Equilibrium Question 739
Question: $ K_1,K_2 $ and $ K_3 $ are the equilibrium constants of the following reactions (I), (II) and (III) respectively:
(I) $ N_2+2O_2\rightarrow 2NO_2 $
(II) $ 2NO_2\rightarrow N_2+2O_2 $
(III) $ NO_2\rightarrow \frac{1}{2}N_2+O_2 $
The correct relation from the following is
Options:
A) $ K_1=\frac{1}{K_2}=\frac{1}{K_3} $
B) $ K_1=\frac{1}{K_2}=\frac{1}{{{(K_3)}^{2}}} $
C) $ K_1=\sqrt{K_2}=K_3 $
D) $ K_1=\frac{1}{K_2}=K_3 $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \begin{matrix} (I) & N_2+2O_2\overset{K_1}{\mathop{\rightarrow }},2NO_2 \\ \end{matrix} $
$ K_1=\frac{{{[NO_2]}^{2}}}{[N_2]{{[O_2]}^{2}}} $ ….(i) $ \begin{matrix} (II) & 2NO_2\overset{K_2}{\mathop{\rightarrow }},N_2+2O_2 \\ \end{matrix} $
$ K_2=\frac{[N_2]{{[O_2]}^{2}}}{{{[NO_2]}^{2}}} $ …(ii) $ \begin{matrix} (III) & NO_2\overset{K_3}{\mathop{\rightarrow }},\frac{1}{2}N_2+O_2 \\ \end{matrix} $
$ K_3=\frac{{{[N_2]}^{1/2}}[O_2]}{[NO_2]} $
$ \therefore {{(K_3)}^{2}}=\frac{[N_2]{{[O_2]}^{2}}}{{{[NO_2]}^{2}}} $ ….(iii)
$ \therefore $ from equation (i), (ii) and (iii) $ K_1=\frac{1}{K_2}=\frac{1}{{{(K_3)}^{2}}} $
BETA
