another artifact of learning by a Chemistry Tampines JC teacher who attended the EJS-OSP Singapore workshop.

http://weelookang.blogspot.com/2014/12/hydrogen-bromine-hydrogen-bromide.html Hydrogen, Bromine, Hydrogen Bromide equilibrium Model run: https://dl.dropboxusercontent.com/u/44365627/lookangEJSworkspace/export/ejss_model_EQB17/EQB17_Simulation.xhtml scr: https://dl.dropboxusercontent.com/u/44365627/lookangEJSworkspace/export/ejss_src_EQB17.zip author: Andy Luo Kangshun, Paco and Wolfgang |

## This simulation emulates the gas phase equilibria between hydrogen, bromine and hydrogen bromide molecules.

$H_{2}$ + $Br_{2}$ ⇌ 2HBr (∆H = -103 kJ mol-1)

## Constants:

$ k_{B} = \frac{R}{ N_{A}} = 1.38060445x10^{23} $

R = 8.314

$ N_{A} = 6.022x10^{-23} $

## Parameters:

Hydrogen are yellow particles. (mass = 2*1.00794 u)

Bromine are red particles. (mass = 2* 79.904 u)

Hydrogen bromide are green particles. (mass = 79.904 + 1.00794 u)

N = 200 molecules

H-H bond: 436 kJ mol-1

Br-Br bond: 193 kJ mol-1

H-Br bond: 366 kJ mol-1

∆H = -103 kJ mol-1

$kf = A \exp{(\frac{-629}{(8.314)(T)})} $

$ kb = A \exp{(\frac{-732}{(8.314)(T)})} $

$ K_{eq} = \frac{kf}{kb} = \exp(\frac{10^{3}}{8.314(T)} $

## Fixed relations

v = Math.sqrt(3*8.314/6.022*10E23*T/mass)

or

$ v = \sqrt( \frac{(3)(8.314)(T)}{6.022x10^{23}m} )$