A mixture of 0.600 mol of bromine and 1.600 mol of iodine is placed into a rigid 1.000-L container at 350°C.
Br2(g) + I2(g) ↔ 2IBr(g)
When the mixture has come to equilibrium, the concentration of iodine monobromide is 1.190 M. What is the equilibrium constant for this reaction at 350°C?
Show step-by step explanation.

A) 3.55 × 10^3
B) 1.24
C) 1.47
D) 282
E) 325

The term equilibrium constant refers to the number that often depict how much the process is able to turn the reactants in to products. In other words, if the reactants are readily turned into products, then it follows that the equilibrium constant will be large and positive.

Concentration of bromine = 0.600 mol /1.000-L = 0.600 M

Concentration of iodine = 1.600 mol/1.000-L = 1.600M

In this case, we must set up the ICE table as shown;

Br2(g) + I2(g) ↔ 2IBr(g)

I 0.6 1.6 0

C -x -x +2x

E 0.6 - x 1.6 - x 1.190

If 2x = 1.190

x = 1.190/2

x = 0.595

The concentrations at equilibrium are;

[Br2] = 0.6 - 0.595 = 0.005

[I2] = 1.6 - 0.595 = 1.005

Hence;

Kc = [IBr]^2/[Br2] [I2]

Kc = ( 1.190)^2/(0.005) (1.005)

Kc = 282

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