(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.
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.
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