The equilibrium constant, K, for the following reaction is 1.80×102 at 698 K. 2HI(g) H2(g) + 12(g) An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.308 M HI, 4.14×102 M H2 and 4.14x102 M 12. What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.199 mol of HI(g) is added to the flask? = M [HI] [H2]= [12] M M Σ Σ Σ

Chemistry: The Molecular Science
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Chapter12: Chemical Equilibrium
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The equilibrium constant, K, for the following reaction is 1.80×102 at 698 K.
2HI(g)
H2(g) + 12(g)
An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.308 M
HI, 4.14×102 M H2 and 4.14x102 M 12. What will be the concentrations of the three
gases once equilibrium has been reestablished, if 0.199 mol of HI(g) is added to the
flask?
=
M
[HI]
[H2]=
[12]
M
M
Σ Σ Σ
Transcribed Image Text:The equilibrium constant, K, for the following reaction is 1.80×102 at 698 K. 2HI(g) H2(g) + 12(g) An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.308 M HI, 4.14×102 M H2 and 4.14x102 M 12. What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.199 mol of HI(g) is added to the flask? = M [HI] [H2]= [12] M M Σ Σ Σ
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