Question

14. Bromine iodide decomposes according to the equation: 2Brl(g) ? Br2(g)+12(g) Keq-00156 A 0.660 mol sample of Brl was injected into a 2.00 L reaction vessel. and 12 at equilibrium. Calculate the concentration of Brl, Br2,

Answer 1

2 BrI (g) Br₂ (g) + I₂ (g)

2 Brl (g) doubleheadarrow Br_2 (g) + l_2 (g) initial [Brl] = 0.660/2.00 = 0.330M [Br_2] = [l_2] = 0.000M at equilibrium [Brl] = (0.330 - 2x)M [Br_2] = [l_2] = xM Equilibrium constant expression can be written as, keq = [Br_2] [l_2]/[Brl]^2 0.0156 = x * x/(0.330 - 2x)^2 0.0156 * (0.1089 + 4x^2 - 1.32 x) = x^2 0.00170 + 0.0624x^2 - 0.00170 = 0 Applying quadartic equation to solve x, x = [- 0.0206 +/- squareroot ((0.0206)^2 - 4 * 0.9376 * (- 0.00170))/(2 * 0.9376) x = 0.0330 M Therefore, at equilibrium, [Brl] = 0.330 - 2 * 0.0330 = 0.264 M [Br_2] = [l_2] = 0.0330

Initial

[BrI] = 0.660 / 2.00 = 0.330 M

[Br₂] = [I₂] = 0.000 M

At equilibirum,

[BrI] = ( 0.330 - 2x ) M

[Br₂] = [I₂] = x M

Equilibrium constant expression can be written as,

Keq = [Br₂][I₂] / [BrI]²

0.0156 = x * x / ( 0.330 - 2x)²

0.0156 * ( 0.1089 + 4x² - 1.32 x ) = x²

0.00170 + 0.0624 x² - 0.0206 x = x²

0.9376 x² + 0.0206 x - 0.00170 = 0

Applying quadartic equation to solve x,

x = [ - 0.0206 +/- sqrt((0.0206)² - 4 * 0.9376 * ( - 0.00170 )) ] / ( 2 * 0.9376 )

x = 0.0330 M

Therefore,

at equilibrium,

[BrI] = 0.330 - 2 * 0.0330 = 0.264 M

[Br₂] = [I₂] = 0.0330