Question

For the equilibrium Br2(g) + Cl2(g) ⇌ 2 BrCl(g) the equilibrium constant Kp is 7.0 at 400 K. If a cylinder is charged with BrCl(g) at an initial pressure of 1.00 atm and the system is allowed to come to equilibrium what is the final (equilibrium) pressure of BrCl? For the equilibrium the equilibrium constant is 7.0 at 400 . If a cylinder is charged with at an initial pressure of 1.00 and the system is allowed to come to equilibrium what is the final (equilibrium) pressure of BrCl?
a. 0.31 atm b. 0.15 atmc. 0.57 atmd. 0.22 atme. 0.45 atm

Answer 1

Answer:

c. 0.57 atm

Explanation:

For the equilibrium:

Br₂(g)    +    Cl₂(g)      ⇄    2BrCl (g)

Let construct the ICE table for this reaction.

                            Br₂(g)          +           Cl₂(g)          ⇄           2BrCl (g)

Initial                     0                              0                                  1

Change                 +x                           +x                                - 2x

Final                      x                               x                                1 - 2x

$K_p = 7$

$K_p= \frac{[BrCl_2]^2}{[Br_2][Cl_2]}$

$7 = \frac{[1-2x]^2}{[x][x]}$

$7 = \frac{[1-2x]^2}{[x]^2}$

$7x^2=(1-2x)(1-2x)$

$7x^2=1-2x-2x+4x^2$

$7x^2=1-4x+4x^2$

$7x^2-4x^2=1-4x$

$3x^2=1-4x$

$3x^2+4x-1 = 0$      ----------(quadratic equation)

using the formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

where ± represent +/-

a =3 ; b = 4 ; c = -1

$\frac{-4+/-\sqrt{4^2-4(3)(1)} }{2*3}$

$\frac{-4+/-\sqrt{16+12} }{6}$

$\frac{-4+/-\sqrt{28} }{6}$

$\frac{-4+\sqrt{28} }{6}$  or $\frac{-4-\sqrt{28} }{6}$

$\frac{-4+5.292}{6}$ or $\frac{-4-5.292}{6}$

x = 0.215 or -1.549

So, we chose the one with a positive integer which is = 0.215

BrCl = 1 - 2(x)

BrCl = 1 - 2(0.215)

BrCl = 1 - 0.43

BrCl = 0.57 atm

∴ the final (equilibrium) pressure of BrCl = 0.57 atm