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

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