Initial rate = k[A][B]2
But [A] = 0.1 M, [B] = 0.2 M and
k = 2.0 × 10-6 mol-2 L2 s-1
∴ Initial rate
= 2.0 × 10-6 mol-2 L2 s-1 × 0.1 M × (0.2 M)2
= 8 × 10-9 M s-1
From the equation 2A + B → A2B 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 , 0.04 M of A and hence 0.02 M of B have reacted.
Thus,
Concentration of A left = [A] = 0.06 M
Concentration of B left = [B] = (0.2 M – 0.02 M) = 0.18 M
Rate = k[A][B]2
= 2.0 × 10-6 mol-2 L2 s-1 × 0.06 M × (0.18 M)2
= 3.89 × 10-9 M s-1