Question

Question 7.22 Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:

2BrCl(g) <---> Br2(g) + Cl2(g)

for which Kc= 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 × 10–3 molL–1, what is its molar concentration in the mixture at equilibrium?

Class XI Equilibrium Page 226

Answer 1

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 2BrCl(g) == Br2(g) + Cl2(g) ; Kc = 32 \: at \: 500K \\ at \: t = 0 \: \: \: 3.3 \times {10}^{ - 3} mol \: \: \: \: \: \: 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0 \\ at \: eqlm \: \: (3.3 \times {10}^{ - 3} - x) \: \: \: \: \: \: \frac{x}{2} \: \: \: \: \: \: \: \: \: \: \frac{x}{2}$
we know,



$K_c=\frac{[Br_2]\:[Cl_2]}{[BrCl_2]^2} \\ \\ = \frac{ \frac{x}{2} \times \frac{x}{2} }{(3.3 \times {10}^{ - 3} - x)^2} = 32 \\ \\ = \frac{ {x}^{2} }{4 \times {(3.3 \times {10}^{ - 3} - x) }^{2} } = 32$
take square root both sides,
x/2(3.3 × 10^-3 - x) = √32
x/(3.3 × 10^-3 - x) = 5.656 × 2 = 11.312
x = 11.312(3.3 × 10^-3 - x )
= 0.03732 - 11.312x
12.312x = 0.03732
x = 3.032 × 10^-3 mol/L

hence, molar concentration of BrCl = (3.3 × 10^-3 - x ) mol/L
= (3.3 × 10^-3 - 3.032 × 10^-3) mol/L
= 2.68 × 10^-4 mol/L