Question

Equilibrium constant $K_{C}$ for the following reaction at $800 \,K$ is, $4$
$NH_{3}\left(g\right)\rightleftharpoons\frac{1}{2}N_{2}\left(g\right)+\frac{3}{2}H_{2}\left(g\right)$
The value of $K_{P}$ for the following reaction will be
$N_{2}\left(g\right)+3H_{2}\left(g\right)\rightleftharpoons2NH_{3}\left(g\right)$

• A. $\left(\frac{800 R}{4}\right)^{2 -}$
• B. $16\times \left(800R\right)^{2}$
• C. $\left[\frac{1}{4 \times 800 R}\right]^{2}$
• D. $\left(800 R\right)^{1 / 2}4$

Answer 1

Answer: (C)

$NH _{3}( g ) \rightleftharpoons \frac{1}{2} N _{2}( g )+\frac{3}{2} H _{2}( g ) ; K _{ C }=4 $
$N _{2}( g )+3 H _{2}( g ) \rightleftharpoons 2 NH _{3}( g ) \Delta n _{ g }=2-4=-2$
So $K_{C}$ for the following reaction will be equal to $\frac{1}{4^{2}}$
$ K _{ p }= K _{ C }( RT )^{\Delta n _{ g }} $
$=\frac{1}{4^{2}} \times(800 \times R )^{-2}$
$=\frac{1}{4 \times 800 R } 2$