In the reaction
2A + B + C → A2B + C,
there is no change in C, therefore its conc. does not affect the rate of the reaction.
Initial rate = k[A] [B]2
But [A] = 0.1 M,
[B] = 0.2 M
and k = 2 x 10–6 M–2 s–1
Therefore initial rate
Rate= [k] x [A] x [B]2
= 2 x 10–6 M–2 s–1 x (0.1 M) (0.2 mol M)2 = 8 x 10–9 ms–1
From the equation:
2A + B + C → A2B + C,
it is clear that when 2 moles of A are used then 1 mol of B is used in the same time. Therefore, when A has been reduced to 0.06 M (due to its 0.04 M has been reacted to 0.02 of B). Thus,
Conc. of A left = [A] = 0.06 M
Conc. of B left = [B] = [0.02 M – 0.02 M]
= 0.018 M
Rate = k[A] [B]2
= 2 x 10–6 M2 S–1 x (0.06 M) (0.18 M)
= 3.89 x 10–9 Ms–1.
t / s |
0 |
30 |
60 |
90 |
Ester / mol L–1 |
0.55 |
0.31 |
0.17 |
|
A/mol L–1 |
0.20 |
0.20 |
0.40 |
B/mol L–1 |
0.30 |
0.10 |
0.05 |
r/mol L–1S–1 |
5.07 x 10–5 |
5.07 x 10–5 |
1.43 x 10–4 |
What is the order of the reaction with respect of A and B?
exp.
|
[A]/ |
[B]/M |
Initial rate of formation |
I |
0.1 |
0.1 |
|
II |
0.3 |
0.2 |
7.2 x 10–2 |
III |
0.3 |
0.4 |
2.88 x 10–1 |
IV |
0.4 |
0.1 |
2.40 x 10–2 |
exp.
|
[A]/ |
[B]/M |
Initial rate of formation |
I |
0.1 |
0.1 |
|
II |
- |
0.2 |
4.0 x 10–2
|
III |
0.4 |
0.4 |
- |
IV |
- |
0.2 |
2.0 x 10–2 |
(a) Plot [N2O5] again t.
(b) Find the half life period for the reaction.
(c) Draw a graph between log [N2O5] and t.
(d) What is rate law?
(e) Calculate the rate constant.
(f) Calculate the half life period from K and compare it with (ii).