Question -
Answer -
the rate = k[A][B]2 with k =2.0 × 10−6 mol−2 L2 s−1.Calculate the initial rate of the reaction when [A] = 0.1 mol L−1, [B] = 0.2 mol L−1.Calculate the rate of reaction after [A] is reduced to 0.06 mol L−1.
Answer
The initial rate of the reaction is
Rate = k [A][B]2
= (2.0 × 10−6 mol−2 L2 s−1) (0.1 mol L−1)(0.2 mol L−1)2
= 8.0 × 10−9 mol−2 L2 s−1
When [A] is reduced from 0.1 mol L−1 to0.06 mol−1, the concentration of A reacted = (0.1 − 0.06)mol L−1 = 0.04 mol L−1
Therefore,concentration of B reacted 
= 0.02 mol L−1
Then, concentration ofB available, [B] = (0.2 − 0.02) mol L−1
= 0.18 mol L−1
After [A] is reduced to 0.06 mol L−1,the rate of the reaction is given by,
Rate = k [A][B]2
= (2.0 × 10−6 mol−2 L2 s−1) (0.06 mol L−1)(0.18 mol L−1)2
= 3.89 mol L−1 s−1