Question

For the reaction ; 2A + B → A2B, the reaction rate = k [A][B]2 with k = 2·0 x 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. Also calculate the reaction rate when [A] is reduced to 0·06 mol L-1.

Answer 1

Answer:

The reaction is 2A+B→A

2

B.

The rate law expression is as follows:

Rate =k[A][B]

2

The rate constant k=2.0×10

−6

mol

−2

L

2

s

−1

.

When [A]=0.1molL

−1

,[B]=0.2 molL

−1

,

The initial rate of the reaction is,

Rate=2.0×10

−6

×(0.1)×(0.2)

2

=8.0×10

−9

molL

−1

s

−1

After [A] is reduced to 0.06 molL

−1

, the rate of reaction is as follows:

Rate =2.0×10

−6

×0.06×(0.18)

2

=3.89×10

−9

molL

−1

s

−1

[Note: A reacted =0.1−0.06=0.04

B reacted =0.02

[B]=0.2−0.02=0.18molL

−1

]