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.

Explanation

For the reaction
`4NH_(3)(g)+5O_(2) (g) rarr 4NO(g)+6H_(2)O(g)`
Rate `=-1/4 (d[NH_(3)])/(dt)`
`=1/5 (d[O_(2)])/(dt)=1/4 (d[NO])/(dt)`
`=1/6 (d[H_(2)O])/(dt)`
(i) Rate of disappearance of ammonia
Rate of disappearance of `NH_(3)`
= Rate of formation of NO
`=3.6xx10^(-3)` mol `L^(-1) S^(-1)`
(ii) Rate of formation of water
Rate of formation of `H_(2)O=(d[H_(2)O])/(dt)`
From rate, equation,
`1/6 (d[H_(2)O)])/(dt)=1/4 (d[NO])/(dt)`
`:. (d[H_(2)O])/(dt)=6/4 xx3.6 xx10^(-3)`
`=5.4xx10^(-3)` mol `L^(-1) S^(-1)`