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

1. `5xx10^(-3) 2.5xx10^(-3)" "3xx10^(-5)`
2. `15xx10^(-3)" "2.5xx10^(-3)" "9xx10^(-5)`
3. `15xx10^(-3)" "10xx10^(-5)" "3.6xx10^(-4)`
(a) Calculate the order of reaction.
(b) Find the rate constant.
(c) Find the initial rate if `[NO] = [H_(2)] = 8.0 xx 10^(–3) M` Assuming rate law can be expressed as follows :
`"rate "=K[NO]^(x) [H_(2)]^(y)`
By analysing the data:
From observation 1 and 2, we see that `[H_(2)]` is constant and when [NO] is tripled, the rate is also tripled.
`rArr" rate "(r) prop [NO]`
`rArr" "x=1`
From observations 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.
Initial rate `=k[NO][H_(2)]=2.4xx(8xx10^(-3))^(2)`
`=1.536xx10^(-4)" mol/L"s.`