Question
For the reaction
: 2A + B + C → A2B
the rate = k[A] [B]2 with k = 2.0 x 10–6 mol–2L2s–1.
Calculate the initial rate of the reaction when [A] = 0.1 mol L-1, [B] = 0.2 mol L-1and [C] = 0.8 M. Calculate the rate of reaction after [A] is reduced to 0.06 mol L–1.
Solution
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.
