Ammonia and oxygen react at high temperature as:
`4NH_(3) (g) + 5O_(2)(g) rarr 4NO(g) + 6H_(2)O (g)`
In an experiment, the rate of formation of `NO` is `3.6 xx 10^(-3) Ms^(-1)`. Calculate (a) the rate of disappearance of ammonia and (b) the rate of formation of water.

Answer

(a) Rate of disappearance of `NH_(2) = 3.6 xx 10^(-3) mol L^(-1) s^(-1)`
( b) Rate of formation of `H_(2)O =5.4 xx 10^(-3) mol L^(-1) s^(-1)`

Explanation

Given : `4NH_(5(g)) +5O_(2(g)) to 4NO_((g)) + 6H_(2)O_((g))`
`(d[NO])/(dt) = 3.6 xx 10^(-3) mol L ^(-1) s^(-1)`
(a) `-(1)/(4) (d[NH_(3)])/(dt) =(1)/(4) (d[NO])/(dt)`
`:.` Rate of disappearance of `NH_(3)`
`=(-d[NH_(3)])/(dt) = (d[NO])/(dt)`
`=3.6 xx 10^(-3) mol L^(-1) s^(-1)`
`:.` Rate of formation of `H_(2)O`
`=(d[H_(2)O])/(dt)=(6)/(4) (d[NO])/(dt)`
`=(6)/(4) xx 3.6 xx10^(-3)`
`=5.4 xx 10^(-3) mol L^(-1) s^(-1)`