The data for the reaction: `A + B overset(k)rarr C`.
`|{:("Experiment",[A]_(0),[B]_(0),"Initial rate"),(1,0.012,0.035,0.10),(2,0.024,0.070,0.80),(3,0.024,0.035,0.10),(4,0.012,0.070,0.80):}|`
The rate law corresponding to the above data is
(a) Rate `= k[B]^(3)` , (b) Rate `= k[B]^(4)`
( c) Rate `= k[A][B]^(3)` , (d) Rate `= k[A]^(2)[B]^(2)`

Options

A. Rate`=k[B]^(3)`
B. Rate`=k[B]^(4)`
C. Rate`=k[A][B]^(3)`
D. Rate`=k[A]^(2)[B]^(2)`

Answer

A

Explanation

From Exp 1
`Rate_(1)=k[A]^(p)[B]^(q)`
`0.1=k(0.012)^(p)(0.035)^(q)`
From Exp 2
`"Rate"_(2)=k[A]^(p)[B]^(q)`
`0.8=k(0.024)^(p)(0.07)^(q)`
`("Rate"_(1))/("Rate"_(2))=(k(0.012)^(p)(0.035)^(q))/(k(0.024)^(p)(0.070)^(q))`
`((0.1)/(0.8))=((1)/(2))^(p)((1)/(2))^(q)`
`((1)/(2))^(3)=((1)/(2))^(p+q)`
order "n"=`p+q=3`
Hence, Rate =`k[B]^(3)`.