Question

The value of Kc for the reaction 2ICl => I2 + Cl2 is it necessarily greater than the value k'c for the reaction ICl (g) => 1/2 I2 (g) + 1/2 Cl2 (g) Explain

Answer 1

2ICl <========> I2 + Cl2 Kc (equilibrium)

ICl (g) <==========> 1/2 I2 (g) + 1/2 Cl2 (g) K'c

If you multiply the first reaction by 1/2 you will get the second one.

Now, what happens with the Constants?

If you multiply by a number, the constant must be raised to the power of that number.

There are the same reactants and products, but we have different coefficients.

Then

`K^(\prime)c\text{ = }\sqrt[]{Kc}`

E.g. if Kc = 12

`K^(\prime)c\text{ = }\sqrt[]{Kc}=\text{ }\sqrt[]{12}=\text{ 3.46 (approx.)}`