`N_(2)` and `O_(2)` combine at a given temperature to produce `NO`. At equilibrium the yield of `NO` is 'x' precent by volume. If `x=sqrt(Ka.b)-(K(a+b))/(4)`, where `K` is the equilibrium constant of the given reaction at the given temperature and `a` and `b` are the volume percentage of `N_(2)` and `O_(2)`, respectively, in the initial state. Report. Report the maximum value of `K` at which `X` is maximum

Explanation

`{:(,N_(2),+,O_(2),hArr,2NO),("Initial volume %",a,,b,,0),("Final volume %",,,,,x):}`
Also `x=sqrt(Ka.b)-(K(a+b))/(4) ...(i)`
The `x` is maximum only when condition of maximum is fulfilled,
i.e., `deltax//deltaa` and `deltax//deltab=0`
By partial differentiation of `x` with respect to a keeping b constant.
From equation (i)
`(deltax)/(deltaa)=(Kb)/(2sqrt(K.a.b))-K/4=sqrt(Kb)/(2sqrt(a))-K/4=0 ...(ii)`
By partial differentiation of x wrt b keeping a constant From equation (i)
`(deltax)/(deltab)=(Ka)/(2sqrt(K.a.b))-K/4=sqrt(Ka)/(2sqrt(b))-K/4=0 ...(iii)`
By equation (ii), `Kab=4b^(2)`
By equation (3),`Kab=4a^(2)`
`:. a=b`
Note: Also this is valid only when `K lt 4` because if `a=b`, Equation (i) yields
`x=sqrt(Ka^(2))-(Ka)/(2)=(sqrt(K)-K/2)`
If `x` is (+ve) `sqrt(K) gt K//2` or `4K gt K^(2)`
or `K(4-K) gt 0` or `0 lt K lt 4`