Asked by Claudia
At 35 degrees C, 2.0 mmol of pure NOCI is introduced into a 2.0-L flask. The NOCI partially decomposes according to the following equilibrium equation.
2 NOCI = 2 NO + CI2
at equilibrium the concentration of NO is 0.032 mol?L. Use an ICE table to determine equilibrium concentrations of NOCI and CI2 at this temperature.
2 NOCI = 2 NO + CI2
at equilibrium the concentration of NO is 0.032 mol?L. Use an ICE table to determine equilibrium concentrations of NOCI and CI2 at this temperature.
Answers
Answered by
DrBob222
(NOCl) = 0.02/2 =0.001 M
................2NOCI ==> 2NO + CI2
I.................0.001............0.......0
C...............-2x................2x......x
E...........0.001-2x...........2x.......x
The problem tells you that 2x = 0.032 M.and that should tell you something is not right because you can't have more NO than NOCl you started with.
................2NOCI ==> 2NO + CI2
I.................0.001............0.......0
C...............-2x................2x......x
E...........0.001-2x...........2x.......x
The problem tells you that 2x = 0.032 M.and that should tell you something is not right because you can't have more NO than NOCl you started with.
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