Question

$N_2$ and $O_2$ combine at a given temperature to produce $NO$. At equilibrium the yield of $NO$ is ${}^{\prime }{x}^{\prime }$ per cent by volume. If $x=\sqrt{K\cdot a\cdot b}-\frac{K(a+b)}{4}$ 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 $O$? respectively in the initial pure mixture, what should be the initial composition of the reacting mixture in order that maximum yield of $NO$ is ensured? Also report the maximum value of $K$ at which ${}^{\prime }{x}^{\prime }$ is maximum.

Answer 1

$N_2+O_2\rightleftharpoons 2NO$

Initial	$a$	$b$	$0$
Final	$a-x$	$b-x$	$2x$

$x=\sqrt{(K\cdot a\cdot b)}-\left[K(a+b)\right]/4$ ....(i)
The $x$ is maximum only when condition of maximum are fulfilled.
i.e., $\frac{\delta x}{\delta a}=0$ and $\frac{\delta x}{\delta b}=0$
By partial differentiation of $x$ with respect to $a$ keeping $b$ constant.
From equation (i)
$\frac{\delta x}{\delta a}=\frac{Kb}{2\sqrt{(K\cdot a\cdot b)}}-\frac{K}{4}=\frac{\sqrt{\left(Kb\right)}}{2\sqrt{a}}-\frac{K}{4}=0$ ...(ii)
By partial differentiation of ${}^{\prime }{x}^{\prime }$ with respect to ${}^{\prime }{b}^{\prime }$ keeping ${}^{\prime }{a}^{\prime }$ constant.
From equation (i)
$\frac{\delta x}{\delta b}=\frac{Ka}{2\sqrt{(K\cdot a\cdot b)}}-\frac{K}{4}=\frac{\sqrt{\left(Ka\right)}}{2\sqrt{b}}-\frac{K}{4}=0$ ...(iii)
By equation (ii) $K\cdot a\cdot b=4b^2$
By equation (iii) $K\cdot a\cdot b=4b^2$
$\therefore a=b$
Also, this is valid only when $K<4$, because if $a=b$, equation (i) yields
$x=\sqrt{\left(K{a}^2\right)}-\frac{Ka}{2}=a(\sqrt{K}-\frac{K}{2})$
If $x$ is +ve $\sqrt{K}>K/2$ or $4K>K^2$
or $K(4-K)>0$ or $0<K<4$