Question

If $K_p$ for the reaction is $1.44\times {10}^{-5}units$, then the value of $K_p$ for the decomposition of ${NH}_3$,

$2{NH}_3\left(g\right)\rightleftharpoons N_2\left(g\right)+3H_2\left(g\right)$, will be:

• A. $\sqrt{1.44\times {10}^{-5}}units$
• B. ${(1.44\times {10}^{-5})}^{2}units$
• C. $\frac{1}{1.44\times {10}^{-5}}units$
• D. $2.88\times {10}^{-5}units$

Answer 1

The correct option is C $\frac{1}{1.44\times {10}^{-5}}units$

The given reaction is :-

$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2{NH}_3\left(g\right)$

$\therefore K_P=\frac{{\left(p_{{NH}_3}\right)}^2}{p_{N_2}.{\left(p_{H_2}\right)}^3}=1.44\times {10}^{-5}\to \left(I\right)$

Now, for the reaction,

$2{NH}_3\left(g\right)\rightleftharpoons N_2\left(g\right)+3H_2\left(g\right)$

$\Rightarrow {K_P}^{{}^{\prime }}=\frac{p_{N_2}.{\left(p_{H_2}\right)}^3}{{\left(p_{{NH}_3}\right)}^2}$

$=\frac{1}{K_P}=\frac{1}{1.44\times {10}^{-5}}units$