A+B→C Let rate
=k[A]x[B]y where order of reaction is
(x+y) .
Putting the values of exp.
1,2, and
3, we get following equations.
0.10=k[0.012]x[0.035]y...(i)
0.80=k[0.024]x[0.070]y ...(ii)
0.10=k[0.024]x[0.035]y ...(iii)
Dividing (ii) by (iii), we get
=()y ⇒2y=8⇒y=3 Keeping
[A] constant,
[B] is doubled, rate becomes 8 times.
Dividing eq. (iii) by eq. (i), we get
=()x ⇒2x=1⇒x=0 Keeping
[B] constant,
[A] is doubled, rate remains unaffected.
Hence, rate is independent of
[A] rate
∝[B]3.