For the equilibrium: 2F2 (g) + O2 (g) + 600 kJ = 2OF2 (g) where ke= 350 The equilibrium law expression for the balanced chemical equation would be: O ([F2][02] )= [OF2]? O [OF,1? = ( [F] [0,] ) [OF ]? - ([02][F2] ) x 600

Chemistry: The Molecular Science
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ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter12: Chemical Equilibrium
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For the equilibrium: 2F, (g) + 02 (g) + 600 kJ = 20F, (g) where ke= 350
%3D
The equilibrium law expression for the balanced chemical equation would be:
O ([F2][02] )- [OF,]
O [OF,1? = ( [F2]² [02] )
OLOF - ([02][F2] ) x 600
O ((F][0] ) + [OF,]?
([F2]+[O3] ) - [OF,]²
Transcribed Image Text:For the equilibrium: 2F, (g) + 02 (g) + 600 kJ = 20F, (g) where ke= 350 %3D The equilibrium law expression for the balanced chemical equation would be: O ([F2][02] )- [OF,] O [OF,1? = ( [F2]² [02] ) OLOF - ([02][F2] ) x 600 O ((F][0] ) + [OF,]? ([F2]+[O3] ) - [OF,]²
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