Compounds 'A' and 'B' react according to the following chemical equation.
`A(g) to 2B(g) to 2C(g)`
Concentration of either 'A' or 'B' were changed Keeping the concentration of one of the reactants constant and rates were measured as a function of initial concentration. Following result were obtained.
Choose the correct option for the rate equations for this reaction.

Options

A. Rate=k`[A]^(2)[B]`
B. Rate=k[A]`[B]^(2)`
C. Rate=k[A]`[B]`
D. Rate=k`[A]^(2)[B]^(0)`

Answer

B

Explanation

Suppose order with respect to A and B are x and y respectively
Rate=`k[A]^(x)[B]^(y)`
For experiment 1,
`0.1=k(0.3)^(x)(0.3)^(y)` . . . (i)
For experiment 2,
`0.4=k(0.3)^(x)(0.6)^(y)` . . . (ii)
For experiment 3,
`0.2=k(0.6)^(x)(0.3)^(y)` . . . (iii)
Dividing equation (ii) and (i)
`(0.4)/(0.1)=((0.6)^(y))/((0.3)^(y))`
`therefore y=2`
Dividing equation (iii) by (i)
`(0.2)/(0.1)=((0.6)^(y))/((0.3)^(y))`
`therefore x=1`
Rate law
Rate=`k[A][B]^(2)`