Suppose order with respect to A is m and with respect to B is n.Then the rate law will be
Rate =k[A]m[B]n
Substituting the value of experiments I to IV, we get
Expt. I: Rate =6.0×10−3=k(0.1)m(0.1)n‌‌‌‌‌‌‌...(i)
Expt. II: Rate =7.2×10−2=k(0.3)m(0.2)n‌‌‌‌‌‌...(ii)
Expt. III: Rate =2.88×10−1=k(0.3)m(0.4)n‌‌‌ ...(iii)
Expt. IV: Rate =2.4×10−2=k(0.4)m(0.1)n‌‌‌‌ ...(iv)
Comparing equation (i) and equation (iv)
∴‌

( ‌Rate)I

(Rate) IV

=

6.0×10−3

2.4×10−2

=

k(0.1)m(0.1)n

k(0.4)m(0.1)n

or,

1

4

=

(0.1)m

(0.4)m

=(

1

4

)m‌‌‌∴m=1
Comparing equation(ii) and equation (iii)

( ‌Rate)II

(Rate)III

=

7.2×10−2

2.88×10−1

=

k(0.3)m(0.2)n

k(0.3)m(0.4)n

or, (

1

2

)2=

(0.2)n

(0.4)n

=(

1

2

)n‌‌∴n=2
Rate law expression is : Rate =k[A][B]2.
The rate constant can be calculated from the given data of any experimentusing expression:
‌‌‌‌‌k=

Rate

[A][B]2

From Expt. I, k=

6.0×10−3

0.1×(0.1)2

=6.0
∴‌Rate constant k=6.0 mol−2 L2min−1
Unit of k,k=

Rate

[A][B]2

=

mol L−1min−1

( mol L−1 ) ( mol L−1 )2

= mol−2 L2min−1