Ammonia and oxygen react at high temperature as `: 4 NH _(3)(g) to 4NO (g) + 6 H _(2) O (g)` in an experiment, the rate of formation of NO is `3.6 xx 10 ^(-3) mol L ^(-1) sec ^(-1).` Calculate the rate on formation of `H _(2)O.`

Explanation

For the Reaction
`4NH_3(g)+5O_2(g) rarr 4NO(g)+6H_2O`
Rate=`-1/4(d[NH_3])/(dt)=1/3(d[O_2])/(dt)=1/4`
`(d[NO])/(dt)=1/6(d[H_2O])/(dt)`
Here `(d[NO])/(dt)=3.6xx10^(-3)"mol L"^(-1)s^(-1)`
`therefore` Rate=`1/4(d[NO])/(dt)`
`=1/4xx3.6xx10^(-3)"mol L"^(-1)s^(-1)`
`=9xx10^(-4)" mol L"^(-1)s^(-1)`
Rate=`1/6(d[H_2O])/(dt)`
`therefore (d[H_2O])/(dt)=6xx"Rate"`
`=6xx9xx10^(-4)" mol L"^(-1)s^(-1)`
`=3.6xx10^(-3)" mol L"^(-1)s^(-1)`