For the reaction `2NO_((g)) +Cl_(2(g)) rarr 2NOCl_((g))` the following data were collected.
All the measurements were taken at `263 K` :

Write the expression for rate law.

Explanation

`{:((a),"Rate"=k[NO]^(lambda)[Cl_(2)]^(beta)),(,"Rate" I = k[0.15]^(lambda)[0.15]^(beta)"........."(i)),(,"Rate" II = k[0.15]^(lambda)[0.30]^(beta)"......"(ii)),(,"Rate" III = k[0.30]^(lambda)[0.15]^(beta)"......"(iii)),(,"Rate" IV = k[0.25]^(lambda)[0.25]^(beta)"......"(iv)):}`
Divide equation (i) and (ii)
`(0.60)/(1.20) = (k[0.15]^(lambda)[0.15]^(beta))/(k[0.15]^(lambda)[0.30]^(beta))`
`rArr 1/2 =[1/2]^(beta) rArr = 1`
Divide equation (i) and (iii)
`(0.60)/(2.40) = (k[0.15]^(lambda)[0.15]^(beta))/(k[0.30]^(lambda)[0.15]^(beta))`
`rArr 1/4 = [1/2]^(beta) rArr lambda = 2`
Rate `= K[NO]^(2) [Cl_(2)]^(1)`
`K = (R)/([NO]^(2)[Cl_(2)]^(1)) = (0.6)/([0.15]^(2)[0.15]^(1)) = (0.6)/((0.15)^(3)) =177.75`
`:.` [Using experiment(1)]
(c ) `R = K [NO]^(2)[Cl_(2)]`
`R = 177.78xx (0.25)^(2) xx 0.25`
`R = 2.78 M//"Min"`.