AP EAMCET Engineering 2018 Apr 23 Shift 2 Paper

First we will compare expt. (1) and expt. (2)
The rate of reaction gets double when [B] is doubled
Therefore, reaction is of first order in B.
Now, we will compare expt. (2) and expt. (3)
The rate of reaction gets double when [C] is doubled
Therefore, reaction is of first order in C.
Now, compare expt. (3) and expt. (4)
The rate of reaction does not change when [A] is doubled
Therefore, reaction is of zero order in A.
The rate law expression can be written as:
Rate law =k[A]0[B]1[C]1
By inserting data of expt. (1) in the above expression, we obtain
2.4×10−6=k[0.02]0[0.1]1[0.003]1
k=

2.4×10−6

0.1×0.03

=8.0×10−4L mol−1s−1