Question

2IBrig), K.-280 at 150°C. Suppose that 0.500 mol IBr in a 1.00 1(g) + Bn(g) 4. For the reaction L flask is allowed to reach equilibrium at equilibrium at 150°C. What are the equilibrium concentrations of IBr, h, and Br2?

Answer 1

Sol.

As initial conc. of IBr = initial moles of IBr / Volume of flask

= 0.5 / 1 = 0.5 M  

Now , Reaction :

I2 + Br2 <----> 2IBr

initial 0 0 0.5

change + x + x - 2x

equilibrium x x (0.5 - 2x)

So , Kc = [IBr]² / ( [I2] × [Br2] )

280 = (0.5 - 2x)² / x²

or , (0.5 - 2x) / x = (280)^1/2 = 16.7332

0.5 - 2x = 16.7332 x

18.7332 x = 0.5

x = 0.5 / 18.7332 = 0.02669

Therefore , equilibrium conc. of all species are :

[I2] = [Br2] = x = 0.02669 M

[IBr] = 0.5 - 2x = 0.5 - 2 × 0.02669 = 0.44662 M