Correct Answer - A::C
`(d)/(dt)[NH_(2)CONH_(2)] = k_(4)[NH_(3)][HNCO]` from (iii)
Applying steady state approximation to `HNCO` or `NH_(3)`
`(d[HNCO])/(dt) = O = k_(3) [NH_(4)NCO]-k_(4)[NH_(3)][HNCO]`
`:. (k_(3))/(k_(4)) = ([NH_(3)][HNCO])/([NH_(4)CNO])`
`(d["urea"])/(dt) = k_(4) xx [NH_(3)][HNCO] = k_(4) xx (k_(3))/(k_(4))[NH_(4)NCO]`
Also, `[NH_(4)NCO] = (k_(1))/(k_(2)) xx [NH_(4)CNO]`
`:. (d["urea"])/(dt) = k_(3) xx (k_(1))/(k_(2)) xxx [NH_(4)CNO]`
`= k[NH_(4)CNO]`