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)])/(d t) = (1)/(5) (-d[O_(2)])/(d t)`
`= (1)/(4) (d[NO])/(d t) = (1)/(6)(d[H_(2)O])/(d t)`
a. Rate of displacement of ammonia
`= "Rate of appearance of NO"`
`= 3.6 xx 10^(-3) mol L^(-1) s^(-1)`
b. Rate of formation of water.
`= (d[H_(2)O])/(d t)`
form the rate equation,
`(1)/(6)(d[H_(2)O])/(d t) = (1)/(4)(d[NO])/(d t)`
`:. (d[H_(2)O])/(d t) = (6)/(4) xx 3.6 xx 10^(-3) = 5.4 xx 10^(-3) mol L^(-1) s^(-1)`