For a reaction
`2NO(g) + 2H_(2)(g) rarr N_(2)(g)+2H_(2)O(g)`
The following data were obtained :
`|{:(,[NO] (mol L^(-1)),[H_(2)] (mol L^(-1)),"Rate" (mol L^(-1) s^(-1))),(1.,5 xx 10^(-3),2.5 xx 10^(-3),3 xx 10^(-5)),(2.,15 xx 10^(-3),2.5 xx 10^(-3),9xx10^(-5)),(3.,15 xx 10^(-3),10 xx 10^(-3),3.6 xx 10^(-4)):}|`
(a) Calculating the order of reactions.
(b) Find the rate constant.
(c) Find the initial rate if `[NO] = [H_(2)] = 8.0 xx 10^(-3) M`

Explanation

(a) Assuming rate law can be expressed as follows:
rate `= k[NO]^(x)[H_(2)]^(y)`
By analyzing the data:
form observations `1` and `2`, we see that `[H_(2)]` is constant and when `[NO]` is tripled, the rate is also tripled.
Rate `prop [NO] rArr x = 1`
form observation `(2)` and `(3)`, we see that `[NO]` is constant, when `[H_(2)]` is increased four times, the rate also increases four times.
Rate `prop [H_(2)] rArr y = 1`
`rArr r = k[NO][H_(2)O]`
The order of reaction w.r.t. `NO` and `H_(2)` is `1` and the overall order of reaction is `1 + 1 = 2`
(b) Rate `= k[NO][H_(2)]` or uisng (1)
`3 xx 10^(-50 = k(5 xx 10^(-3)) (2.5 xx 10^(-3)) or k = 2.4`
( c) Initial rate `= k[NO][H_(2)] = 2.4 xx (8 xx 10^(-3)) (8 xx 10^(-3))`
`= 1.536 xx 10^(-4) mol L^(-1) s^(-1)`