Rate of a reaction `A + B rarr` Product, is given as a function of different initial concentration of `A` and `B`.
`|{:([A] (mol L^(-1)),(B) (mol L^(-1)),"Initial rate" (mol L^(-1) min^(-1)),),(0.01,0.01,0.005,),(0.02,0.01,0.010,),(0.01,0.02,0.005,):}|`
Determine the order of the reaction with respect to `A` and with respect to `B`. What is the half life of `A` in the reaction ?

Explanation

Let the rate of reaction be
Rate `= k[A]^(x)[B]^(y)`
form the data given, it is clear that by doubling the concentration of `A`, the rate also becomes double when `B` is kept constant. Thus, the rate is directly proportional to concentration of `A`.
Rate `prop [A]`, i.e., `x = 1`
or the order of reaction w.r.t. to `A` is
When the concentration of `A` is kept constant and the concentration of `B` is doubled, the rate does not change, i.e., `y = 0`, or the order of reaction w.r.t. to `B` is zero.
Thus, reaction rate, `-(dx)/(dt) = k[A]`
Again `k = (0.005)/(0.01) = 0.5 min^(-1)`
Half life of `A = (0.693)/(k) = (0.693)/(0.5) = 1.386 min`