Question

The following equilibrium constants are given $N _{2}+3 H _{2} \rightleftharpoons 2 NH _{3} ; K _{1}$ $N _{2}+ O _{2} \rightleftharpoons 2 NO ; K _{2}$ $H _{2}+1 / 2 O _{2} \rightleftharpoons H _{2} O ; K _{3}$ The equilibrium constant for the oxidation of the $NH _{3}$ by oxygen to give $NO$ is:

• A. $\frac{ K _{1} \,K _{2}}{ K _{3}}$
• B. $\frac{ K _{2} \,K _{3}^{3}}{ K _{1}}$
• C. $\frac{ K _{2} \,K _{3}^{2}}{ K _{1}}$
• D. $\frac{ K _{2}^{2} \,K _{3}}{ K _{1}}$

Answer 1

Answer: (B)

$\frac{ K _{2} \,K _{3}^{3}}{ K _{1}}$

Answer 2

(I) $N _{2}+3 H _{2} \rightleftharpoons 2 NH _{3} ; K _{1}=\frac{\left[ NH _{3}\right]^{2}}{\left[ N _{2}\right]\left[ H _{2}\right]^{3}}$

(II) $N _{2}+ O _{2} \rightleftharpoons 2 NO ; K _{2}=\frac{[ NO ]^{2}}{\left[ N _{2}\right]\left[ O _{2}\right]}$

(III) $H _{2}+\frac{1}{2} O _{2} \longrightarrow H _{2} O ; K _{3}=\frac{\left[ H _{2} O \right]}{\left[ H _{2}\right]\left[ O _{2}\right]^{1 / 2}}$

(II $+3 \times$ III $-$ II) will give

$2 NH _{3}+\frac{5}{2} O _{2} \stackrel{ K }{\rightleftharpoons} 2 NO +3 H _{2} O$
$\therefore K = K _{2} \times K _{3}^{3} / K _{1}$