For the reaction `4NH_(3) + 5O_(2) to 4NO + 6H_(2) O`, if the rate of disappearance of `NH_(3)` is `3.6 xx 10^(-3)` mol `L^(-1) s^(-1)`, what is the rate of formation of `H_(2) O`

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)`