In a reaction between A and B, the initial rate of reaction was measured for different initial concentrations of A and B as given below:

What is the order of reaction with respect to A and B ?

Explanation

Let the order of reaction w.rt. A is x and w.r.t. B is y. Then the rate of reaction is given by :
rate `=k[A]^x [B]^y`
For experiment I,
` 5.07 xx 10^(-5) = ( 0.2 )^x (0.30 )^y`
For experiment II,
`5.07 xx 10^(-5) = ( 0.2 )^x (0.10 )^y`
For experiment III,
`7.6 xx 10^(-5) = ( 0.40 )^x ( 0.05)^y`
Dividing (ii) by (i), we get
` ( 5.07 xx 10^(-5))/( 5.07 xx 10^(-5)) = 1 = ( (0.2 )^x (0.1 )^y)/( 0.2)^x ( 0.30)^y) = ((0.10)/(0.30))^y`
` therefore y=0`
Dividing (iii) by (i), we get
`( 7.6 xx 10^(-5) )/(5.07 xx 10^(-5)) = 1.499 = 1.5 = (( 0.40 )^x ( 0.05 )^y)/( ( 0.2 )^x ( 0.30 )^y)`
`=( 2)^x ((0.05 )/(0.30 ))^y`
Substituting y = 0, we get 1.5=`(2)^x`
On solving, log 1.5=x log 2 or 0.1761 =x `xx` 0.3010
`x= ( 0.1761 )/( 0.3010 )=0.58 = 0.5`
Thus the order of reaction with respect to A is 0.5 and w.r.t. B is zero.