2x )²

On solving,

0.092195 = x / [(0.021 -2x)]

12.846 x = 0.021

Therefore,

x = 0.021/12.846 = 0.00163

On substituting the value of x,

P(IBr) = 2.1 x 10^-2 - 2x = 0.021 –[2 x 0.00163] = 0.01773 atm

P(IBr) = 0.01773 atm

P(Br₂) = 0 + x = 0.00163 atm

P(I₂) = 0 + x = 0.00163 atm