The data given below are for the reaction of `NO` and `Cl_(2)` to from `NOCl` at `295 K`.
`{:([Cl_(2)],[NO],,,"Initial rate" (mol litre^(-1) sec^(-1))),(0.05,0.05,,,1xx10^(-3)),(0.15,0.05,,,3xx10^(-3)),(0.05,0.15,,,9xx10^(-3)):}`
(a) What is the order with respect to `NO` and `Cl_(2)` in the reaction?
(b) Write the rate expression.
(c ) Calculate the rate constant.
(d) Determine the reaction rate when conc. of `Cl_(2)` and `NO` are `0.2 M` and `0.4 M` respectively.

Explanation

For the reaction, `2NO+Cl_(2)rarr 2NOCl`
Rate=`K[Cl_(2)]^(m)[NO]^(n) …(1)`
Where, m and n are order of reaction w.r.t. `Cl_(2)` and `NO`, respectively. From the given data:
`1xx10^(-3)= K[0.05]^(m)[0.05]^(n) …(2)`
`3xx10^(-3)= K[0.15]^(m)[0.05]^(n) ...(3)`
`9xx10^(-3)= K[0.05]^(m)[0.15]^(n) ...(4)`
By eqs. (2) and (3), `m=1`
By eqs. (2) and (4), `n=2`
(a) `:.` Order w.r.t. `NO` is `2` and w.r.t. to `Cl_(2)` is `1`.
(b) Also, rate expression `r=K[Cl_(2)]^(1) [NO]^(2)`
(c ) And rate constant, `K=(r)/([Cl_(2)]^(1)[NO]^(2))`
`=(1xx10^(-3))/([0.05]^(1)[0.05]^(2))`
`=8 litre^(2) mol^(-2) sec^(-1)`
(d) Further, `r=K[Cl_(2)]^(1)[NO]^(2)=8[0.2]^(1)[0.4]^(2)`
`=0.256 mol litre^(-1) sec^(-1)`