Reaction Reversibility

Sample problem (similar to 11 & 12)
Concentration
(mol/L)
N2O4  2NO2. N2O4: Start = 1.0, finish = 0.20
2
NO2 (1.60)
1
N2O4 (0.20)
Time
N2O4 : first find start and finish, then draw curve
Start = 1.0, finish = 0.20
NO2 : first find start and finish, then draw curve
Start = 0, finish = 1.60 …if 0.8 N2O4 is used
then 1.6 NO2 must be produced (N2O4  2NO2)
Question 1 - 3
1. N2O4  2NO2
2. With a double arrow ()
3.
Decomposition of N2O4
Conc. (mol/L)
2
1.6
1.2
N2O4
NO2
0.8
0.4
0
0
2
4
6
Time (min)
8
10
12
Question 4 - 8
4. No, not all N2O4 was used up. On the graph
N2O4 does not go to zero.
5. 1.6 mol NO2 were produced
6. 2 mol NO2 should by produced (according to
the balanced equation )
7. Think back to Ep graphs. Both forward and
reverse reactions occur. As N2O4 breaks
down the concentration of NO2 increases.
This increased [ ] increases the rate of the
reaction of NO2 combining to form N2O4
8. [N2O4] = [N2O4]initial – 1/2 [NO2] or in words
(note: 2 is from the balanced equation)
Question 9 - 10
9. The equilibrium concentrations end up being
the same whether we start from pure
reactants or pure products (assuming the
number of atoms for each element is the
same).
10. We can reach the same equilibrium if we
start with pure NO2 (this is the idea behind
reaction reversibility). We would have to
start with twice as much NO2 as N2O4 to
have the same number of atoms. Thus we
would have to start with 2 mol of NO2.
Concentration
(mol/L)
Question 11
2
HI (1.56)
1
H2, I2 (0.22)
Time
HI: first find start and finish, then draw curve
Start = 0, finish = 1.56
H2: first find start and finish, then draw curve
Start = 1, finish = 0.22 …
Since H2 + I2  2HI, if 1.56 HI is produced,
0.78 H2 must have been used (1 - 0.78 = 0.22)
Concentration
(mol/L)
2
1
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Question 12
H2 (0.9)
NH3 (0.4)
N2 (0.3)
Time
N2: Start = 0.5, finish = 0.3
H2: Start = 1.5, finish = 0.9 …
Since N2 + 3H2  2NH3, 3x as much H2 is
used compared to N2 (1.5 - (0.2 x 3) = 0.9)
NH3: Start = 0, finish = 0.4 …
Since N2 + 3H2  2NH3, 2x as much NH3 is
produced than N2 used (0 + (0.2 x 2) = 0.4)