Compound 'A' and B react according to the following chemical equation.
`A_((g)) + 2B_((g)) to 2C_((g))`
Concentration of either A or B were changed keeping the concentrations of one of the reactants constant and rates were mesured as a function of initial concentration. Following results were obtained. Choose the correct option for the rate equations for this reaction.
`{:("Experiment","Initial concentration of"[A]//mol L^(-1),"Initial concentration of"[B]mol L^(-1),"Initial rate of formation of"[c]//mol L^(-1)s^(-1)),(1,0.30,0.30,0.10),(2,0.30,0.60,0.40),(3,0.60,0.30,0.20):}`

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

Let order with respect to A and B are x and y respectively.
`:.` Rate = `k (A)^(x) (B)^(y)`
`0.1 = k (0.3)^(x) (0.3)^(y)`
`0.4 = k (0.3)^(x) (0.6)^(y)`
`0.2 = k (0.6)^(x) (0.3)^(y)`
`(0.4)/(0.1) = ((0.6)^(y))/((0.3)^(y))`
`:. y = 1`
Dividing (iii) by (i) `(0.2)/(0.1) = ((0.6)^(x))/((0.3)^(x))`
`:. x = 1`
Rate law will be:
Rate `= k [A]^(1) [B]^(2)`