For the reaction :
`2H_(2) +2NO hArr 2H_(2)O +N_(2)`, the following rate data was obtained :

Calculate the following:
(1) The overall order of reaction.
(2) The rate law.
(3) The value of rate constant (k).

Explanation

Initial rate expression can be written as
Rate = `k[NO]^(a) [H_(2)]^(b)`
Substituting the values of concentration of NO and `H_(2)` from experimental data 1 and 2
`("Rate")_(1)=k(0.40)^(a) (0.40)^(b)" "...(i)`
`("Rate")_(2)=k(0.80)^(a)(0.40)^(b)" "...(ii)`
Dividing equation (ii) by equation (i)
`(("Rate")_(2))/(("Rate")_(1))=(k(0.80)^(a)(0.40)^(b))/(k(0.40)^(a)(0.40)^(b))`
`=(18.4xx10^(-3))/(4.6xx10^(-3))`
`(2)^(a)=4`
`a=2`
Thus, the order of reaction with respect to NO is 2.
On comparing experiment data (1) and (3)
`("Rate")_(3)=k(0.40)^(a)(0.80)^(b)" " ...(iii)`
`("Rate")_(1)=k(0.40)^(a)(0.40)^(b)" "...(iv)`
Dividing equation (iii) by equation (iv)
`(("Rate")_(3))/(("Rate")_(1))=(k(0.40)^(a)(0.80)^(b))/(k(0.40)^(a)(0.40)^(b))`
`=(9.2xx10^(-3))/(4.6xx10^(-3))`
`(2)^(b)=2`
`b=1`
Thus, order of reaction with respect to `H_(2)` is 1
(1) Overall order of the reaction `= 2+1=3`
(2) The rate law for the reaction
Rate `=k[NO]^(2)[H_(2)]`
(3) Rate constant can be calculated by substituting the values of rate of [NO] and `[H_(2)]` for any experiment
`k=("Rate")/([NO]^(2)[H_(2)])`
`=(4.6xx10^(-3))/((0.40)^(2)(0.40))`
`=(4.6xx10^(-3))/(0.064)`
`=71.875xx10^(-3)`
Thus, the value of rate constant
`k=71.875xx10^(-3)mol^(-2)L^(2)s^(-1)`