In: Chemistry
Ans. Initial [I2] = Moles / Volume of reaction flask
= 0.45 mol/ 2.0 L
= 0.225 M
Initial [Cl2] = 0.45 mol/ 2.0 L = 0.225 M
# Since there is no ICl initially, the reaction would go to the left to from ICl so restore equilibrium.
That is, we have to proceed with I2 + Cl2 --------> 2ICl
When a reaction is reversed, the equilibrium constant is reciprocated.
So, new Keq’ for backward reaction = 1 / Keq for forward reaction
Or, Keq’ = 1 / 0.11 = 9.091
# Create an ICE table with the initial concentrations as shown in picture-
Now,
Equilibrium constant, keq’ = [ICl]2 / ([I2] [Cl2])
Or, 9.091 = (2X)2 / [(0.225 - X) (0.225 - X)]
Or, 9.091 x (0.050625 + X2 - 0.45X) = 4X2
Or, 0.460231875 + 9.091X2 - 4.09095X - 4X2 = 0
Or, 5.091X2 - 4.09095X + 0.460231875 = 0
Solving the quadratic equation we get following two roots-
X1 = 0.669 ; X2 = 0.135
Since X can’t be greater than 0.225, reject X1.
Therefore, X = 0.135
# [ICl] at equilibrium = 2X = 1 x 0.135 = 0.270
Hence, equilibrium [ICl] = 0.270 M