The following rate data were obtained at 303 K for the following reaction :
`2A +B to C+D`

What is the rate law? What is the order with respect to each reactant and the overall order? What are the units of rate constant?

Explanation

For the given reaction rate `=k[A]^x [B]^Y`
After substituting the values of initial rates and initial concentration, the value of x and y can be calculated. Then x + y will give the order of the reaction.
For experiment `1, 6 xx 10^(-3) = (0.1)^x (0.1)^y` ...(i)
For experiment II, `7.2 xx 10^(-2) =(0.3)^x (0.2)^y`
For experiment III, `2.88 xx 10^(-1) = (0.3)^x (0.4)^y `
For experiment IV, `2.4 xx 10^(-2) =(0.4)^x (0.1)^y`
Determination of x. In order to get the value of x, divide equation (iv) by (i)
` ( 2.4 xx 10^(-2) )/( 6.0 xx 10^(-3)) = (( 0.4 )^x ( 0.1 )^y)/((0.1)^x (0.1 )^y) , (4)^1 = (4)^x`,
` therefore x=1`
Determination of y. In order to get the value of y, divide equation (ii) by (ii).
` 2.88 xx 10^(-1))/( 7.2 xx 10^(-2)) = ( ( 0.3 )^x ( 0.4 )^y)/( (0.3 )^x ( 0.2 )^y) ,4=(2)^y`
or ` (2)^2 =(2)^y therefore y=2`
Order of reaction with respect to A and B are 1 and 2 respectively.
The over all order of the reaction = 1 + 2 = 3.
` therefore` Rate law `=k[A]^1 [B]^2 " " [ :. x=1 ,y=2]`
units of rate constant . since rate ` =k [A]^1 [B]^2`
` therefore k=( " rate " )/([A]^1 [B]^2) = ( 6.0 xx 10^(-3) mol L^(-1) min^(-))/( ( 0.1 mol L^(-1)) (0.1 mol L^(-1) )^2)`
`=6.0 mol^(-2) L^(+2) min^(-)`
` therefore ` unit of rate constant ,` k is mol^(-2) L^(+2) min^(-)`