The following rate data were obtained at 300 K for the reaction
` 2A +B to C +D `

Calculate the rate of formation of D when : [A]=0.5 mol `L^(-)` and [B]=0.2 mol `L^(-)`

Answer

Rate ` =k [A][B]^2; 0.1 mol L^(-) min^(-)`

Explanation

Rate of reaction is given as: Rate `= k[A]^x [B]^y` Rates of reaction in experiments 1 to 4 are:
`(i) 5.0 xx 10^(-3) =k (0.1)^x (0.1)^y`
`(ii) 6.0 xx 10^(-2) =k(0.3)^x (0.2)^y`
` (iii) 2.4 xx 10^(-1) =k (0.3)^x (0.4)^y`
`(iv) 2 xx 10^(-2) =k(0.4)^x (0.1)^y`
Dividing eq. (i) by (iv), we get,
` ( 5.0 xx 10^(-3))/( 2.0 xx 10^(-2)) = (k (0.1 )^x ( 0.1 )^y )/( k( 0.4 )^x (0.1 )^y) or (1/4)^1 = (1/4)^x` ,
hence ` x=1`
Dividing equation (ii) by (iii), we get :
` ( 6 xx 10^(-2))/( 2.4 xx 10^(-1)) = (k ( 0.3 )^x (0.2 )^y)/(k ( 0.3 )^x ( 0.4 )^y) or 1/4 = (1/2)^2 = (1/2)^y` ,
hence `y=2`
Rate equation `=k[A][B]^2`
from experiment (i) ` 5 xx 10^(-3) = k ( 0.1 )^1 ( 0.1 )^2`
thus ` ,k =( 5 xx 10^(-3))/((0.1 ) (0.1 )^2) = 5 `
hence, rate ` = 5 xx ( 0.5 )^1 ( 0.2 )^2 = 0.1 mol L^(-) min""^(-1)`