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
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 concentra...
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