Chemistry equilibrium - Remove H2

[il postino]

Homework Statement

It is desired to remove the molecular hydrogen present in a flask. What will be better to introduce? Cl2 or Br2?

Relevant Equations

kc

It is desired to remove the molecular hydrogen present in a flask. What will be better to introduce, $Cl_2$ or $Br_2$? Why?

$2 HCl <-> H_2 + Cl_2$ (1)
$k = 3,2.10^{-14}$

$H_2 + Br_2 <->2 HBr$ (2)
$k = 2.10^9$

I thought about turning equation (1) to make the comparison better:

$H_2 + Cl_2 <-> 2 HCl$ (1)
$k = 1/3,2.10^{-14} = 3,1.10^{13}$

$H_2 + Br_2 <->2 HBr$ (2)
$k = 2.10^9$

Then I reasoned the following:
If equation 1 has a higher value of Kc, then it will be the most favorable, therefore it is better to enter $Cl_2$

Is the reasoning correct?
What do you think?
Thank you!

 

[Borek]

Sounds OK to me.

 

[shjacks45]

Cl2. Because Cl2 will react with H2 at STP and Bromine will not. Light induces Cl2 to become Cl free radicals which initiates chain reaction. Boom. Hydrogen molecules have 104 kcal/mole bond energy which is pretty high activation energy to start reaction.