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.
(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.
To determine the equilibrium concentrations of NOCI and CI2 at this temperature, we can use an ICE table, which stands for Initial, Change, and Equilibrium. Here's how you can do it:
1. Write down the balanced equation for the reaction:
2 NOCI ⇌ 2 NO + CI2
2. Set up the ICE table:
| 2 NOCI ⇌ 2 NO + CI2
---------------------------------
I | 2.0 mmol (Initial concentration)
C | -x mmol +2x mmol +x mmol (Change in concentration)
E | (2.0 - x) mmol 0.032 mol/L x mmol (Equilibrium concentration)
In this table, x represents the change in concentration for each species at equilibrium. We assume the initial concentration of NOCI is 2.0 mmol, and at equilibrium, it becomes (2.0 - x) mmol. The initial concentration of NO is zero, and it increases by 2x mmol at equilibrium. The initial concentration of CI2 is also zero, and it increases by x mmol at equilibrium.
3. Use the given information that the equilibrium concentration of NO is 0.032 mol/L. From the ICE table, we know that the concentration of NO at equilibrium is 2x mmol. Since 1 mol = 1000 mmol, we can convert the concentration of NO to mol/L:
0.032 mol/L = 2x mmol / 1000
Solving for x:
x = (0.032 mol/L) * (1000 mmol / 2) = 16 mmol
4. Substitute the value of x back into the ICE table to find the equilibrium concentrations:
| 2 NOCI ⇌ 2 NO + CI2
---------------------------------------------
I | 2.0 mmol (Initial concentration)
C | -16 mmol +32 mmol +16 mmol (Change in concentration)
E | (2.0 - 16) mmol 0.032 mol/L 16 mmol (Equilibrium concentration)
Therefore, at equilibrium, the concentration of NOCI is (2.0 - 16) mmol = -14 mmol (negative sign due to a decrease in concentration), and the concentration of CI2 is 16 mmol.