Question

1. Consider the following reaction: Br2(g)+Cl2(g)⇌2BrCl(g) Kp=1.11×10−4 at 150 K. A reaction mixture initially contains a Br2 partial pressure of 750. torr and a Cl2 partial pressure of 735 torr at 150 K.

Calculate the equilibrium partial pressure of BrCl in atm and torr. (State any assumptions).

Answer 1

Br2(g)         +         Cl2(g)        ⇌         2BrCl(g)
I:           750                                735                           0
C:         - x                      - x                            2x
E:      750 – x                 735 – x                      2x

So

Kp = (2x)² / [ (750 - x)*(735 - x) ]

1.11 x 10⁻⁴ = 4x² / [ (750 - x)*(735 - x) ]

Now, since Kp is small, let's initially assume that x will be small compared to the numbers in the denominator. Then,

1.11 x 10⁻⁴ = 4x² / (750 * 735)

4x² = (1.11 x 10⁻⁴) (750 * 735)

4x² = 61.19

x² = 15.3

x = 3.91

So, equilibrium partial pressure of BrCl = 2x

                                                               = 2 x 3.91

                                                               = 7.82 torr

Now,

1 torr = 0.00131579 atm

7.82 torr = 7.82 x 0.00131579 atm

= 0.0102 atm