Consider the following reaction for which K = 1.60 x
10-7 at some temperature:
2 NOCl (g) 2 NO (g) + Cl2(g)
In a given experiment, 0.795 moles of NOCl(g) were placed in an
otherwise empty 1.93 L vessel. Complete the following table by
entering numerical values in the Initial row and
values containing the variable "x" in
the Change and Equilibrium rows.
Define 2x as the amount (mol/L) of NOCl that reacts to reach
equilibrium. Include signs in the Change column
to indicate a gain or loss of concentration. (Omit units, use 3
sig.fig. and write concentrations less than 1 as 0.###, not as
.###. If nothing is present initially, enter 0 for the
molarity.)
[NOCl] (M)
[NO] (M)
[Cl2] (M)
Initial
0.412
0
0
Change
-2x
+2x
+x
Equilibrium
0.412-2x
2x
x
Since K is very small, K = 1.60 x 10-7, very little
product will be present at equilibrium, which tells us that very
little NOCl will react to reach equilibrium. Because of this, the
equilibrium concentration of NOCl will remain, essentially, at its
initial concentration. Making this assumption, calculate
the equilibrium concentration of NO. Note:
use the ICE table from the previous problem to help you solve this
problem. (Scientific notation can be entered using the following
convention: 1.93 x 10-3 = 1.93E-3.)
[NO]eq =